Chapter 8 Application of Integrals (Additional Questions)

To find the area of the region bounded by the curve $y = x^2$, the x-axis, and the lines $x=1$ and $x=3$, we need to integrate the function $y = x^2$ with respect to $x$ from $x=1$ to $x=3$.

The area under a curve $y = f(x)$ from $x=a$ to $x=b$ is given by the definite integral $\int_a^b f(x) dx$.

In this case, $f(x) = x^2$, $a=1$, and $b=3$.

Therefore, the area is given by:

Let's evaluate this integral to verify:

Comparing this with the given options, option (A) matches our derived integral expression.

The correct option is (A).

To find the area of the region bounded by the curve $x = y^2$, the y-axis, and the lines $y=0$ and $y=2$, we need to integrate the function $x = y^2$ with respect to $y$ from $y=0$ to $y=2$.

The area under a curve $x = g(y)$ from $y=c$ to $y=d$ is given by the definite integral $\int_c^d g(y) dy$.

In this case, $g(y) = y^2$, $c=0$, and $d=2$.

Therefore, the area is given by:

Let's evaluate this integral to verify:

Comparing this with the given options, option (A) matches our derived integral expression.

The correct option is (A).

To find the area of the region bounded by the line $y = 2x+1$, the x-axis, and the ordinates $x=-1$ and $x=1$, we first need to determine where the line intersects the x-axis. This occurs when $y=0$.

Setting $y=0$ in the equation $y = 2x+1$, we get:

Solving for $x$:

This means the line crosses the x-axis at $x = -\frac{1}{2}$. We are considering the region between $x=-1$ and $x=1$.

For the area calculation, we need to consider the absolute value of the integral for parts of the region where the curve is below the x-axis. The function $y = 2x+1$ is:

• Negative when $2x+1 < 0 \implies x < -\frac{1}{2}$. In our interval, this is from $x=-1$ to $x=-\frac{1}{2}$.
• Positive when $2x+1 > 0 \implies x > -\frac{1}{2}$. In our interval, this is from $x=-\frac{1}{2}$ to $x=1$.

Therefore, the total area is the sum of the absolute value of the integral from $x=-1$ to $x=-\frac{1}{2}$ and the integral from $x=-\frac{1}{2}$ to $x=1$.

The area is given by:

Let's evaluate these integrals to confirm:

First integral:

Absolute value of the first integral is $\left|-\frac{1}{4}\right| = \frac{1}{4}$.

Second integral:

Total Area $= \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = \frac{5}{2}$.

The expression for the area matches option (D).

The correct option is (D).

The equation of a circle centered at the origin with radius $a$ is $x^2 + y^2 = a^2$.

We can express $y$ in terms of $x$ from this equation: $y^2 = a^2 - x^2$, which gives $y = \pm \sqrt{a^2 - x^2}$.

The equation $y = \sqrt{a^2 - x^2}$ represents the upper semi-circle, and $y = -\sqrt{a^2 - x^2}$ represents the lower semi-circle.

The area of the circle can be found by integrating the function representing the upper semi-circle from $x=-a$ to $x=a$ and then doubling it (to account for the lower semi-circle), or by considering the symmetry of the circle.

The area of the upper semi-circle is given by $\int_{-a}^a \sqrt{a^2-x^2} dx$.

Since the circle is symmetric about the x-axis, the total area of the circle is twice the area of the upper semi-circle.

Alternatively, due to symmetry about the y-axis as well, we can consider the area in the first quadrant, which is $\int_0^a \sqrt{a^2-x^2} dx$. Multiplying this by 4 will give the total area of the circle.

Let's evaluate the integral $\int_0^a \sqrt{a^2-x^2} dx$ using trigonometric substitution. Let $x = a \sin \theta$. Then $dx = a \cos \theta d\theta$. When $x=0$, $\sin \theta = 0 \implies \theta = 0$. When $x=a$, $\sin \theta = 1 \implies \theta = \frac{\pi}{2}$.

Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$:

So, the total area of the circle is $4 \times \frac{\pi a^2}{4} = \pi a^2$.

Both options (B) and (C) represent valid integral forms for the area of the circle. However, option (B) is a more common and direct way to represent the area using integration by considering the symmetry in four quadrants.

The correct option is (B).

The equation of the parabola is given as $y^2 = 4ax$.

The latus rectum of a parabola $y^2 = 4ax$ is a line segment passing through the focus perpendicular to the axis of symmetry, with endpoints on the parabola. The focus of this parabola is at $(a, 0)$. The latus rectum is the line $x = a$.

We need to find the area of the region bounded by the parabola $y^2 = 4ax$ and the line $x=a$.

From the equation of the parabola, we can express $y$ in terms of $x$: $y = \pm \sqrt{4ax} = \pm 2\sqrt{a}\sqrt{x}$.

The area can be calculated by integrating $y$ with respect to $x$ from the vertex ($x=0$) to the latus rectum ($x=a$). Due to the symmetry of the parabola about the x-axis, we can calculate the area of the upper half and multiply it by 2.

The area of the upper half is given by $\int_0^a y dx = \int_0^a 2\sqrt{a}\sqrt{x} dx$.

The total area is:

Now, let's evaluate the integral:

The area of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum is $\frac{8a^2}{3}$.

The correct option is (A).

The area of the region bounded by the curve $y = \sin x$, the x-axis, from $x=0$ to $x=\pi$ is given by the definite integral of $\sin x$ from $0$ to $\pi$.

The function $y = \sin x$ is non-negative for $x$ in the interval $[0, \pi]$. Therefore, the area is directly given by the integral without needing absolute values.

The area is calculated as:

Now, let's evaluate the integral:

The area is $2$. Option (A) represents the correct integral form, and option (C) gives the calculated value.

The correct option is (C).

The area of the region between two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$, where $f(x) \ge g(x)$ for all $x$ in the interval $[a, b]$, is found by integrating the difference between the upper curve and the lower curve over that interval.

The area is mathematically represented as:

This formula directly computes the area by integrating the difference of the two functions.

Using the properties of definite integrals, we can also write this as:

This second form is derived from the first by using the linearity property of integrals, which states that $\int_a^b (h(x) \pm k(x)) dx = \int_a^b h(x) dx \pm \int_a^b k(x) dx$.

Both expressions (B) and (C) correctly represent the area of the region bounded by the two curves under the given condition.

The correct option is (D).

To find the points of intersection of the curves $y = x^2$ and $y = x$, we need to set the expressions for $y$ equal to each other.

Setting the two equations equal:

To solve for $x$, we can rearrange the equation:

Factor out $x$:

This gives us two possible values for $x$:

$x = 0$ or $x - 1 = 0 \implies x = 1$.

Now, we find the corresponding $y$ values by substituting these $x$ values into either of the original equations. Using $y = x$ is simpler:

• When $x = 0$, $y = 0$. So, one point of intersection is $(0, 0)$.
• When $x = 1$, $y = 1$. So, the other point of intersection is $(1, 1)$.

These points are $(0, 0)$ and $(1, 1)$.

The correct option is (C).

First, let's find the points of intersection of the curves $y = x^2$ and $y = 4$.

Setting $x^2 = 4$, we get $x = \pm 2$. So the points of intersection are $(-2, 4)$ and $(2, 4)$.

We can calculate the area in different ways:

Method 1: Integrating with respect to $x$.

The region is bounded above by $y=4$ and below by $y=x^2$. The limits of integration for $x$ are from $-2$ to $2$.

Area $= \int_{-2}^2 (\text{upper curve} - \text{lower curve}) dx = \int_{-2}^2 (4 - x^2) dx$.

This matches option (A).

Let's evaluate this integral:

Method 2: Integrating with respect to $y$.

We need to express $x$ in terms of $y$. From $y = x^2$, we get $x = \pm \sqrt{y}$. The right half of the parabola is $x = \sqrt{y}$, and the left half is $x = -\sqrt{y}$.

The region is bounded on the right by $x = \sqrt{y}$ and on the left by $x = -\sqrt{y}$. The limits of integration for $y$ are from $0$ to $4$.

Area $= \int_0^4 (\text{right curve} - \text{left curve}) dy = \int_0^4 (\sqrt{y} - (-\sqrt{y})) dy = \int_0^4 2\sqrt{y} dy$.

This matches option (B).

Let's evaluate this integral:

Method 3: Using symmetry with respect to the y-axis.

The region is symmetric about the y-axis. We can calculate the area in the first quadrant (where $x \ge 0$) and multiply it by 2. For $x \ge 0$, the parabola is $x = \sqrt{y}$ and the region is bounded by $y=x^2$, $y=4$, and the y-axis. The integration is with respect to $x$ from $0$ to $2$.

Area of the right half $= \int_0^2 (4 - x^2) dx$.

Total Area $= 2 \times \int_0^2 (4 - x^2) dx$.

This matches option (C).

Let's evaluate this integral:

All three options represent the correct area calculation for the given region.

The correct option is (D).

The equation of the ellipse is given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

We can find the area of the ellipse by integrating. From the equation, we can express $y$ in terms of $x$ for the upper half of the ellipse:

The upper half of the ellipse is given by $y = \frac{b}{a} \sqrt{a^2 - x^2}$. The ellipse extends from $x = -a$ to $x = a$.

The area of the ellipse is twice the area of the upper half:

We can factor out the constants $\frac{2b}{a}$:

The integral $\int_{-a}^a \sqrt{a^2 - x^2} dx$ represents the area of a semi-circle of radius $a$, which is $\frac{1}{2} \pi a^2$.

Substituting this value:

Alternatively, we can use the substitution $x = a \sin \theta$, $dx = a \cos \theta d\theta$. When $x=-a$, $\theta = -\frac{\pi}{2}$. When $x=a$, $\theta = \frac{\pi}{2}$.

The area of the ellipse is $\pi ab$.

The correct option is (A).

The region is bounded by the parabola $x = y^2$ and the line $x=4$.

First, let's find the points of intersection. Setting $y^2 = 4$, we get $y = \pm 2$. The points of intersection are $(-2, 4)$ and $(2, 4)$ in terms of $(y, x)$ if we consider $y$ as the independent variable, or $(4, -2)$ and $(4, 2)$ if we consider $x$ as the independent variable.

Evaluating Option (A): $\int_0^4 \sqrt{x} dx$

This integral represents the area under the curve $y = \sqrt{x}$ from $x=0$ to $x=4$. This is the area between the curve $y = \sqrt{x}$ and the x-axis. This does not represent the area between $x=y^2$ and $x=4$.

Evaluating Option (B): $\int_{-2}^2 (4 - y^2) dy$

This integral represents the area between the curves $x = 4$ (as the "upper" function of $y$) and $x = y^2$ (as the "lower" function of $y$), integrated with respect to $y$. The limits of integration for $y$ are from $-2$ to $2$, which are the y-coordinates of the intersection points.

Area $= \int_{-2}^2 (4 - y^2) dy$. This is a correct representation of the area.

Let's calculate its value: $\left[ 4y - \frac{y^3}{3} \right]_{-2}^2 = \left( 8 - \frac{8}{3} \right) - \left( -8 - \frac{-8}{3} \right) = \frac{16}{3} - (-\frac{16}{3}) = \frac{32}{3}$.

Evaluating Option (C): $2 \int_0^2 (4 - y^2) dy$

This integral uses the symmetry of the region about the x-axis. The parabola $x=y^2$ is symmetric with respect to the x-axis. We can calculate the area in the upper half (from $y=0$ to $y=2$) and multiply it by 2. The limits for $y$ are from $0$ to $2$, and the integrand is the difference between the bounding lines $x=4$ and $x=y^2$.

Area $= 2 \int_0^2 (4 - y^2) dy$. This is also a correct representation of the area.

Let's calculate its value: $2 \left[ 4y - \frac{y^3}{3} \right]_0^2 = 2 \left( (8 - \frac{8}{3}) - (0) \right) = 2 \left( \frac{16}{3} \right) = \frac{32}{3}$.

Evaluating Option (D): $\int_0^4 (4 - x) \frac{dx}{2\sqrt{x}}$

This integral appears to be an attempt to integrate with respect to $x$. The area between $x=y^2$ and $x=4$ can also be written as integrating $x_{right} - x_{left}$ with respect to $y$. If we were to integrate with respect to $x$, it would be the area bounded by $y = \sqrt{x}$, $y = -\sqrt{x}$ and $x=4$. The area would be $\int_0^4 (\sqrt{x} - (-\sqrt{x})) dx = \int_0^4 2\sqrt{x} dx$. This option's integrand is $(4-x) / (2\sqrt{x})$, which does not directly correspond to a standard area calculation for this region. Let's evaluate it to see if it yields the correct area.

Let $u = \sqrt{x}$. Then $du = \frac{1}{2\sqrt{x}} dx$ and $x = u^2$. When $x=0$, $u=0$. When $x=4$, $u=2$.

The integral becomes $\int_0^2 (4 - u^2) du$. This is the same as option (C) with variable $u$ instead of $y$. So this option also represents the area.

Options (B), (C), and (D) all represent the area of the region bounded by $x = y^2$ and $x=4$. Option (A) does not.

The correct options are (B), (C), and (D).

Let's analyze the Assertion (A) and the Reason (R) separately.

Assertion (A): The area of the region bounded by $y=x^3$, the x-axis, and $x=-1, x=1$ is $\int_{-1}^1 x^3 dx = 0$.

To find the area bounded by a curve and the x-axis, we need to consider the absolute value of the function if it dips below the x-axis. The integral $\int_{-1}^1 x^3 dx$ calculates the net signed area. Let's evaluate it:

So, the statement $\int_{-1}^1 x^3 dx = 0$ is true.

However, the *area* of the region is not $0$. The function $y=x^3$ is negative for $x \in [-1, 0)$ and positive for $x \in (0, 1]$. The actual area requires integrating the absolute value of the function:

Area $= \int_{-1}^1 |x^3| dx = \int_{-1}^0 |x^3| dx + \int_0^1 |x^3| dx$.

For $x \in [-1, 0)$, $|x^3| = -x^3$. For $x \in [0, 1]$, $|x^3| = x^3$.

Area $= \int_{-1}^0 (-x^3) dx + \int_0^1 x^3 dx$

Since the calculated area is $1/2$, and the Assertion claims the area is $0$, Assertion (A) is false in its statement about the area being $0$. The integral value is $0$, but that's not the area.

Reason (R): The function $f(x) = x^3$ is an odd function, and the integral of an odd function over a symmetric interval $[-a, a]$ is zero. However, area is always non-negative.

First, let's check if $f(x) = x^3$ is an odd function. An odd function satisfies $f(-x) = -f(x)$.

$f(-x) = (-x)^3 = -x^3 = -f(x)$. So, $f(x) = x^3$ is indeed an odd function.

The integral of an odd function over a symmetric interval $[-a, a]$ is zero. This is a known property of definite integrals.

Also, the statement "area is always non-negative" is true. Area is a measure of spatial extent, which cannot be negative.

Therefore, Reason (R) is true.

Relationship between Assertion (A) and Reason (R):

Assertion (A) incorrectly equates the value of the integral of $x^3$ over $[-1, 1]$ with the area. Reason (R) correctly explains why the integral is zero (due to $x^3$ being odd) and correctly states that area must be non-negative, thus implying that the integral result of $0$ does not represent the actual area. Reason (R) highlights the distinction between signed area (integral) and geometric area, which is precisely what makes Assertion (A) false in its conclusion about the area.

Since Assertion (A) is false and Reason (R) is true, the correct option is (D).

The correct option is (D).

To find the area of the region bounded by the curve $y = \cos x$, the x-axis, from $x=0$ to $x=\pi/2$, we need to evaluate the definite integral of $\cos x$ over this interval.

The function $y = \cos x$ is non-negative in the interval $[0, \pi/2]$. Therefore, the area is directly given by the integral.

The area is calculated as:

Now, let's evaluate this integral:

The area of the region is $1$. Option (A) represents the correct integral form, and option (B) gives the calculated value of the area.

The correct option is (B).

To find the area of the region bounded by the parabola $y=x^2$ and the line $y=x$, we first need to find the points of intersection.

Setting $x^2 = x$, we get $x^2 - x = 0$, which factors as $x(x-1) = 0$. The points of intersection are at $x=0$ and $x=1$. The corresponding y-values are $y=0$ and $y=1$. So the intersection points are $(0,0)$ and $(1,1)$.

In the interval $[0, 1]$, we need to determine which function is greater. Let's test a value, say $x=0.5$.

For the line $y=x$, $y=0.5$.

For the parabola $y=x^2$, $y=(0.5)^2 = 0.25$.

Since $0.5 > 0.25$, the line $y=x$ is above the parabola $y=x^2$ in the interval $[0, 1]$.

The area between two curves $f(x)$ and $g(x)$ where $f(x) \ge g(x)$ on $[a, b]$ is given by $\int_a^b (f(x) - g(x)) dx$.

In this case, $f(x) = x$ (the upper curve), $g(x) = x^2$ (the lower curve), $a=0$, and $b=1$.

Therefore, the area is:

Let's evaluate this integral:

The calculated area is $\frac{1}{6}$.

Comparing this with the options:

• Option (A) is $\int_0^1 x^2 dx - \int_0^1 x dx = \int_0^1 (x^2 - x) dx$. This would give a negative area since $x^2 < x$ in this interval.
• Option (B) is $\int_0^1 (x - x^2) dx$, which is the correct form.
• Option (C) is $\int_0^1 (x^2 - x) dx$, which would result in a negative value.
• Option (D) is $|\int_0^1 (x^2 - x) dx|$. This would be $|-\frac{1}{6}| = \frac{1}{6}$, which is the correct area.

However, the question asks for the integral that *represents* the area. Option (B) directly represents the area calculation by subtracting the lower curve from the upper curve.

The correct option is (B).

The region is in the first quadrant, bounded by the line $y=x$ and the circle $x^2+y^2=1$.

First, let's find the point of intersection of the line $y=x$ and the circle $x^2+y^2=1$ in the first quadrant. Substitute $y=x$ into the circle's equation:

Since we are in the first quadrant, $x > 0$, so $x = \frac{1}{\sqrt{2}}$. Since $y=x$, the point of intersection is $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.

We can calculate this area by integrating with respect to $x$. The region is split into two parts at the point of intersection:

• From $x=0$ to $x=\frac{1}{\sqrt{2}}$, the region is bounded above by the line $y=x$ and below by the x-axis ($y=0$).
• From $x=\frac{1}{\sqrt{2}}$ to $x=1$, the region is bounded above by the circle $y=\sqrt{1-x^2}$ and below by the x-axis ($y=0$).

So, the total area is the sum of these two integrals:

This matches option (A).

Let's consider the geometric interpretation for option (D).

The region bounded by the line $y=x$, the circle $x^2+y^2=1$, and the x-axis in the first quadrant can be viewed as a sector of the circle minus a triangle.

The line $y=x$ makes an angle of $\theta = \frac{\pi}{4}$ with the positive x-axis. The point of intersection $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ corresponds to this angle, as $\cos(\pi/4) = \sin(\pi/4) = \frac{1}{\sqrt{2}}$.

The area of the sector of the circle with radius $r=1$ and angle $\theta = \frac{\pi}{4}$ is given by $\frac{1}{2} r^2 \theta = \frac{1}{2} (1)^2 \left(\frac{\pi}{4}\right) = \frac{\pi}{8}$.

This sector is bounded by the x-axis, the line $y=x$, and the arc of the circle. The region described in the question is actually the area *between* the line $y=x$ and the circle arc, from their intersection point to where the circle intersects the y-axis (or alternatively, the area under the circle arc minus the area under the line from $x=0$ to $x=1$, but this is not how the options are structured).

Let's re-read the question: "The area of the region bounded by the line $y=x$ and the circle $x^2+y^2=1$ in the first quadrant". This implies the area enclosed *between* these two curves.

If the question means the area enclosed between the line $y=x$ and the arc of the circle, from the intersection point up to where the circle meets the y-axis, or from the intersection point to where the circle meets the x-axis, the formulation can be different.

Let's reconsider the wording: "bounded by the line $y=x$ and the circle $x^2+y^2=1$ in the first quadrant". This means the area enclosed by these two curves and the axes. It seems the question intends to mean the area enclosed by the arc of the circle and the line segment connecting the origin to the intersection point. This interpretation leads to the area being a sector minus a triangle.

Let's assume the question is asking for the area enclosed by the arc of the circle from the x-axis to the intersection point, and the line segment from the origin to the intersection point. This is the area of a sector of the circle. The angle of the sector made by the line $y=x$ with the x-axis is $\pi/4$. The radius of the circle is $1$.

Area of the sector $= \frac{1}{2} r^2 \theta = \frac{1}{2} (1)^2 (\frac{\pi}{4}) = \frac{\pi}{8}$.

The area of the triangle formed by the origin, $(1/\sqrt{2}, 0)$, and $(1/\sqrt{2}, 1/\sqrt{2})$ is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{4}$.

The area under the line $y=x$ from $0$ to $1/\sqrt{2}$ is $\int_0^{1/\sqrt{2}} x dx = [\frac{x^2}{2}]_0^{1/\sqrt{2}} = \frac{(1/\sqrt{2})^2}{2} = \frac{1/2}{2} = \frac{1}{4}$.

The area under the circle arc from $1/\sqrt{2}$ to $1$ is $\int_{1/\sqrt{2}}^1 \sqrt{1-x^2} dx$. This integral represents the area of a segment of the circle.

Option (A) represents the area under the line from $0$ to $1/\sqrt{2}$ plus the area under the circle arc from $1/\sqrt{2}$ to $1$. This is the area under the *upper boundary* of the region formed by the x-axis, the line $y=x$, and the circle $x^2+y^2=1$. This is a valid way to calculate the area of the described region.

Let's evaluate Option (A):

First part: $\int_0^{1/\sqrt{2}} x dx = \left[ \frac{x^2}{2} \right]_0^{1/\sqrt{2}} = \frac{(1/\sqrt{2})^2}{2} - 0 = \frac{1/2}{2} = \frac{1}{4}$.

Second part: $\int_{1/\sqrt{2}}^1 \sqrt{1-x^2} dx$. To evaluate this, let $x = \sin \theta$, $dx = \cos \theta d\theta$. When $x=1/\sqrt{2}$, $\theta=\pi/4$. When $x=1$, $\theta=\pi/2$. $\sqrt{1-x^2} = \cos \theta$.

$\int_{\pi/4}^{\pi/2} \cos \theta (\cos \theta d\theta) = \int_{\pi/4}^{\pi/2} \cos^2 \theta d\theta = \int_{\pi/4}^{\pi/2} \frac{1+\cos 2\theta}{2} d\theta = \frac{1}{2} \left[ \theta + \frac{1}{2} \sin 2\theta \right]_{\pi/4}^{\pi/2}$

= $\frac{1}{2} \left[ \left(\frac{\pi}{2} + \frac{1}{2} \sin \pi \right) - \left(\frac{\pi}{4} + \frac{1}{2} \sin \frac{\pi}{2} \right) \right] = \frac{1}{2} \left[ \frac{\pi}{2} - (\frac{\pi}{4} + \frac{1}{2}) \right] = \frac{1}{2} \left[ \frac{\pi}{4} - \frac{1}{2} \right] = \frac{\pi}{8} - \frac{1}{4}$.

Total Area = $\frac{1}{4} + \left(\frac{\pi}{8} - \frac{1}{4}\right) = \frac{\pi}{8}$.

Option (D) suggests a geometric calculation: Area of sector - Area of triangle.

The sector is bounded by the x-axis, the line $y=x$, and the arc of the circle from the x-axis to the point of intersection $(1/\sqrt{2}, 1/\sqrt{2})$. The angle of this sector is $\pi/4$. The radius is $1$.

Area of Sector $= \frac{1}{2} r^2 \theta = \frac{1}{2} (1)^2 (\frac{\pi}{4}) = \frac{\pi}{8}$.

The triangle is formed by the origin $(0,0)$, the intersection point $(1/\sqrt{2}, 1/\sqrt{2})$, and the projection of the intersection point on the x-axis $(1/\sqrt{2}, 0)$. The area of this triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{4}$.

The area of the region bounded by the line $y=x$ and the circle $x^2+y^2=1$ in the first quadrant is the area of the sector minus the area of the triangle formed by the origin and the intersection points on the axes, which is not what is asked.

The area enclosed between the line $y=x$ and the circle $x^2+y^2=1$ from their intersection point $(1/\sqrt{2}, 1/\sqrt{2})$ to the y-axis or x-axis is what the options are trying to represent.

Let's re-examine Option (D). The region bounded by $y=x$ and the circle in the first quadrant refers to the area enclosed by the arc of the circle and the line segment from origin to the intersection point. This area is precisely the sector of the circle with angle $\pi/4$. The area of this sector is $\frac{\pi}{8}$.

Option (A) calculation gives $\frac{\pi}{8}$.

Option (D) states "Area of sector - Area of triangle". This would be the area of the circular segment between the chord and the arc. However, the region described is typically interpreted as the area of the sector defined by the line $y=x$ and the x-axis, and the circle arc. If this area is $\pi/8$. Then if we subtract a triangle of area $1/4$, we get $\pi/8 - 1/4$. This does not seem right.

Let's assume the question meant the area enclosed by the line segment from origin to the intersection point, the arc of the circle from the intersection point to the y-axis, and the y-axis. This area can be calculated by integrating with respect to y from 0 to $1/\sqrt{2}$ for the line $x=y$ and from $1/\sqrt{2}$ to 1 for the circle $x=\sqrt{1-y^2}$.

Area $= \int_0^{1/\sqrt{2}} y dy + \int_{1/\sqrt{2}}^1 \sqrt{1-y^2} dy$. This is analogous to option (A) but with y as the integration variable.

Let's consider the wording again: "The area of the region bounded by the line $y=x$ and the circle $x^2+y^2=1$ in the first quadrant". This implies the area enclosed by these two curves. This area is the segment of the circle cut by the line $y=x$, from the intersection point to the x-axis, and the area under the line from origin to the intersection point.

The area of the sector is $\frac{\pi}{8}$. The area of the triangle formed by the origin and the intersection point $(1/\sqrt{2}, 1/\sqrt{2})$ with the x-axis is $1/4$. The area under the line $y=x$ from $0$ to $1/\sqrt{2}$ is $1/4$. The area under the circle from $1/\sqrt{2}$ to $1$ is $\pi/8 - 1/4$. So the total area under the upper boundary is $1/4 + (\pi/8 - 1/4) = \pi/8$. This is the area of the sector.

Option (D) "Area of sector - Area of triangle". The sector is formed by the origin, the intersection point and the x-axis, so its area is $\pi/8$. The triangle is formed by the origin and the intersection point projected on the x-axis $(1/\sqrt{2},0)$, area is $1/4$. So this option describes $\pi/8 - 1/4$, which is the area of the circular segment.

The question asks for the area bounded by $y=x$ and the circle $x^2+y^2=1$ in the first quadrant. This typically means the area enclosed between the arc of the circle and the line segment from the origin to the intersection point. This area is the sector of the circle defined by the line $y=x$ and the x-axis. This area is $\frac{\pi}{8}$.

Option (A) correctly calculates this area using integration.

Option (D) states "Area of sector - Area of triangle". If the sector is the one with angle $\pi/4$ (area $\pi/8$) and the triangle is formed by the origin and the intersection point, its area is $1/4$. The difference is $\pi/8 - 1/4$, which is the area of the circular segment. This is not the region described.

Let's re-interpret option (D): perhaps it means the area of the sector defined by the origin, the intersection point, and the x-axis, MINUS the area of the triangle formed by the origin and the intersection point. If the area of the sector is $\pi/8$, and the area of the triangle formed by origin and intersection point is $1/4$, then $\pi/8 - 1/4$ is the area of the segment. This is not the intended region.

Let's assume the question intends to ask for the area of the sector of the circle defined by the origin, the intersection point $(1/\sqrt{2}, 1/\sqrt{2})$, and the x-axis. This area is $\pi/8$. Option (A) calculates this area. Option (D) seems to be a statement about how to calculate a circular segment, not the region itself.

Considering the integration in option (A), it sums the area under the line $y=x$ from $0$ to $1/\sqrt{2}$ and the area under the circle $y=\sqrt{1-x^2}$ from $1/\sqrt{2}$ to $1$. This correctly sums up to the area of the sector of the circle defined by the line $y=x$ and the x-axis.

The correct option is (A).

The region is bounded by the curve $y = |x|$ and the line $y=1$. The curve $y=|x|$ forms a V-shape with its vertex at the origin. The line $y=1$ is a horizontal line.

The points of intersection are where $|x|=1$, which means $x=1$ or $x=-1$. So the intersection points are $(-1, 1)$ and $(1, 1)$.

We can calculate the area in a couple of ways:

Method 1: Using integration with respect to x.

The region is bounded above by $y=1$ and below by $y=|x|$. The integration limits for $x$ are from $-1$ to $1$.

Area $= \int_{-1}^1 (\text{upper curve} - \text{lower curve}) dx = \int_{-1}^1 (1 - |x|) dx$.

This matches option (C).

Let's evaluate this integral:

Since $|x|$ is an even function and $(1-|x|)$ is also an even function, we can write:

This matches option (B).

Let's evaluate the integral:

The area is $1$. Both options (B) and (C) correctly represent the calculation of this area.

Method 2: Geometric approach.

The region bounded by $y=|x|$ and $y=1$ forms a triangle. The base of the triangle lies on the line $y=1$ from $x=-1$ to $x=1$. The length of the base is $1 - (-1) = 2$. The height of the triangle is the perpendicular distance from the vertex (origin) to the line $y=1$, which is $1$.

Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 1 = 1$.

Option (A) $\int_0^1 x dx$ calculates the area of the triangle in the first quadrant bounded by $y=x$, the x-axis, and the line $x=1$. This area is $1/2$, which is only half of the total area.

Both options (B) and (C) correctly represent the integral calculation for the area.

The correct option is (D).

To find the intersection points of the curve $x = y^2$ and the line $x = 2-y$, we set the expressions for $x$ equal to each other.

Equating the two expressions for $x$:

Rearrange the equation to form a quadratic equation in $y$:

Factor the quadratic equation:

This gives us two possible values for $y$:

$y+2=0 \implies y = -2$

$y-1=0 \implies y = 1$

Now, we find the corresponding $x$ values by substituting these $y$ values into either of the original equations. Using $x = y^2$ is convenient:

• When $y = 1$: $x = (1)^2 = 1$. The intersection point is $(1, 1)$.
• When $y = -2$: $x = (-2)^2 = 4$. The intersection point is $(4, -2)$.

The intersection points are $(1, 1)$ and $(4, -2)$. The question asks for the limits of integration, which are the y-coordinates of the intersection points, i.e., $1$ and $-2$.

Looking at the options, option (A) provides the points $(1,1)$ and $(4,-2)$. Option (D) states the same points and correctly identifies the y-coordinates as 1 and -2, which are indeed the limits of integration when integrating with respect to y.

The correct option is (A) (and also (D), but (A) is more direct in listing the points). Option (A) lists the intersection points correctly. Option (D) also lists the points correctly and then specifies the y-coordinates which are the limits of integration.

Given the options, option (A) is the most direct answer to finding the intersection points.

The correct option is (A).

To find the area of the region bounded by the parabola $y = x^2$ and the line $y = 2x+3$, we first need to find the points of intersection.

Set the equations equal to each other:

Rearrange into a quadratic equation:

Factor the quadratic equation:

The solutions are $x=3$ and $x=-1$. These are the limits of integration.

Next, we need to determine which function is the upper boundary and which is the lower boundary in the interval $[-1, 3]$. Let's pick a test value within this interval, for example, $x=0$.

For the parabola $y = x^2$, when $x=0$, $y = 0^2 = 0$.

For the line $y = 2x+3$, when $x=0$, $y = 2(0)+3 = 3$.

Since $3 > 0$, the line $y = 2x+3$ is above the parabola $y = x^2$ in the interval $[-1, 3]$.

The area between two curves $f(x)$ and $g(x)$, where $f(x) \ge g(x)$ on $[a, b]$, is given by $\int_a^b (f(x) - g(x)) dx$.

Here, $f(x) = 2x+3$ (upper curve) and $g(x) = x^2$ (lower curve), with $a=-1$ and $b=3$.

So, the area is:

This matches option (B).

Option (A) has the integrand in the wrong order, leading to a negative area.

Option (C) sums the functions, which is incorrect for finding the area between curves.

Option (D) is $\int_{-1}^3 (2x+3) dx - \int_{-1}^3 x^2 dx$, which by the properties of definite integrals is equal to $\int_{-1}^3 ((2x+3) - x^2) dx$. So, option (D) also represents the correct area calculation.

Since the question asks "Which integral represents...", and both (B) and (D) are mathematically equivalent and correct representations, we should choose the option that most directly shows the difference. Option (B) is the combined form, which is generally preferred for simplicity.

However, if the question allows for multiple correct representations, then (D) is also correct as it is a valid application of integral properties to arrive at the same result as (B).

Let's check the common interpretation of such multiple-choice questions. Option (B) is the direct representation of the area calculation. Option (D) is a mathematically equivalent form.

Let's evaluate the integral to be sure: $\int_{-1}^3 (2x+3 - x^2) dx = [x^2 + 3x - \frac{x^3}{3}]_{-1}^3$.

$= (3^2 + 3(3) - \frac{3^3}{3}) - ((-1)^2 + 3(-1) - \frac{(-1)^3}{3})$

$= (9 + 9 - 9) - (1 - 3 - (-\frac{1}{3}))$

$= 9 - (-2 + \frac{1}{3}) = 9 - (-\frac{6}{3} + \frac{1}{3}) = 9 - (-\frac{5}{3}) = 9 + \frac{5}{3} = \frac{27+5}{3} = \frac{32}{3}$.

Both (B) and (D) represent this calculation.

The most direct representation of the area calculation is option (B). Option (D) is also mathematically correct and represents the same calculation. If only one option can be selected, the question implies a single integral form. If multiple can be selected, then both (B) and (D) would be valid.

However, often in such questions, the combined integral form is preferred as the primary answer.

The correct option is (B).

The area of the region bounded by the curve $y = e^x$, the x-axis, and the lines $x=0$ and $x=1$ is found by integrating the function $y=e^x$ with respect to $x$ from $x=0$ to $x=1$.

The function $y = e^x$ is always positive, so the area is directly given by the definite integral:

Now, we evaluate the integral:

Substitute the limits of integration:

Since $e^1 = e$ and $e^0 = 1$, we have:

The area of the region is $e-1$.

The correct option is (A).

To find the area of the region bounded by the curve $y = \sin x$, the x-axis, from $x=0$ to $x=2\pi$, we need to consider that the area is always non-negative. The function $\sin x$ is positive in the interval $(0, \pi)$ and negative in the interval $(\pi, 2\pi)$.

Therefore, we need to calculate the area in two parts:

1. The area from $x=0$ to $x=\pi$, where $\sin x \ge 0$. Area$_1 = \int_0^\pi \sin x dx$.
2. The area from $x=\pi$ to $x=2\pi$, where $\sin x \le 0$. We take the absolute value of the integral. Area$_2 = \left| \int_\pi^{2\pi} \sin x dx \right|$.

Let's calculate these:

Area$_1 = \int_0^\pi \sin x dx = [-\cos x]_0^\pi = (-\cos \pi) - (-\cos 0) = (-(-1)) - (-1) = 1 + 1 = 2$.

Area$_2 = \left| \int_\pi^{2\pi} \sin x dx \right| = \left| [-\cos x]_\pi^{2\pi} \right| = \left| (-\cos 2\pi) - (-\cos \pi) \right| = \left| (-1) - (-(-1)) \right| = |-1 - 1| = |-2| = 2$.

The total area is the sum of Area$_1$ and Area$_2$:

Note that if we were to simply calculate $\int_0^{2\pi} \sin x dx$, the result would be $[-\cos x]_0^{2\pi} = -\cos(2\pi) - (-\cos 0) = -1 - (-1) = 0$, which represents the net signed area, not the geometric area.

The correct option is (C).

Let's analyze each statement:

Statement (A): Area under $y=f(x)$ from $a$ to $b$ is $\int_a^b f(x) dx$ regardless of whether $f(x)$ is above or below the x-axis.

This statement is false. The integral $\int_a^b f(x) dx$ calculates the *net signed area*. If $f(x)$ is below the x-axis, the integral yields a negative value, whereas the geometric area is always non-negative. For example, $\int_0^{2\pi} \sin x dx = 0$, but the area between $y=\sin x$ and the x-axis from $0$ to $2\pi$ is $4$. To find the geometric area, we must use the absolute value of the function.

Statement (B): If $f(x) \le 0$ for $x \in [a, b]$, the area is $|\int_a^b f(x) dx|$.

This statement is true. If $f(x)$ is non-positive over the interval $[a, b]$, then $\int_a^b f(x) dx$ will be non-positive. The geometric area is the absolute value of this integral, which makes it positive. For instance, if $f(x)=-2$ for $x \in [0,1]$, the area is $|\int_0^1 -2 dx| = |-2| = 2$. This is correct.

Statement (C): If a region is bounded by $x=g(y)$, the y-axis, $y=c$ and $y=d$, the area is $\int_c^d g(y) dy$ assuming $g(y) \ge 0$ on $[c, d]$.

This statement is true. This is analogous to finding the area under a curve $y=f(x)$ above the x-axis. When integrating with respect to $y$, if the curve $x=g(y)$ is to the right of the y-axis (i.e., $g(y) \ge 0$) and the interval of integration for $y$ is $[c, d]$, the integral $\int_c^d g(y) dy$ correctly gives the area bounded by the curve, the y-axis, and the horizontal lines $y=c$ and $y=d$. For example, the area bounded by $x=y^2$ and the y-axis from $y=0$ to $y=2$ is $\int_0^2 y^2 dy$, which is correct since $y^2 \ge 0$ for $y \in [0,2]$.

Statement (D): The area between $y=f(x)$ and $y=g(x)$ from $a$ to $b$ is $\int_a^b |f(x)-g(x)| dx$.

This statement is true. This is the general formula for the area between two curves. The absolute value ensures that the integrand is always non-negative, accounting for cases where $g(x) > f(x)$ in some subintervals. This correctly captures the geometric area between the curves.

The true statements are (B), (C), and (D).

The correct options are (B), (C), and (D).

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): The area of the region bounded by the curve $y^2 = x$ and the line $x=1$ is $4/3$ square units.

To find the area, we first identify the boundaries. The curve is $y^2 = x$. The line is $x=1$. The region is bounded by these two curves. From $y^2 = x$, we have $y = \pm \sqrt{x}$. The line $x=1$ intersects the curve $y^2=x$ when $y^2=1$, so $y=\pm 1$. The points of intersection are $(1, 1)$ and $(1, -1)$.

We can integrate with respect to $x$. The upper boundary is $y = \sqrt{x}$ and the lower boundary is $y = -\sqrt{x}$. The integration limits are from $x=0$ (the vertex of the parabola $x=y^2$) to $x=1$.

The area is given by $\int_0^1 (\sqrt{x} - (-\sqrt{x})) dx = \int_0^1 2\sqrt{x} dx$.

Let's evaluate this integral:

So, the area is indeed $4/3$ square units. Assertion (A) is true.

Reason (R): The area is given by $\int_0^1 2\sqrt{x} dx$.

As shown in the evaluation of Assertion (A), the integral $\int_0^1 2\sqrt{x} dx$ correctly represents the area bounded by $y^2=x$ and $x=1$. The function $y=\sqrt{x}$ is the upper half of the parabola, and $y=-\sqrt{x}$ is the lower half. The difference $2\sqrt{x}$ represents the vertical width of the region at a given $x$. The integration from $x=0$ to $x=1$ covers the entire region bounded by the parabola and the line $x=1$. Thus, Reason (R) is true.

Relationship between Assertion (A) and Reason (R):

Reason (R) provides the correct integral setup that leads to the calculation of the area mentioned in Assertion (A). The integral in (R) is precisely the one used to derive the area value in (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains how to calculate the area stated in Assertion (A).

The correct option is (A).

Let's match the descriptions in Column I with the correct integral forms in Column II:

(i) Area bounded by $y=f(x)$, x-axis, $x=a, x=b$ ($f(x)\ge 0$)

When $f(x)$ is non-negative, the area under the curve $y=f(x)$ above the x-axis, between $x=a$ and $x=b$, is given directly by the integral $\int_a^b f(x) dx$. This corresponds to option (c).

(ii) Area bounded by $x=g(y)$, y-axis, $y=c, y=d$ ($g(y)\ge 0$)

When $g(y)$ is non-negative, the area bounded by the curve $x=g(y)$ to the right of the y-axis, between $y=c$ and $y=d$, is given by the integral $\int_c^d g(y) dy$. This corresponds to option (a).

(iii) Area between $y=f(x)$ and x-axis from $a$ to $b$ where $f(x)$ crosses x-axis

When $f(x)$ changes sign in the interval $[a, b]$, the total geometric area is found by integrating the absolute value of $f(x)$, which is $\int_a^b |f(x)| dx$. This ensures that areas below the x-axis are counted as positive. This corresponds to option (b).

(iv) Area between $y=f(x)$ and $y=g(x)$ from $a$ to $b$ ($f(x) \ge g(x)$)

The area between two curves $y=f(x)$ and $y=g(x)$, where $f(x)$ is the upper curve and $g(x)$ is the lower curve, is given by integrating the difference between the upper and lower functions: $\int_a^b (f(x)-g(x)) dx$. This corresponds to option (d).

The correct matches are:

• (i) $\rightarrow$ (c)
• (ii) $\rightarrow$ (a)
• (iii) $\rightarrow$ (b)
• (iv) $\rightarrow$ (d)

The equation of the parabola is given as $x^2 = 4ay$. This is a parabola opening upwards with its vertex at the origin $(0,0)$.

The latus rectum is a line segment passing through the focus and perpendicular to the axis of symmetry. For the parabola $x^2 = 4ay$, the focus is at $(0, a)$ and the axis of symmetry is the y-axis. The latus rectum is the horizontal line $y = a$.

We need to find the area of the region bounded by the parabola $x^2 = 4ay$ and its latus rectum $y=a$.

To find the area, we can integrate with respect to $y$. From the equation of the parabola, we can express $x$ in terms of $y$: $x^2 = 4ay \implies x = \pm \sqrt{4ay} = \pm 2\sqrt{a}\sqrt{y}$.

The region is bounded on the right by $x = 2\sqrt{a}\sqrt{y}$ and on the left by $x = -2\sqrt{a}\sqrt{y}$. The integration limits for $y$ are from the vertex ($y=0$) to the latus rectum ($y=a$).

The area is given by the integral of the difference between the right and left boundaries with respect to $y$:

Simplify the integrand:

Now, we evaluate the integral:

The area of the region bounded by the parabola $x^2 = 4ay$ and its latus rectum is $\frac{8a^2}{3}$.

The correct option is (A).

The region is bounded by the line $y=3x$, the x-axis ($y=0$), and the vertical line $x=2$. This forms a triangle with its vertex at the origin $(0,0)$.

We can find the area of this region by integration or by using the formula for the area of a triangle.

Method 1: Integration

The area is given by the integral of the function $y=3x$ from $x=0$ to $x=2$, since the function is above the x-axis in this interval.

Evaluate the integral:

The area is $6$ square units.

Method 2: Geometric formula

The region is a triangle with vertices at $(0,0)$, $(2,0)$, and $(2, 3 \times 2) = (2,6)$.

The base of the triangle lies along the x-axis from $x=0$ to $x=2$, so the base length is $2$.

The height of the triangle is the perpendicular distance from the vertex $(2,6)$ to the base on the x-axis, which is $6$.

Area of a triangle $= \frac{1}{2} \times \text{base} \times \text{height}$

Area $= \frac{1}{2} \times 2 \times 6 = 6$ square units.

Both methods yield an area of $6$ square units.

The correct option is (B).

First, let's find the points of intersection of the two curves $y^2 = 4x$ and $x^2 = 4y$.

From $x^2 = 4y$, we have $y = \frac{x^2}{4}$.

Substitute this into the first equation:

This gives $x=0$ or $x^3 = 64$, which means $x=4$.

If $x=0$, then $y = \frac{0^2}{4} = 0$. So, one intersection point is $(0,0)$.

If $x=4$, then $y = \frac{4^2}{4} = \frac{16}{4} = 4$. So, the other intersection point is $(4,4)$.

Now let's consider the integrals representing the area.

From $y^2 = 4x$, we get $y = \pm \sqrt{4x} = \pm 2\sqrt{x}$. Since we are in the region of intersection where $y \ge 0$, we consider $y = 2\sqrt{x}$.

From $x^2 = 4y$, we get $y = \frac{x^2}{4}$.

To find the area between these curves from $x=0$ to $x=4$, we need to determine which function is the upper boundary. For $x \in (0, 4)$, let's test $x=1$: $y = 2\sqrt{1} = 2$ and $y = \frac{1^2}{4} = \frac{1}{4}$. Since $2 > 1/4$, $y = 2\sqrt{x}$ is the upper curve.

The area is given by $\int_0^4 (\text{upper curve} - \text{lower curve}) dx$.

Area $= \int_0^4 (2\sqrt{x} - \frac{x^2}{4}) dx$.

This matches option (B).

Let's check option (A): $\int_0^4 (\sqrt{4x} - x^2/4) dx$. Since $\sqrt{4x} = 2\sqrt{x}$, option (A) is identical to option (B).

Now let's evaluate the integral to check option (C):

So, the area is $\frac{16}{3}$ square units. This matches option (C).

Since options (A) and (B) are equivalent integral representations of the area, and option (C) gives the correct value of that area, option (D) "All of the above describe or provide the area" is the most appropriate answer.

The correct option is (D).

To find the area of the region bounded by the curve $y = x^3$, the y-axis, and the lines $y=1$ and $y=8$, we need to integrate with respect to $y$. This means we need to express $x$ in terms of $y$.

From the equation $y = x^3$, we can solve for $x$ by taking the cube root of both sides:

This can also be written as $x = y^{1/3}$.

The area is bounded by this curve ($x=y^{1/3}$), the y-axis ($x=0$), and the horizontal lines $y=1$ and $y=8$.

The area is calculated by integrating $x$ with respect to $y$ over the given interval for $y$. Since $x = y^{1/3}$ is non-negative for $y \in [1, 8]$, the area is:

This matches option (C).

Option (B) is $\int_1^8 \sqrt[3]{y} dy$, which is the same as $\int_1^8 y^{1/3} dy$. Therefore, option (B) is also correct.

Option (A) $\int_1^8 x^3 dy$ would be incorrect because we need to integrate $x$ with respect to $y$, not $x^3$. Also, the limits of integration are for $y$, so the integrand should be in terms of $y$. If we were integrating with respect to $x$, the integral would be for the area bounded by $y=x^3$ and the x-axis, which is not the case here.

Since options (B) and (C) are identical and correctly represent the area calculation, option (D) is the correct choice.

The correct option is (D).

The region is bounded by the curve $y = 4 - x^2$ and the x-axis. The curve is above the x-axis between its x-intercepts.

The x-intercepts are found by setting $y=0$: $4 - x^2 = 0 \implies x^2 = 4 \implies x = \pm 2$.

Option (A): $\int_{-2}^2 (4-x^2) dx$

This integral represents the area under the curve $y = 4-x^2$ and above the x-axis from $x=-2$ to $x=2$. Since $y = 4-x^2$ is non-negative in this interval, this integral correctly represents the area.

Option (B): $2 \int_0^2 (4-x^2) dx$

The function $y = 4-x^2$ is an even function because $f(-x) = 4 - (-x)^2 = 4 - x^2 = f(x)$. Therefore, the area from $x=-2$ to $x=2$ is twice the area from $x=0$ to $x=2$. This integral also correctly represents the area.

Option (C): $32/3$ square units

Let's evaluate the integral from option (A) or (B) to check this value:

So, the area is indeed $32/3$ square units. Option (C) is also correct.

Option (D): All of the above are relevant to finding the area.

Since options (A), (B), and (C) all correctly describe or provide the area, this option is the most comprehensive answer.

The correct option is (D).

The region is bounded by the curve $y = |x|$, the x-axis ($y=0$), and the vertical lines $x=-2$ and $x=2$.

Option (A): $\int_{-2}^2 |x| dx$

This integral represents the area under the curve $y=|x|$ from $x=-2$ to $x=2$. Since $|x| \ge 0$ for all $x$, this integral correctly calculates the geometric area.

Option (B): $2 \int_0^2 x dx$

The function $y=|x|$ is an even function ($f(-x) = |-x| = |x| = f(x)$). Therefore, the area from $x=-2$ to $x=2$ is twice the area from $x=0$ to $x=2$. For $x \ge 0$, $|x|=x$. So, $\int_0^2 |x| dx = \int_0^2 x dx$. Thus, $2 \int_0^2 x dx$ correctly represents the total area.

Option (C): $4$ square units

Let's evaluate the integral from option (A) or (B) to check this value.

Using option (B):

Alternatively, using geometry: The region is a triangle with vertices at $(0,0)$, $(-2,2)$, and $(2,2)$. The base lies on the line $y=2$ from $x=-2$ to $x=2$, so the base length is $2 - (-2) = 4$. The height is the perpendicular distance from the vertex at the origin to the line $y=2$, which is $2$.

Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 2 = 4$ square units.

Option (C) is also correct.

Option (D): All of the above describe or provide the area.

Since options (A), (B), and (C) all correctly represent or calculate the area, this option is the most fitting.

The correct option is (D).

When calculating the area between two curves using integration, we consider elements of area. If we are integrating with respect to $x$ (using vertical strips), the elements of area are thin rectangles with height $(f(x)-g(x))$ and width $dx$. The axis parallel to these elements of area is the y-axis, and the integration is along the x-axis.

If we are integrating with respect to $y$ (using horizontal strips), the elements of area are thin rectangles with width $(g(y)-h(y))$ and height $dy$. The axis parallel to these elements of area is the x-axis, and the integration is along the y-axis.

The statement asks about the variable of integration corresponding to the axis ____ to the elements of area. The elements of area are typically rectangles. If the width is $dx$, the height is $(f(x)-g(x))$, and the element of area is approximately $(f(x)-g(x))dx$. The "axis" here refers to the axis along which the integration is performed.

Let's rephrase: the variable of integration (e.g., $x$ or $y$) represents a distance along an axis. The elements of area (dA) are typically thought of as being formed by differentials along the axes. For integration with respect to $x$, we have elements of area $dA = (f(x)-g(x))dx$. The $dx$ is a differential along the x-axis. The y-axis is parallel to the height of these strips $(f(x)-g(x))$.

Consider the wording "variable corresponding to the axis ____ to the elements of area". The elements of area are strips. If we integrate with respect to $x$, the strips are vertical, and $dx$ is along the x-axis. The y-axis is parallel to the height of these strips. The axis of integration is the x-axis.

Let's think about the orientation of the elements of area. For integration with respect to $x$, we have vertical strips of width $dx$. The height is $(f(x)-g(x))$. The y-axis is parallel to the height of these strips.

For integration with respect to $y$, we have horizontal strips of height $dy$. The width is $(g(y)-h(y))$. The x-axis is parallel to the width of these strips.

The statement is asking about the relationship between the variable of integration and the elements of area relative to an axis.

To find the area between two curves $y=f(x)$ and $y=g(x)$, we integrate with respect to $x$. The elements of area are vertical strips of width $dx$. The axis parallel to these elements of area (which are essentially lines parallel to the y-axis) is the y-axis. The integration is along the x-axis.

The phrasing is a bit ambiguous. Let's consider what is meant by "axis ____ to the elements of area." The elements of area are thin rectangles. If we integrate with respect to $x$, these rectangles have their base along the x-axis (with width $dx$). Their height is $(f(x)-g(x))$. The y-axis is parallel to the height of these rectangles.

However, if we interpret "axis" as the axis of integration, then the variable of integration corresponds to a distance along that axis. The elements of area are perpendicular to the axis of integration.

Let's re-read: "integrate the absolute difference of the functions with respect to the variable corresponding to the axis ____ to the elements of area."

If we integrate with respect to $x$, the elements of area are vertical strips. These strips are perpendicular to the x-axis (the axis of integration). The y-axis is parallel to the height of these strips.

The phrasing "variable corresponding to the axis ____ to the elements of area" implies a relationship between the axis along which the integration variable measures distance and the elements of area themselves.

Consider a vertical strip of area $dA = (f(x)-g(x))dx$. The $dx$ is along the x-axis. The elements of area are "aligned" along the x-axis. The y-axis is perpendicular to the x-axis. The height of the strip is along the y-axis direction.

If the elements of area are vertical strips, they are perpendicular to the axis of integration (x-axis). The y-axis is parallel to the height of these strips.

The question is about the variable of integration and its relation to the axis and elements of area. The variable of integration (e.g., $x$) measures distance along an axis (e.g., the x-axis). The elements of area are formed by this variable's differential ($dx$) and the difference in function values.

The elements of area (dA) are usually thought of as being formed by a differential along the axis of integration. If we integrate with respect to $x$, we have $dA \approx (f(x)-g(x))dx$. The $dx$ is a segment along the x-axis. The elements of area are arranged along the x-axis.

Let's think about the structure of the integral $\int_a^b h(x) dx$. The $dx$ represents a small segment along the x-axis. The elements of area are like thin rectangles whose width is $dx$ and height is $h(x)$. These elements are arranged along the x-axis.

The axis that is **perpendicular** to these elements of area (meaning perpendicular to the base $dx$ and parallel to the height $h(x)$) is the y-axis. However, the statement asks for the axis that is ____ to the elements of area.

Let's reconsider the statement: "integrate with respect to the variable corresponding to the axis ____ to the elements of area." The variable of integration ($x$) corresponds to the x-axis. The elements of area are roughly rectangles with base $dx$ and height $(f(x)-g(x))$. These rectangles are oriented such that their base is along the x-axis, and their height is parallel to the y-axis. The axis of integration (x-axis) is parallel to the bases of these elements of area.

If we integrate with respect to $y$, the elements are horizontal strips with base $dy$ along the y-axis. Their width is $(g(y)-h(y))$ along the x-axis. The axis of integration (y-axis) is parallel to the bases of these elements of area.

Therefore, the variable of integration corresponds to the axis that is **parallel** to the elements of area (specifically, their bases).

The correct option is (A).

To find the area of the region bounded by $y^2 = x$ and $x = 2y$, we first find the points of intersection.

Substitute $x = 2y$ into $y^2 = x$:

Rearrange the equation:

Factor out $y$:

This gives $y=0$ or $y=2$.

Now find the corresponding $x$ values using $x=2y$:

• When $y=0$, $x = 2(0) = 0$. Intersection point is $(0,0)$.
• When $y=2$, $x = 2(2) = 4$. Intersection point is $(4,2)$.

We can calculate the area by integrating with respect to $y$. The limits of integration for $y$ are from $0$ to $2$. We need to express $x$ in terms of $y$ for both curves.

From $y^2 = x$, we have $x = y^2$.

From $x = 2y$, we have $x = 2y$.

To determine which function is the right boundary and which is the left boundary, we test a value between $y=0$ and $y=2$, say $y=1$.

For $x = y^2$, when $y=1$, $x = 1^2 = 1$.

For $x = 2y$, when $y=1$, $x = 2(1) = 2$.

Since $2 > 1$, the line $x = 2y$ is to the right of the curve $x = y^2$ in the interval $y \in (0, 2)$.

The area is given by $\int_c^d (\text{right boundary} - \text{left boundary}) dy$.

Area $= \int_0^2 (2y - y^2) dy$. This matches option (B).

Let's evaluate this integral:

The area is $4/3$ square units, which matches option (C).

Option (A) integrates with respect to $x$. From $y^2=x$, we get $y=\sqrt{x}$ (upper branch). From $x=2y$, we get $y=x/2$. The area is $\int_0^4 (\sqrt{x} - x/2) dx$. This is a correct representation of the area.

Let's evaluate option (A) to confirm:

Both option (A) and (B) represent the area, and option (C) gives the correct value. Therefore, option (D) is the most appropriate answer.

The correct option is (D).

The curve $y = \sqrt{a^2 - x^2}$ represents the upper semi-circle of radius $a$ centered at the origin, since $y \ge 0$. The x-axis is $y=0$. The domain for $x$ where $y$ is real is $-a \le x \le a$.

Option (A): $\int_{-a}^a \sqrt{a^2 - x^2} dx$

This integral directly represents the area under the curve $y = \sqrt{a^2 - x^2}$ from $x=-a$ to $x=a$, which is the area of the upper semi-circle. This is a correct representation of the area.

Option (B): $2 \int_0^a \sqrt{a^2 - x^2} dx$

The function $y = \sqrt{a^2 - x^2}$ is an even function because the domain $[-a, a]$ is symmetric about $x=0$, and $\sqrt{a^2 - (-x)^2} = \sqrt{a^2 - x^2}$. Therefore, the area from $x=-a$ to $x=a$ is twice the area from $x=0$ to $x=a$. This integral also correctly represents the area of the semi-circle.

Option (C): $\frac{1}{2} \pi a^2$

The area of a semi-circle with radius $a$ is half the area of a full circle ($\pi a^2$). So, the area of the semi-circle is indeed $\frac{1}{2} \pi a^2$. This is the calculated value of the area.

Option (D): All of the above are related to the area.

Since options (A) and (B) are correct integral representations, and option (C) is the correct calculated value of the area, all these statements are related to finding the area of the semi-circle.

The correct option is (D).

The region is bounded by the curve $y=x^2$, the x-axis ($y=0$), and the line $x=-2$. Since the parabola $y=x^2$ is symmetric about the y-axis and is always non-negative ($x^2 \ge 0$), the region bounded by $y=x^2$, the x-axis, and the line $x=-2$ implies the area from the y-axis (where $x=0$) to the line $x=-2$.

The area under a curve $y=f(x)$ above the x-axis from $x=a$ to $x=b$ is given by $\int_a^b f(x) dx$, provided $f(x) \ge 0$ in that interval. In this case, $f(x) = x^2$, which is always non-negative.

The boundaries are the curve $y=x^2$, the x-axis ($y=0$), and the line $x=-2$. The natural extent of the parabola to the left of the y-axis and bounded by $x=-2$ implies the region is between $x=-2$ and $x=0$.

Therefore, the area is given by the integral of $x^2$ from $x=-2$ to $x=0$.

Area $= \int_{-2}^0 x^2 dx$. This matches option (A).

Let's examine the other options:

• Option (B): $\int_0^{-2} x^2 dx$. The order of limits in a definite integral is important. $\int_a^b f(x) dx = -\int_b^a f(x) dx$. While this integral is mathematically related, the standard representation for area from a lower limit to an upper limit is with the lower limit first.
• Option (C): $|\int_{-2}^0 x^2 dx|$. Since $x^2$ is non-negative on $[-2, 0]$, $\int_{-2}^0 x^2 dx$ will be non-negative. So, $|\int_{-2}^0 x^2 dx| = \int_{-2}^0 x^2 dx$. This is also a correct way to express the area.
• Option (D): $\int_{-2}^0 |x^2| dx$. Since $x^2$ is always non-negative, $|x^2| = x^2$. So, this integral is $\int_{-2}^0 x^2 dx$, which is also correct.

All options (A), (C), and (D) are correct representations of the area. However, the most direct and standard representation when $f(x) \ge 0$ is $\int_a^b f(x) dx$. The question asks for "the area", implying a single best representation among the choices.

Given that $y=x^2$ is always non-negative, options (A), (C), and (D) are all mathematically correct ways to express the area. Option (A) is the most direct form. Option (C) and (D) use absolute values which are redundant here but don't make the expression incorrect.

Let's evaluate the integral: $\int_{-2}^0 x^2 dx = [\frac{x^3}{3}]_{-2}^0 = \frac{0^3}{3} - \frac{(-2)^3}{3} = 0 - \frac{-8}{3} = \frac{8}{3}$. The area is $8/3$ square units.

Since the question asks which integral *represents* the area, and all (A), (C), and (D) correctly represent it, and (A) is the most straightforward representation, we should choose the option that encompasses all correct forms if available, or the most direct one.

The question structure seems to imply selecting one best option. Option (A) is the most direct integral. Options (C) and (D) are also correct representations. If only one can be chosen, (A) is the most fundamental. However, often in "select all that apply" type questions, multiple can be correct. This is not stated here.

Let's assume the question is asking for a correct integral formulation. All (A), (C), and (D) are correct integral formulations. If this were a multiple correct answers question, then A, C, and D would be chosen. However, it appears to be a single choice question.

In the context of finding area, when the function is non-negative, the integral $\int_a^b f(x) dx$ is the standard representation. The use of absolute value or splitting the integral is for when the function changes sign.

Option (A) is the most direct and standard way to write the integral for the area when the function is non-negative over the interval.

The correct option is (A).

The region is in the first quadrant and is bounded by the line $y=x$, the horizontal line $y=1$, and the y-axis ($x=0$).

We can find this area by integrating with respect to $x$ or with respect to $y$. Let's consider both:

Method 1: Integrating with respect to $x$.

The region is bounded below by $y=x$ and above by $y=1$. The intersection of $y=x$ and $y=1$ is at $x=1$. The region is also bounded by the y-axis ($x=0$). So the limits for $x$ are from $0$ to $1$.

The area is given by:

This matches option (B).

Method 2: Integrating with respect to $y$.

We need to express $x$ in terms of $y$. From $y=x$, we have $x=y$. The region is bounded on the right by $x=y$ and on the left by the y-axis ($x=0$). The limits for $y$ are from $y=0$ (implied by the first quadrant and the intersection with the y-axis) up to $y=1$ (the horizontal line). However, the region is bounded by $y=x$, $y=1$, and the y-axis. This forms a triangle with vertices $(0,0)$, $(0,1)$, and $(1,1)$.

If we integrate with respect to $y$, the right boundary is $x=y$ and the left boundary is $x=0$. The limits for $y$ are from $0$ to $1$.

The area is given by:

This matches option (C).

Let's evaluate the integral from option (B):

Let's evaluate the integral from option (C):

Both options (B) and (C) correctly represent the area, and both yield the correct value of $1/2$.

Option (A) $\int_0^1 x dx$ represents the area under $y=x$ from $0$ to $1$, which is the same as option (C), so (A) is also a correct representation of the area.

Since the question asks for "the area", and multiple options correctly represent it, we should select the option that reflects this. However, if we must choose only one that represents the area calculation based on the typical presentation of such problems:

• Option (A) is the area under $y=x$ from $0$ to $1$.
• Option (B) is the area between $y=1$ and $y=x$ from $x=0$ to $x=1$.
• Option (C) is the area between $x=y$ and $x=0$ from $y=0$ to $y=1$.

Geometrically, the region described is a triangle with vertices $(0,0)$, $(0,1)$, and $(1,1)$.

Option (A) $\int_0^1 x dx$ represents the area of the triangle with vertices $(0,0)$, $(1,0)$, and $(1,1)$. This is not the described region.

Option (B) $\int_0^1 (1-x) dx$ represents the area of the triangle described, as it is the area under $y=1$ from $0$ to $1$ minus the area under $y=x$ from $0$ to $1$. This correctly describes the area of the triangle with vertices $(0,0)$, $(0,1)$, and $(1,1)$.

Option (C) $\int_0^1 y dy$ represents the area between the curve $x=y$ and the y-axis from $y=0$ to $y=1$. This is also the area of the triangle with vertices $(0,0)$, $(0,1)$, and $(1,1)$.

Since both (B) and (C) correctly represent the area, and (A) does not directly represent the described region (it represents a different triangle), we need to be careful.

Let's re-visualize the region. The boundaries are $y=x$, $y=1$, and the y-axis ($x=0$).

Vertices are: * Intersection of $y=x$ and y-axis ($x=0$): $(0,0)$. * Intersection of $y=1$ and y-axis ($x=0$): $(0,1)$. * Intersection of $y=x$ and $y=1$: $(1,1)$. This is indeed a triangle with vertices $(0,0), (0,1), (1,1)$.

Area using integration with respect to $x$: Upper curve is $y=1$, lower curve is $y=x$. Limits are $x=0$ to $x=1$. So, $\int_0^1 (1-x) dx$. This is option (B).

Area using integration with respect to $y$: Right curve is $x=y$, left curve is $x=0$. Limits are $y=0$ to $y=1$. So, $\int_0^1 (y-0) dy = \int_0^1 y dy$. This is option (C).

Option (A) represents the area of the triangle with vertices $(0,0), (1,0), (1,1)$.

Since both (B) and (C) are correct representations of the area, and the question doesn't specify integration variable preference, both are valid.

However, the question asks for "the area...is:", implying a specific integral form. Typically, if multiple integral forms are correct, the options would reflect that.

Let's check if there is any subtle difference or preference.

Option (B) integrates with respect to $x$, as described by the bounds $x=0$ and the implied $x=1$.

Option (C) integrates with respect to $y$, as described by the bounds $y=0$ and $y=1$.

Both are perfectly valid ways to represent the area.

If the question were "Which of the following integrals represents the area...", and both (B) and (C) were options, and no other option like "Both B and C" existed, then there might be a preferred method based on context or typical examples.

However, if we have to choose the "best" fit based on how the problem is stated (bounded by $y=x$, $y=1$, y-axis), integrating with respect to $y$ might seem more direct as $y=1$ and the y-axis are given as explicit boundaries for $y$ and $x$ respectively.

Let's consider the wording carefully: "bounded by $y=x$, $y=1$, and the y-axis".

When integrating with respect to $y$, the boundaries $y=0$ and $y=1$ are explicit. The bounding curves are $x=y$ and $x=0$. So $\int_0^1 (y-0) dy$ is a direct representation.

When integrating with respect to $x$, the boundaries are $x=0$ and $x=1$. The bounding curves are $y=1$ and $y=x$. So $\int_0^1 (1-x) dx$ is also a direct representation.

Both are equally valid. If we must pick one, it's usually context-dependent or a flaw in the question if both are listed without a "both" option.

Let's re-evaluate the options and the problem statement.

The problem statement doesn't give a preference for the integration variable.

If this is a single-choice question, and both (B) and (C) are correct integral representations, there might be an implicit preference. Typically, when boundaries are given in both $x$ and $y$, either method is acceptable.

Let's assume there might be a preferred way based on typical problem structures.

The description of the region can be visualized as a triangle. The vertices are $(0,0)$, $(0,1)$, $(1,1)$.

Option (B) integrates with respect to $x$. The integrand is $(1-x)$, which is the difference in y-values. The limits are $x=0$ to $x=1$. This is correct.

Option (C) integrates with respect to $y$. The integrand is $y$, which is the difference in x-values (since $x=y$ and $x=0$). The limits are $y=0$ to $y=1$. This is correct.

Option (A) $\int_0^1 x dx$ represents the area of a triangle with vertices $(0,0), (1,0), (1,1)$. This is a different region.

Both (B) and (C) are correct representations. Since there is no "Both B and C" option, we need to consider if one is more appropriate or if there's a standard convention. Usually, if both are correct, they would be grouped. If only one can be chosen, it could be that one form is more directly derived from the bounding curves as given.

Given $y=x$, $y=1$, and the y-axis ($x=0$).

Integrating with respect to $y$ uses the bounds $y=0$ and $y=1$ directly, and the curves $x=y$ and $x=0$. This seems very direct.

Integrating with respect to $x$ uses bounds $x=0$ and $x=1$. The bounds for $x$ are derived from the intersection of $y=x$ and $y=1$. The curves are $y=1$ and $y=x$. This is also direct.

Let's assume the question intends for us to pick one valid integral. Both (B) and (C) are valid.

In absence of further constraints, both (B) and (C) are correct. However, in multiple choice questions of this type, if two options are equally valid integral representations, there might be a slight preference or a mistake in the question design if only one answer is expected. Given the options, and that both (B) and (C) are correct integral representations, let's consider the possibility that the question is looking for a specific formulation.

Often, problems present boundaries in terms of $y$ (like $y=1$) and the y-axis ($x=0$), making integration with respect to $y$ a natural first step.

Let's re-examine the wording. "bounded by $y=x$, $y=1$, and the y-axis". This clearly defines the region.

Both (B) and (C) correctly calculate this area.

If forced to choose one, the phrasing might subtly hint at one over the other. "bounded by $y=x$, $y=1$" suggests a difference in $y$ values. "and the y-axis" might suggest integration with respect to $y$.

However, the structure of (B) using $x$ as the variable and the limits $0$ to $1$ derived from the intersection, and the integrand $(1-x)$ also fits perfectly.

Let's consider the typical way such a question might be posed in a textbook or exam. Both are common ways to set up the integral. It's possible that the question designers intended one form, or that it's poorly designed if both are valid and equally presented.

If we consider the boundaries as given: $y=x$ (which is $x=y$), $y=1$, and $x=0$. Integrating with respect to $y$ uses these directly: $\int_0^1 (y-0) dy$. This is option (C).

If we consider the boundaries in terms of $x$: $y=x$ and $y=1$. The intersection gives $x=1$. The y-axis gives $x=0$. So limits are $x=0$ to $x=1$. The integrand is $(1-x)$. This is option (B).

Both seem equally valid.

Given the commonality of both approaches, and no explicit preference stated, this is a difficult choice without context. However, the phrasing "bounded by $y=x$, $y=1$, and the y-axis" highlights $y=1$ and the y-axis ($x=0$) as key boundaries. If we think of horizontal strips, bounded by $x=y$ and $x=0$, integrated from $y=0$ to $y=1$, option (C) is a direct fit.

The correct option is (C).

Let's analyze the given statement.

The area of the region bounded by $y=x$ and $y=x^2$ is calculated by integrating the difference between the upper curve ($y=x$) and the lower curve ($y=x^2$) from their intersection points. The intersection points are $x=0$ and $x=1$, and the area is $\int_0^1 (x - x^2) dx = \frac{1}{6}$.

The area of the region bounded by $x=y$ and $x=y^2$. The intersection points are $y=0$ and $y=1$. The area is calculated by integrating with respect to $y$ the difference between the right curve ($x=y$) and the left curve ($x=y^2$). The area is $\int_0^1 (y - y^2) dy$. This integral is structurally identical to the first one, just with the variable $y$ instead of $x$, and it also evaluates to $\frac{1}{6}$.

Let's consider the options:

(A) Symmetry

The line $y=x$ is the line of symmetry for the graphs of a function and its inverse. The parabola $y=x^2$ (for $x \ge 0$) and the curve $x=y^2$ (which is $y=\sqrt{x}$ for $y \ge 0$) are reflections of each other across the line $y=x$. The fact that the area is the same arises from this symmetry.

(B) Coincidence

Coincidence implies that two things are identical. While the areas are the same, the statement isn't just about the numerical value being identical, but about the underlying reason why they are identical.

(C) A general property of functions and their inverses

The relationship between $y=x^2$ (for $x \ge 0$) and $x=y^2$ (which is $y=\sqrt{x}$) is that they are inverse functions (within appropriate domains). The fact that the area calculated using $y$ as the integration variable and $x=y$ and $x=y^2$ is the same as when using $x$ as the integration variable and $y=x$ and $y=x^2$ is related to the geometric interpretation of integrating with respect to $x$ versus $y$ for inverse functions.

The property that the area enclosed by a curve, its inverse, and the line $y=x$ exhibits certain symmetries is a general property related to inverse functions.

(D) Both (A) and (C)

The reason the areas are the same is due to the symmetry of the graphs of $y=x^2$ and $x=y^2$ (or $y=\sqrt{x}$) with respect to the line $y=x$. This symmetry is a manifestation of them being inverse functions. Therefore, both symmetry and the properties of inverse functions are relevant explanations.

The correct option is (D).

To find the area of the region bounded by the curve $y = x^2+1$, the x-axis ($y=0$), and the lines $x=-1$ and $x=2$, we need to integrate the function $y=x^2+1$ with respect to $x$ over the interval $[-1, 2]$.

First, let's check the sign of the function $f(x) = x^2+1$ over the interval $[-1, 2]$.

For any real value of $x$, $x^2 \ge 0$. Therefore, $x^2+1 \ge 0+1 = 1$. This means the function $y=x^2+1$ is always positive and is always above the x-axis.

When the function is non-negative over the entire interval of integration, the area is directly given by the definite integral of the function over that interval.

The area is calculated as:

This matches option (A).

Let's evaluate this integral:

The area is $6$ square units.

Now let's look at the options:

• Option (A) is the correct integral form.
• Option (B) is $|\int_{-1}^2 (x^2+1) dx|$. Since the integral's value is $6$ (positive), $|\int_{-1}^2 (x^2+1) dx| = |6| = 6$. So, this is also a correct representation of the area.
• Option (C) is $\int_{-1}^2 (x^2+1) dx$ (since $x^2+1 \ge 0$). This is essentially the same as option (A) but adds the justification that the function is non-negative, making the absolute value in (B) redundant. This is also a correct statement.

Since options (A) and (C) are essentially the same correct integral form, and option (B) is also a correct representation (though redundant in its absolute value), and the question asks what "is" the area, and options (A) and (C) are the integral that calculates it, option (D) "Both (A) and (C)" is the most appropriate choice as it covers both correct integral representations.

The correct option is (D).

To find the area of the region bounded by the curves $y = \sin x$ and $y = \cos x$ from $x=0$ to $x=\pi/2$, we first need to find their intersection point(s) in this interval.

Set $\sin x = \cos x$:

In the interval $[0, \pi/2]$, the solution is $x = \pi/4$.

Now we need to determine which function is greater in the intervals $[0, \pi/4]$ and $[\pi/4, \pi/2]$.

• In the interval $[0, \pi/4]$, $\cos x \ge \sin x$. For example, at $x=0$, $\cos 0 = 1$ and $\sin 0 = 0$.
• In the interval $[\pi/4, \pi/2]$, $\sin x \ge \cos x$. For example, at $x=\pi/2$, $\sin (\pi/2) = 1$ and $\cos (\pi/2) = 0$.

The area between two curves $f(x)$ and $g(x)$ from $a$ to $b$ is given by $\int_a^b |f(x) - g(x)| dx$. In this case, we need to split the integral at the intersection point $x=\pi/4$.

Area $= \int_0^{\pi/4} |\cos x - \sin x| dx + \int_{\pi/4}^{\pi/2} |\sin x - \cos x| dx$.

Since $\cos x \ge \sin x$ on $[0, \pi/4]$, $|\cos x - \sin x| = \cos x - \sin x$.

Since $\sin x \ge \cos x$ on $[\pi/4, \pi/2]$, $|\sin x - \cos x| = \sin x - \cos x$.

So, the area is:

This matches option (D).

Let's examine why the other options are incorrect:

• Option (A): $\int_0^{\pi/2} (\sin x - \cos x) dx$. This assumes $\sin x \ge \cos x$ over the entire interval, which is not true. This would calculate the signed area and might not be the correct geometric area.
• Option (B): $\int_0^{\pi/2} (\cos x - \sin x) dx$. This assumes $\cos x \ge \sin x$ over the entire interval, which is also not true.
• Option (C): $|\int_0^{\pi/2} (\sin x - \cos x) dx|$. This takes the absolute value of the net signed area over the entire interval, which is not necessarily the correct geometric area if the integrand changes sign within the interval.

Option (D) correctly accounts for the change in the upper and lower functions by splitting the integral at the intersection point and using the absolute difference.

The correct option is (D).

The region is bounded by the parabola $y=x^2$, the horizontal lines $y=1$ and $y=4$. We are asked to integrate with respect to $y$.

First, we need to express $x$ in terms of $y$. From $y=x^2$, we get $x = \pm \sqrt{y}$.

Since we are in the first quadrant (implicitly, or considering the full region bounded by the parabola and horizontal lines, which is symmetric about the y-axis), the right boundary is $x = \sqrt{y}$ and the left boundary is $x = -\sqrt{y}$.

The area is calculated by integrating the difference between the right boundary and the left boundary with respect to $y$, from $y=1$ to $y=4$.

Area $= \int_1^4 (\text{right boundary} - \text{left boundary}) dy$

Simplifying the integrand:

This matches option (C).

Let's examine the other options:

• Option (A): $\int_1^4 (4-1) dy$. This represents the area of a rectangle with height $(4-1)=3$ and width $dy$, integrated over $y$. This is not correct as it doesn't involve the curve $y=x^2$.
• Option (B): $\int_1^4 \sqrt{y} dy$. This represents the area between the curve $x=\sqrt{y}$ and the y-axis from $y=1$ to $y=4$, but it doesn't account for the left boundary ($x=-\sqrt{y}$).
• Option (D): $\int_1^4 (4-1) dx$. This integral is with respect to $x$, but the region is defined by bounds on $y$, and we are asked to integrate with respect to $y$. Also, $(4-1)$ is a constant, not the difference of functions of $x$.

Therefore, option (C) is the correct representation of the area when integrating with respect to $y$. If we were to calculate the value:

The area is $\frac{28}{3}$ square units.

The correct option is (C).

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): The area of the region bounded by $y=|x|$ and the x-axis from $x=-1$ to $x=1$ is 1 square unit.

The region is bounded by the V-shaped graph of $y=|x|$, the x-axis, and the vertical lines $x=-1$ and $x=1$. This forms a triangle with vertices at $(0,0)$, $(-1,1)$, and $(1,1)$. The base of this triangle lies on the line $y=1$ from $x=-1$ to $x=1$, so the base length is $1 - (-1) = 2$. The height of the triangle is the perpendicular distance from the vertex $(0,0)$ to the line $y=1$, which is $1$.

The area of this triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 1 = 1$ square unit.

Thus, Assertion (A) is true.

Reason (R): The area is $\int_{-1}^1 |x| dx = \int_{-1}^0 (-x) dx + \int_0^1 x dx = [\frac{-x^2}{2}]_{-1}^0 + [\frac{x^2}{2}]_0^1 = (0 - (-\frac{1}{2})) + (\frac{1}{2} - 0) = \frac{1}{2} + \frac{1}{2} = 1$.

This reason correctly sets up the integral for the area. The function $y=|x|$ needs to be split into two parts because of the absolute value: $y=-x$ for $x<0$ and $y=x$ for $x \ge 0$.

• For the interval $[-1, 0]$, $|x| = -x$. So, $\int_{-1}^0 (-x) dx = [\frac{-x^2}{2}]_{-1}^0 = (\frac{-0^2}{2}) - (\frac{-(-1)^2}{2}) = 0 - (-\frac{1}{2}) = \frac{1}{2}$.
• For the interval $[0, 1]$, $|x| = x$. So, $\int_0^1 x dx = [\frac{x^2}{2}]_0^1 = (\frac{1^2}{2}) - (\frac{0^2}{2}) = \frac{1}{2} - 0 = \frac{1}{2}$.

The total area is the sum of these two parts: $\frac{1}{2} + \frac{1}{2} = 1$. The calculations in Reason (R) are correct.

Thus, Reason (R) is true.

Relationship between Assertion (A) and Reason (R):

Reason (R) provides the step-by-step integral calculation that precisely leads to the area value stated in Assertion (A). Therefore, Reason (R) is the correct explanation for Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A).

The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. We need to find the area of the region in the first quadrant.

Option (A): $\int_0^a \frac{b}{a}\sqrt{a^2-x^2} dx$

From the ellipse equation, we solve for $y$: $\frac{y^2}{b^2} = 1 - \frac{x^2}{a^2} \implies y^2 = b^2(1 - \frac{x^2}{a^2}) = \frac{b^2}{a^2}(a^2 - x^2)$.

For the first quadrant, $y \ge 0$, so $y = \frac{b}{a}\sqrt{a^2-x^2}$.

The integral $\int_0^a y dx = \int_0^a \frac{b}{a}\sqrt{a^2-x^2} dx$ represents the area under this curve from $x=0$ to $x=a$. This is indeed the area of the quarter ellipse in the first quadrant.

Let's evaluate this integral. Using the substitution $x = a \sin \theta$, $dx = a \cos \theta d\theta$. When $x=0$, $\theta=0$. When $x=a$, $\theta=\pi/2$. $\sqrt{a^2-x^2} = a \cos \theta$.

So, option (A) is a correct representation of the area.

Option (B): $\int_0^b \frac{a}{b}\sqrt{b^2-y^2} dy$

Similarly, solving for $x$ in terms of $y$: $\frac{x^2}{a^2} = 1 - \frac{y^2}{b^2} \implies x^2 = a^2(1 - \frac{y^2}{b^2}) = \frac{a^2}{b^2}(b^2 - y^2)$.

For the first quadrant, $x \ge 0$, so $x = \frac{a}{b}\sqrt{b^2-y^2}$.

The integral $\int_0^b x dy = \int_0^b \frac{a}{b}\sqrt{b^2-y^2} dy$ represents the area bounded by this curve, the y-axis, and the lines $y=0$ and $y=b$. This is also the area of the quarter ellipse in the first quadrant.

This integral will also evaluate to $\frac{\pi ab}{4}$. So, option (B) is also a correct representation.

Option (C): $\frac{1}{4}\pi ab$

This is the calculated value of the area of the quarter ellipse, which we confirmed by evaluating the integrals in (A) and (B). So, option (C) is also correct.

Option (D): All of the above describe or provide the area.

Since options (A), (B), and (C) are all correct representations or the value of the area of the region in the first quadrant bounded by the ellipse, this option is the most appropriate.

The correct option is (D).

The region is bounded by the parabola $y=x^2$, the y-axis ($x=0$), and the horizontal line $y=1$. To find the area by integrating with respect to $y$, we need to express $x$ in terms of $y$.

From $y=x^2$, since we are considering the region bounded by the y-axis and the parabola, we take the positive root for $x$ (as the y-axis is $x=0$ and the parabola extends to the right in the first quadrant for $y>0$). So, $x = \sqrt{y}$.

The area is obtained by integrating $x$ with respect to $y$ from the lower y-bound to the upper y-bound. The y-bounds are given as $y=0$ (the x-axis, or the start of the parabola) and $y=1$ (the given line).

The area is given by:

This matches option (B).

Let's examine the other options:

• Option (A): $\int_0^1 x^2 dy$. This integral is incorrect because it uses $x^2$ (which is $y$) as the integrand when integrating with respect to $y$. The integrand should be the expression for $x$ in terms of $y$.
• Option (C): $2/3$ square units. Let's evaluate the integral from option (B):
$\int_0^1 \sqrt{y} dy = \int_0^1 y^{1/2} dy = \left[ \frac{y^{3/2}}{3/2} \right]_0^1 = \left[ \frac{2}{3} y^{3/2} \right]_0^1 = \frac{2}{3} (1)^{3/2} - \frac{2}{3} (0)^{3/2} = \frac{2}{3}$. So, this is the correct calculated area.
• Option (D): $1$ square unit. This is incorrect.

Both option (B) and option (C) are correct. Option (B) is the integral representation, and option (C) is the calculated value of the area.

Since the question asks for "The area... is:", and option (C) provides the numerical value of the area, it is the direct answer to what the area is. Option (B) is the integral form that leads to it.

However, it's also possible the question is asking for the integral representation. In such cases, if both the integral and the value are provided as separate correct options, the intended answer might be the value.

Let's assume the question is asking for either the correct integral or the correct area value. Both (B) and (C) are correct.

Typically, if the calculated value is an option, and it is correct, it is the primary answer.

The correct option is (C).

The area of the region bounded by the curve $y = \tan x$, the x-axis, from $x=0$ to $x=\pi/4$. In this interval, $\tan x \ge 0$.

Option (A): $\int_0^{\pi/4} \tan x dx$

This integral correctly represents the area under the curve $y = \tan x$ from $x=0$ to $x=\pi/4$, as $\tan x$ is non-negative in this interval.

Option (B): $\log |\sec x| \Big|_0^{\pi/4}$

The integral of $\tan x$ is $\int \tan x dx = \log |\sec x| + C$ (or $-\log |\cos x| + C$). Evaluating this definite integral gives:

This is the correct antiderivative evaluated at the limits. So, option (B) is also a correct representation.

Option (C): $\log(\sqrt{2}) - \log(1) = \frac{1}{2}\log 2$

Let's evaluate the definite integral from option (B):

We know that $\sec(\pi/4) = \frac{1}{\cos(\pi/4)} = \frac{1}{1/\sqrt{2}} = \sqrt{2}$.

And $\sec(0) = \frac{1}{\cos(0)} = \frac{1}{1} = 1$.

So, the evaluation becomes:

Using logarithm properties, $\log(\sqrt{2}) = \log(2^{1/2}) = \frac{1}{2}\log 2$. And $\log(1) = 0$.

So, the result is $\frac{1}{2}\log 2 - 0 = \frac{1}{2}\log 2$. This matches option (C).

Option (D): All of the above describe or provide the area.

Since options (A), (B), and (C) are all correct representations or the calculated value of the area, this option is the most appropriate.

The correct option is (D).

Given:

The curve is $y^2 = x$.

The bounding lines are $x = 1$, $x = 4$, and the x-axis.

The region is in the first quadrant.

To Find:

The area of the region bounded by the given curve and lines.

Solution:

The equation of the curve is $y^2 = x$.

Since the region is in the first quadrant, we consider the positive square root of $x$, which is $y = \sqrt{x}$.

The area of the region bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by the integral:

In this case, $a = 1$ and $b = 4$, and $y = \sqrt{x}$.

So, the area is:

We can rewrite $\sqrt{x}$ as $x^{1/2}$.

Now, we integrate $x^{1/2}$ with respect to $x$:

Now, we evaluate the definite integral from 1 to 4:

Substitute the upper and lower limits:

Calculate the terms:

Substitute these values back into the area equation:

The area of the region is $\frac{14}{3}$ square units.

Given:

The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

To Find:

The area of the region bounded by the ellipse.

Solution:

The standard equation of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Comparing the given equation with the standard form:

Here, $a$ is the semi-major axis and $b$ is the semi-minor axis.

The area of an ellipse is given by the formula:

Substituting the values of $a$ and $b$:

The area of the region bounded by the ellipse is $12\pi$ square units.

Alternate Method using Integration:

From the equation of the ellipse, we can express $y$ in terms of $x$ for the upper half of the ellipse (in the first quadrant):

The area of the ellipse can be found by integrating $y$ from $x = -4$ to $x = 4$ and multiplying by 2 (for the upper and lower halves), or by integrating from $x = 0$ to $x = 4$ and multiplying by 4 (for all four quadrants). Let's use the latter approach for simplicity.

Area of the ellipse = $4 \times \int_{0}^{4} y \, dx$

To solve this integral, we can use a trigonometric substitution. Let $x = 4 \sin \theta$. Then $dx = 4 \cos \theta \, d\theta$.

When $x = 0$, $4 \sin \theta = 0 \implies \sin \theta = 0 \implies \theta = 0$.

When $x = 4$, $4 \sin \theta = 4 \implies \sin \theta = 1 \implies \theta = \frac{\pi}{2}$.

Substitute these into the integral:

So the integral becomes:

Using the identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$:

Integrate with respect to $\theta$:

Evaluate at the limits:

Both methods yield the same result.

Given:

The curve is $y = \sin x$.

The bounding lines are $x = 0$ and $x = \pi$.

The region is bounded by the x-axis.

To Find:

The area of the region bounded by the given curve and lines.

Solution:

The area of the region bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by the integral:

In this case, $a = 0$ and $b = \pi$, and $y = \sin x$. Since $\sin x \ge 0$ for $x \in [0, \pi]$, the area is directly given by the integral.

So, the area is:

Now, we integrate $\sin x$ with respect to $x$:

Now, we evaluate the definite integral from 0 to $\pi$:

Substitute the upper and lower limits:

Calculate the cosine values:

Substitute these values back into the area equation:

The area of the region bounded by the curve $y = \sin x$ and the x-axis from $x = 0$ to $x = \pi$ is 2 square units.

Given:

The curve is the parabola $y = x^2$.

The bounding line is $y = 4$.

To Find:

The area of the region bounded by the given parabola and line.

Solution:

First, find the points of intersection between the parabola $y = x^2$ and the line $y = 4$.

Set the equations equal to each other:

Taking the square root of both sides:

The points of intersection are $(-2, 4)$ and $(2, 4)$.

The area of the region bounded by two curves $y = f(x)$ and $y = g(x)$, where $f(x) \ge g(x)$ on the interval $[a, b]$, is given by:

In this case, the upper curve is $y = 4$ and the lower curve is $y = x^2$. The limits of integration are from $x = -2$ to $x = 2$. So, $f(x) = 4$ and $g(x) = x^2$, $a = -2$, and $b = 2$.

Therefore, the area is:

Now, we integrate the expression $(4 - x^2)$ with respect to $x$:

Now, we evaluate the definite integral from -2 to 2:

Substitute the upper and lower limits:

Simplify the terms:

The area of the region bounded by the parabola $y = x^2$ and the line $y = 4$ is $\frac{32}{3}$ square units.

Given:

The equation of the circle is $x^2 + y^2 = 16$.

To Find:

The area of the region bounded by the circle.

Solution:

The standard equation of a circle centered at the origin is $x^2 + y^2 = r^2$, where $r$ is the radius of the circle.

Comparing the given equation with the standard form:

The radius of the circle is 4 units.

The area of a circle is given by the formula:

Substituting the value of $r$:

The area of the region bounded by the circle $x^2 + y^2 = 16$ is $16\pi$ square units.

Given:

The bounding lines are $y = x$, $x = 1$, $x = 2$, and the x-axis (which is $y = 0$).

To Find:

The area of the region bounded by the given lines.

Solution:

The region is bounded by the function $y = x$ from $x = 1$ to $x = 2$, and the x-axis.

The area of the region bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by the integral:

In this case, $a = 1$, $b = 2$, and $y = x$. Since $y = x$ is positive in the interval $[1, 2]$, the area is directly given by the integral.

So, the area is:

Now, we integrate $x$ with respect to $x$:

Now, we evaluate the definite integral from 1 to 2:

Substitute the upper and lower limits:

Simplify the terms:

The area of the region bounded by the lines $y = x$, $x = 1$, $x = 2$, and the x-axis is $\frac{3}{2}$ square units.

Given:

The curve is $y = \cos x$.

The bounding lines are $x = 0$ and $x = \pi/2$.

The region is bounded by the x-axis.

To Find:

The area of the region bounded by the given curve and lines.

Solution:

The area of the region bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by the integral:

In this case, $a = 0$, $b = \pi/2$, and $y = \cos x$. Since $\cos x \ge 0$ for $x \in [0, \pi/2]$, the area is directly given by the integral.

So, the area is:

Now, we integrate $\cos x$ with respect to $x$:

Now, we evaluate the definite integral from 0 to $\pi/2$:

Substitute the upper and lower limits:

Calculate the sine values:

Substitute these values back into the area equation:

The area of the region bounded by the curve $y = \cos x$ and the x-axis from $x = 0$ to $x = \pi/2$ is 1 square unit.

Given:

The curve is the parabola $x^2 = 4y$.

The bounding line is $y = 2$.

To Find:

The area of the region bounded by the given parabola and line.

Solution:

First, we need to express $x$ in terms of $y$ for the parabola $x^2 = 4y$.

The bounding line is $y = 2$.

To find the points of intersection, substitute $y=2$ into the parabola equation:

The points of intersection are $(-2\sqrt{2}, 2)$ and $(2\sqrt{2}, 2)$.

The area of the region bounded by two curves $x = f(y)$ and $x = g(y)$, where $f(y) \ge g(y)$ on the interval $[c, d]$, is given by:

In this case, the right curve is $x = 2\sqrt{y}$ and the left curve is $x = -2\sqrt{y}$. The limits of integration are from $y = 0$ (the vertex of the parabola) to $y = 2$. So, $f(y) = 2\sqrt{y}$ and $g(y) = -2\sqrt{y}$, $c = 0$, and $d = 2$.

Therefore, the area is:

We can rewrite $\sqrt{y}$ as $y^{1/2}$.

Now, we integrate $4y^{1/2}$ with respect to $y$:

Now, we evaluate the definite integral from 0 to 2:

Substitute the upper and lower limits:

Simplify the terms:

The area of the region bounded by the parabola $x^2 = 4y$ and the line $y = 2$ is $\frac{16\sqrt{2}}{3}$ square units.

Alternate Method using Integration with respect to x:

From $x^2 = 4y$, we get $y = \frac{x^2}{4}$.

The upper boundary is $y = 2$ and the lower boundary is $y = \frac{x^2}{4}$.

The limits of integration are the x-values where the curves intersect, which we found to be $x = -2\sqrt{2}$ and $x = 2\sqrt{2}$.

The area is given by:

Integrate with respect to $x$:

Evaluate the definite integral:

Substitute the limits:

Simplify:

Both methods yield the same result.

Given:

The equation of the circle is $x^2 + y^2 = 4$. This is a circle centered at the origin with radius $r = 2$.

The line is $x = 1$.

To Find:

The area of the smaller part of the circle cut off by the line $x = 1$.

Solution:

The line $x = 1$ cuts the circle into two parts. We need to find the area of the smaller part, which is the segment of the circle to the left of the line $x=1$.

The area of a circular segment can be found by subtracting the area of the triangle formed by the center and the chord from the area of the sector subtended by the chord.

First, let's find the points of intersection of the circle $x^2 + y^2 = 4$ and the line $x = 1$.

Substitute $x = 1$ into the circle equation:

The points of intersection are $(1, \sqrt{3})$ and $(1, -\sqrt{3})$.

We can find the area by integration. The smaller part is to the left of $x = 1$. So we integrate from $x = -2$ (leftmost point of the circle) to $x = 1$.

From $x^2 + y^2 = 4$, we have $y = \pm \sqrt{4 - x^2}$. The upper half is $y = \sqrt{4 - x^2}$ and the lower half is $y = -\sqrt{4 - x^2}$.

The area of the smaller part is the integral of the difference between the upper and lower curves from $x=-2$ to $x=1$.

To evaluate $\int \sqrt{4 - x^2} \, dx$, we use the trigonometric substitution $x = 2 \sin \theta$. Then $dx = 2 \cos \theta \, d\theta$.

$\sqrt{4 - x^2} = \sqrt{4 - (2 \sin \theta)^2} = \sqrt{4 - 4 \sin^2 \theta} = \sqrt{4(1 - \sin^2 \theta)} = \sqrt{4 \cos^2 \theta} = 2 \cos \theta$.

When $x = -2$, $2 \sin \theta = -2 \implies \sin \theta = -1 \implies \theta = -\frac{\pi}{2}$.

When $x = 1$, $2 \sin \theta = 1 \implies \sin \theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}$.

So the integral becomes:

Using the identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$:

Integrate:

Evaluate at the limits:

We know that $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\sin(-\pi) = 0$.

Combine the $\pi$ terms:

So,

Alternative Method using Geometry:

The area of the smaller segment can be calculated as the area of the sector minus the area of the triangle.

The radius of the circle is $r = 2$. The chord is at $x = 1$. The points are $(1, \sqrt{3})$ and $(1, -\sqrt{3})$.

Let $\theta$ be half the angle subtended by the chord at the center. In the right-angled triangle formed by the center $(0,0)$, $(1,0)$, and $(1, \sqrt{3})$, we have:

This means $\theta = \frac{\pi}{3}$ radians (or 60 degrees).

The full angle subtended by the chord at the center is $2\theta = \frac{2\pi}{3}$ radians.

Area of the sector = $\frac{1}{2}r^2 (2\theta)$ (where $2\theta$ is the angle in radians)

Area of the triangle formed by the center and the chord:

The base of the triangle is the distance between $(1, \sqrt{3})$ and $(1, -\sqrt{3})$, which is $2\sqrt{3}$.

The height of the triangle is the perpendicular distance from the center $(0,0)$ to the chord $x=1$, which is $1$.

Area of the smaller segment = Area of the sector - Area of the triangle

The area of the smaller part is $\frac{4\pi}{3} - \sqrt{3}$ square units.

Given:

The curves are $y = x^2$ (a parabola) and $y = x$ (a straight line).

To Find:

The area of the region bounded by the given curves.

Solution:

First, find the points of intersection of the two curves by setting their equations equal to each other:

Rearrange the equation:

Factor out $x$:

This gives us two solutions for $x$: $x = 0$ and $x = 1$.

When $x = 0$, $y = 0^2 = 0$ and $y = 0$. So, one intersection point is $(0, 0)$.

When $x = 1$, $y = 1^2 = 1$ and $y = 1$. So, the other intersection point is $(1, 1)$.

The area of the region bounded by two curves $y = f(x)$ and $y = g(x)$, where $f(x) \ge g(x)$ on the interval $[a, b]$, is given by:

In the interval $[0, 1]$, the line $y = x$ is above the parabola $y = x^2$. So, $f(x) = x$ and $g(x) = x^2$. The limits of integration are $a = 0$ and $b = 1$.

Therefore, the area is:

Now, we integrate the expression $(x - x^2)$ with respect to $x$:

Now, we evaluate the definite integral from 0 to 1:

Substitute the upper and lower limits:

Simplify the terms:

The area of the region bounded by the curve $y = x^2$ and the line $y = x$ is $\frac{1}{6}$ square units.

Given:

The curve is $y = e^x$.

The bounding lines are $x = 0$ and $x = 1$.

The region is bounded by the x-axis ($y = 0$).

To Find:

The area of the region bounded by the given curve and lines.

Solution:

The area of the region bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by the integral:

In this case, $a = 0$, $b = 1$, and $y = e^x$. Since $e^x$ is always positive for all real $x$, the function is above the x-axis in the given interval.

So, the area is:

Now, we integrate $e^x$ with respect to $x$:

Now, we evaluate the definite integral from 0 to 1:

Substitute the upper and lower limits:

Calculate the values:

Substitute these values back into the area equation:

The area of the region bounded by the curve $y = e^x$, the x-axis, and the lines $x = 0$ and $x = 1$ is $e - 1$ square units.

Given:

The bounding lines are $y = 2x$, $x = 0$ (the y-axis), and $x = 1$.

To Find:

The area of the region bounded by the given lines.

Solution:

The region is bounded by the function $y = 2x$ between the vertical lines $x = 0$ and $x = 1$. Since the problem doesn't explicitly mention the x-axis as a boundary, we assume it's the area under the curve $y=2x$ down to the x-axis within the specified x-limits.

The area of the region bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by the integral:

In this case, $a = 0$, $b = 1$, and $y = 2x$. Since $2x \ge 0$ for $x \in [0, 1]$, the area is directly given by the integral.

So, the area is:

Now, we integrate $2x$ with respect to $x$:

Now, we evaluate the definite integral from 0 to 1:

Substitute the upper and lower limits:

Simplify:

The area of the region bounded by the lines $y = 2x$, $x = 0$, and $x = 1$ is 1 square unit.

Given:

The curve is $y = \sqrt{x}$.

The bounding lines are the x-axis ($y = 0$) and $x = 4$.

To Find:

The area of the region bounded by the given curve and lines.

Solution:

The area of the region bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by the integral:

In this case, $a = 0$ (since $\sqrt{x}$ is defined for $x \ge 0$ and the bounding line is $x=4$), $b = 4$, and $y = \sqrt{x}$. Since $\sqrt{x} \ge 0$ for $x \in [0, 4]$, the area is directly given by the integral.

So, the area is:

We can rewrite $\sqrt{x}$ as $x^{1/2}$.

Now, we integrate $x^{1/2}$ with respect to $x$:

Now, we evaluate the definite integral from 0 to 4:

Substitute the upper and lower limits:

Calculate the terms:

Substitute these values back into the area equation:

The area of the region bounded by the curve $y = \sqrt{x}$, the x-axis, and the line $x = 4$ is $\frac{16}{3}$ square units.

Given:

The equation of the parabola is $y^2 = 4ax$.

The latus rectum of the parabola $y^2 = 4ax$ is the line segment perpendicular to the axis of symmetry passing through the focus, with endpoints on the parabola.

To Find:

The area of the region bounded by the parabola and its latus rectum.

Solution:

The standard form of the parabola is $y^2 = 4ax$.

The focus of this parabola is at $(a, 0)$.

The latus rectum is a vertical line passing through the focus, so its equation is $x = a$.

To find the endpoints of the latus rectum, substitute $x = a$ into the parabola's equation:

The endpoints of the latus rectum are $(a, 2a)$ and $(a, -2a)$.

We need to find the area between the parabola $y^2 = 4ax$ and the line $x = a$. We can solve for $y$ in terms of $x$ for the parabola:

The area can be calculated by integrating with respect to $x$. The parabola is symmetric about the x-axis. We can find the area of the upper half and double it. The upper half is bounded by $y = 2\sqrt{ax}$ and the x-axis, from $x = 0$ to $x = a$.

Area of the upper half = $\int_{0}^{a} 2\sqrt{ax} \, dx$

We can pull out the constant $2\sqrt{a}$:

Rewrite $\sqrt{x}$ as $x^{1/2}$:

Integrate $x^{1/2}$:

Evaluate the definite integral:

The total area bounded by the parabola and its latus rectum is twice the area of the upper half:

The area of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum is $\frac{8}{3} a^2$ square units.

Given:

The curve is $y = x^3$.

The bounding lines are the x-axis ($y = 0$), $x = -2$, and $x = 1$.

To Find:

The area of the region bounded by the given curve and lines.

Solution:

The area bounded by the curve $y = f(x)$ and the x-axis from $x = a$ to $x = b$ is given by $\int_{a}^{b} |f(x)| \, dx$.

In this problem, the interval is $[-2, 1]$. The function $y = x^3$ is negative for $x \in [-2, 0)$ and positive for $x \in (0, 1]$. We need to consider the absolute value of the function in the integration.

The total area is the sum of the areas in the intervals $[-2, 0]$ and $[0, 1]$.

Area $= \int_{-2}^{1} |x^3| \, dx = \int_{-2}^{0} |-x^3| \, dx + \int_{0}^{1} |x^3| \, dx$

For $x \in [-2, 0]$, $x^3$ is negative, so $|x^3| = -x^3$.

For $x \in [0, 1]$, $x^3$ is positive, so $|x^3| = x^3$.

Area $= \int_{-2}^{0} (-x^3) \, dx + \int_{0}^{1} x^3 \, dx$

Now, we integrate $-x^3$ and $x^3$:

$\int -x^3 \, dx = -\frac{x^4}{4} + C$

$\int x^3 \, dx = \frac{x^4}{4} + C$

Evaluate the first integral:

Evaluate the second integral:

The total area is the sum of these two areas:

The area of the region bounded by the curve $y = x^3$, the x-axis, and the ordinates $x = -2$ and $x = 1$ is $\frac{17}{4}$ square units.

Given:

The curve is the parabola $y^2 = 9x$.

The bounding vertical lines are $x = 1$ and $x = 4$.

To Find:

The area of the region bounded by the given parabola and lines.

Solution:

From the equation $y^2 = 9x$, we can express $y$ in terms of $x$. Since the area bounded by the curve and the lines is usually taken as the area between the curve and the x-axis, we consider both the positive and negative parts of $y$.

Solving for $y$: $y = \pm \sqrt{9x} = \pm 3\sqrt{x}$.

The region is bounded by the curve $y = 3\sqrt{x}$ (upper half) and $y = -3\sqrt{x}$ (lower half) between $x = 1$ and $x = 4$.

The area of the region is given by the integral:

Here, $a = 1$, $b = 4$, $y_{upper} = 3\sqrt{x}$, and $y_{lower} = -3\sqrt{x}$.

So, the area is:

Rewrite $\sqrt{x}$ as $x^{1/2}$:

Now, integrate $6x^{1/2}$ with respect to $x$:

Now, evaluate the definite integral from 1 to 4:

Substitute the upper and lower limits:

Calculate the terms:

Substitute these values back into the area equation:

The area of the region bounded by the curve $y^2 = 9x$ and the lines $x = 1$ and $x = 4$ is 28 square units.

Given:

The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

To Find:

The area enclosed by the ellipse using integration.

Solution:

From the equation of the ellipse, we can express $y$ in terms of $x$ for the upper half of the ellipse:

The ellipse is symmetric about both the x-axis and the y-axis. We can find the area of the ellipse by calculating the area in the first quadrant (where $x \ge 0$ and $y \ge 0$) and multiplying it by 4.

In the first quadrant, the ellipse is bounded by $y = \frac{b}{a}\sqrt{a^2 - x^2}$, the x-axis ($y=0$), and the y-axis ($x=0$) up to $x = a$.

Area of the ellipse = $4 \times \int_{0}^{a} y \, dx$

We can pull out the constant $\frac{4b}{a}$:

To solve the integral $\int_{0}^{a} \sqrt{a^2 - x^2} \, dx$, we can use the trigonometric substitution $x = a \sin \theta$. Then $dx = a \cos \theta \, d\theta$.

When $x = 0$, $a \sin \theta = 0 \implies \sin \theta = 0 \implies \theta = 0$.

When $x = a$, $a \sin \theta = a \implies \sin \theta = 1 \implies \theta = \frac{\pi}{2}$.

Also, $\sqrt{a^2 - x^2} = \sqrt{a^2 - (a \sin \theta)^2} = \sqrt{a^2(1 - \sin^2 \theta)} = \sqrt{a^2 \cos^2 \theta} = a \cos \theta$ (since $a > 0$ and $\cos \theta \ge 0$ for $\theta \in [0, \pi/2]$).

Substitute these into the integral:

Pull out the constant $a^2$:

Using the identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$:

Integrate with respect to $\theta$:

Evaluate at the limits:

The area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\pi ab$ square units.

Given:

The curve is $y = x^2 + 1$.

The bounding lines are the x-axis ($y = 0$), $x = -1$, and $x = 2$.

To Find:

The area of the region bounded by the given curve and lines.

Solution:

The area of the region bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by the integral:

In this case, $a = -1$, $b = 2$, and $y = x^2 + 1$. Since $x^2 + 1$ is always positive for all real values of $x$, the curve is above the x-axis in the interval $[-1, 2]$.

So, the area is:

Now, we integrate $(x^2 + 1)$ with respect to $x$:

Now, we evaluate the definite integral from -1 to 2:

Substitute the upper and lower limits:

Simplify the terms:

Combine terms within the parentheses:

The area of the region bounded by the curve $y = x^2 + 1$, the x-axis, and the lines $x = -1$ and $x = 2$ is 6 square units.

Given:

The curve is the parabola $y^2 = 4x$.

The bounding vertical line is $x = 3$.

To Find:

The area of the region bounded by the given parabola and line.

Solution:

From the equation $y^2 = 4x$, we can express $y$ in terms of $x$. The parabola is symmetric about the x-axis.

Solving for $y$: $y = \pm \sqrt{4x} = \pm 2\sqrt{x}$.

The region is bounded by the parabola $y = 2\sqrt{x}$ (upper half) and $y = -2\sqrt{x}$ (lower half) between $x = 0$ (vertex of the parabola) and $x = 3$.

The area of the region is given by the integral:

Here, $a = 0$, $b = 3$, $y_{upper} = 2\sqrt{x}$, and $y_{lower} = -2\sqrt{x}$.

So, the area is:

Rewrite $\sqrt{x}$ as $x^{1/2}$:

Now, integrate $4x^{1/2}$ with respect to $x$:

Now, evaluate the definite integral from 0 to 3:

Substitute the upper and lower limits:

Calculate the terms:

Substitute these values back into the area equation:

The area of the region bounded by the parabola $y^2 = 4x$ and the line $x = 3$ is $8\sqrt{3}$ square units.

Given:

The circle is $x^2 + y^2 = 4$. This is a circle centered at the origin with radius $r = 2$.

The bounding lines are $x = 0$ (the y-axis) and $x = 2$.

The region is in the first quadrant.

To Find:

The area of the region in the first quadrant bounded by the circle and the given lines.

Solution:

From the equation of the circle, $x^2 + y^2 = 4$, we can express $y$ in terms of $x$ for the first quadrant:

The area of the region in the first quadrant bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by the integral:

In this case, $a = 0$, $b = 2$, and $y = \sqrt{4 - x^2}$.

So, the area is:

To solve this integral, we use the trigonometric substitution $x = 2 \sin \theta$. Then $dx = 2 \cos \theta \, d\theta$.

When $x = 0$, $2 \sin \theta = 0 \implies \sin \theta = 0 \implies \theta = 0$.

When $x = 2$, $2 \sin \theta = 2 \implies \sin \theta = 1 \implies \theta = \frac{\pi}{2}$.

Also, $\sqrt{4 - x^2} = \sqrt{4 - (2 \sin \theta)^2} = \sqrt{4 - 4 \sin^2 \theta} = \sqrt{4(1 - \sin^2 \theta)} = \sqrt{4 \cos^2 \theta} = 2 \cos \theta$ (since $\cos \theta \ge 0$ for $\theta \in [0, \pi/2]$).

Substitute these into the integral:

Using the identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$:

Integrate with respect to $\theta$:

Evaluate at the limits:

Alternatively, since the limits of integration $x=0$ to $x=2$ cover exactly one quarter of the circle (from the y-axis to the positive x-axis), and the circle has radius $r=2$, the area of the full circle is $\pi r^2 = \pi (2^2) = 4\pi$. The area in the first quadrant is therefore $\frac{1}{4}$ of the total area, which is $\frac{1}{4}(4\pi) = \pi$.

The area of the region in the first quadrant bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ is $\pi$ square units.

Given:

The curve is $y = \log x$. It's implied that this is the natural logarithm, $\ln x$.

The bounding lines are the x-axis ($y = 0$), $x = 1$, and $x = e$.

To Find:

The area of the region bounded by the given curve and lines.

Solution:

The area of the region bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by the integral:

In this case, $a = 1$, $b = e$, and $y = \log x$ (which we interpret as $\ln x$).

For $x \in [1, e]$, $\ln x \ge 0$, so the curve is above the x-axis.

So, the area is:

To integrate $\log x$ (or $\ln x$), we use integration by parts. Let $u = \log x$ and $dv = dx$.

Then $du = \frac{1}{x} \, dx$ and $v = x$.

The integration by parts formula is $\int u \, dv = uv - \int v \, du$.

Now, we evaluate the definite integral from 1 to $e$:

Substitute the upper and lower limits:

We know that $\log e = \ln e = 1$ and $\log 1 = \ln 1 = 0$.

The area bounded by the curve $y = \log x$, the x-axis, and the lines $x = 1$ and $x = e$ is 1 square unit.

Given:

The curve is $y = |x|$. This is a V-shaped graph with its vertex at the origin. For $x \ge 0$, $y = x$, and for $x < 0$, $y = -x$.

The bounding line is $y = 1$.

To Find:

The area of the region bounded by the curve $y = |x|$ and the line $y = 1$.

Solution:

First, find the points of intersection between $y = |x|$ and $y = 1$.

Case 1: $x \ge 0$. Then $y = x$. Setting $y = 1$, we get $x = 1$. So, one intersection point is $(1, 1)$.

Case 2: $x < 0$. Then $y = -x$. Setting $y = 1$, we get $-x = 1$, which means $x = -1$. So, the other intersection point is $(-1, 1)$.

The region is a triangle with vertices at $(0, 0)$, $(1, 1)$, and $(-1, 1)$.

We can find the area of this triangle. The base of the triangle lies on the line $y = 1$, from $x = -1$ to $x = 1$. The length of the base is $1 - (-1) = 2$.

The height of the triangle is the perpendicular distance from the vertex $(0, 0)$ to the base line $y = 1$. The height is $1 - 0 = 1$.

The area of a triangle is given by $\frac{1}{2} \times base \times height$.

Alternatively, using integration:

The area can be found by integrating the difference between the upper curve ($y = 1$) and the lower curve ($y = |x|$). The limits of integration are from $x = -1$ to $x = 1$.

Area $= \int_{-1}^{1} (1 - |x|) \, dx$

Since the integrand $(1 - |x|)$ is an even function (i.e., $f(-x) = f(x)$), we can integrate from 0 to 1 and multiply by 2.

Area $= 2 \int_{0}^{1} (1 - |x|) \, dx$

For $x \in [0, 1]$, $|x| = x$. So the integral becomes:

Area $= 2 \int_{0}^{1} (1 - x) \, dx$

Integrate $(1 - x)$:

Evaluate the definite integral:

Substitute the limits:

The area bounded by the curve $y = |x|$ and the line $y = 1$ is 1 square unit.

Given:

Two parabolas: $y = x^2$ and $y^2 = x$.

To Find:

The area of the region bounded by these two parabolas.

Solution:

First, find the points of intersection of the two parabolas.

Substitute $y = x^2$ from the first equation into the second equation:

Rearrange the equation:

Factor out $x$:

This gives two possibilities: $x = 0$ or $x^3 - 1 = 0$.

If $x = 0$, then from $y = x^2$, $y = 0^2 = 0$. So, one intersection point is $(0, 0)$.

If $x^3 - 1 = 0$, then $x^3 = 1$, which implies $x = 1$.

If $x = 1$, then from $y = x^2$, $y = 1^2 = 1$. So, the other intersection point is $(1, 1)$.

Now, we need to determine which curve is above the other in the interval $[0, 1]$.

Let's test a value between 0 and 1, for example, $x = 0.5$.

For $y = x^2$: $y = (0.5)^2 = 0.25$.

For $y^2 = x$, $y = \sqrt{x}$ (considering the upper part of the parabola $y^2=x$ in the first quadrant): $y = \sqrt{0.5} \approx 0.707$.

Since $0.707 > 0.25$, the curve $y = \sqrt{x}$ is above $y = x^2$ in the interval $[0, 1]$.

The area between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$, where $f(x) \ge g(x)$ on $[a, b]$, is given by:

Here, $f(x) = \sqrt{x}$ (from $y^2 = x$) and $g(x) = x^2$. The limits of integration are $a = 0$ and $b = 1$.

So, the area is:

Rewrite $\sqrt{x}$ as $x^{1/2}$:

Now, integrate term by term:

So, the integral is:

Evaluate at the limits:

Simplify:

The area of the region bounded by the two parabolas $y = x^2$ and $y^2 = x$ is $\frac{1}{3}$ square units.

Given:

A parabola: $y^2 = 4ax$.

A line: $y = 2ax$.

To Find:

The area of the region bounded by the parabola and the line.

Solution:

First, find the points of intersection of the parabola $y^2 = 4ax$ and the line $y = 2ax$.

Substitute $y = 2ax$ into the equation of the parabola:

Rearrange the equation:

Factor out $4ax$:

This gives two possibilities:

1. $4ax = 0 \implies x = 0$ (assuming $a \neq 0$).

2. $ax - 1 = 0 \implies ax = 1 \implies x = \frac{1}{a}$.

Now find the corresponding $y$ values:

If $x = 0$, then $y = 2a(0) = 0$. So, one intersection point is $(0, 0)$.

If $x = \frac{1}{a}$, then $y = 2a\left(\frac{1}{a}\right) = 2$. So, the other intersection point is $\left(\frac{1}{a}, 2\right)$.

Now, we need to determine which curve is above the other in the interval $[0, \frac{1}{a}]$.

From $y^2 = 4ax$, we get $y = \pm \sqrt{4ax} = \pm 2\sqrt{ax}$. Since the intersection points are in the first quadrant (assuming $a > 0$), we consider $y = 2\sqrt{ax}$ for the upper part of the parabola.

The line is $y = 2ax$.

Let's test a value between $0$ and $\frac{1}{a}$, for example, $x = \frac{1}{2a}$.

For the line $y = 2ax$: $y = 2a\left(\frac{1}{2a}\right) = 1$.

For the parabola $y = 2\sqrt{ax}$: $y = 2\sqrt{a\left(\frac{1}{2a}\right)} = 2\sqrt{\frac{1}{2}} = 2 \frac{1}{\sqrt{2}} = \sqrt{2}$.

Since $\sqrt{2} \approx 1.414 > 1$, the parabola $y = 2\sqrt{ax}$ is above the line $y = 2ax$ in the interval $[0, \frac{1}{a}]$.

The area between the curves is given by:

Rewrite $\sqrt{ax}$ as $\sqrt{a}x^{1/2}$:

Now, integrate term by term:

So, the integral is:

Evaluate at the limits:

Simplify the first part:

So, the expression becomes:

The area bounded by the curve $y^2 = 4ax$ and the line $y = 2ax$ is $\frac{1}{3a}$ square units.

Given:

The bounding line is $y = 3x + 2$.

The other boundaries are the x-axis ($y = 0$) and the ordinates $x = -1$ and $x = 1$.

To Find:

The area of the region bounded by the given line and boundaries.

Solution:

The area bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by $\int_{a}^{b} |f(x)| \, dx$.

In this problem, $a = -1$, $b = 1$, and $y = 3x + 2$. We need to determine if the line $y = 3x + 2$ crosses the x-axis within the interval $[-1, 1]$.

Set $y = 0$ to find the x-intercept:

Since $x = -\frac{2}{3}$ is within the interval $[-1, 1]$, the line crosses the x-axis. We need to split the integral into two parts: from $x = -1$ to $x = -\frac{2}{3}$, and from $x = -\frac{2}{3}$ to $x = 1$.

For $x \in [-1, -\frac{2}{3}]$, $3x + 2 \le 0$, so $|3x + 2| = -(3x + 2) = -3x - 2$.

For $x \in [-\frac{2}{3}, 1]$, $3x + 2 \ge 0$, so $|3x + 2| = 3x + 2$.

The total area is the sum of the areas in these two intervals:

Area $= \int_{-1}^{1} |3x + 2| \, dx = \int_{-1}^{-2/3} (-3x - 2) \, dx + \int_{-2/3}^{1} (3x + 2) \, dx$

Now, integrate each part:

For the first integral: $\int (-3x - 2) \, dx = -\frac{3x^2}{2} - 2x + C$

For the second integral: $\int (3x + 2) \, dx = \frac{3x^2}{2} + 2x + C$

The total area is the sum of these two parts:

The area of the region bounded by the line $y = 3x + 2$, the x-axis, and the ordinates $x = -1$ and $x = 1$ is $\frac{13}{3}$ square units.

Given:

A parabola: $y = x^2$.

A line: $y = x+2$.

To Find:

The area of the region bounded by the parabola and the line.

Solution:

First, find the points of intersection of the curve $y = x^2$ and the line $y = x+2$.

Set the equations equal to each other:

Rearrange the equation:

Factor the quadratic equation:

This gives the x-coordinates of the intersection points: $x = 2$ and $x = -1$.

Now, find the corresponding y-coordinates:

If $x = 2$, then $y = 2+2 = 4$. The point is $(2, 4)$.

If $x = -1$, then $y = -1+2 = 1$. The point is $(-1, 1)$.

To find the area between the curves, we integrate the difference between the upper curve and the lower curve with respect to $x$, from the smaller x-value to the larger x-value.

In the interval $[-1, 2]$, let's check which function is greater.

Test a value, for instance $x = 0$:

For $y = x^2$: $y = 0^2 = 0$.

For $y = x+2$: $y = 0+2 = 2$.

Since $2 > 0$, the line $y = x+2$ is above the parabola $y = x^2$ in the interval $[-1, 2]$.

The area is given by:

Here, $a = -1$, $b = 2$, $y_{upper} = x+2$, and $y_{lower} = x^2$.

So, the area is:

Now, integrate the expression $(x+2-x^2)$:

Evaluate the definite integral from -1 to 2:

Substitute the upper and lower limits:

Simplify:

The area of the region bounded by the curve $y = x^2$ and the line $y = x+2$ is $\frac{9}{2}$ square units.

Given:

A circle: $x^2 + y^2 = 8x$.

A parabola: $y^2 = 4x$.

We need to find the area in the first quadrant (lying above the x-axis) bounded by these curves.

To Find:

The area of the specified region.

Solution:

First, let's analyze the given curves.

The parabola is $y^2 = 4x$. This is a standard parabola opening to the right, with its vertex at the origin $(0, 0)$. In the first quadrant, $y = \sqrt{4x} = 2\sqrt{x}$.

The circle is $x^2 + y^2 = 8x$. We can rewrite this equation by completing the square:

This is the equation of a circle with center $(4, 0)$ and radius $r = 4$.

Now, find the points of intersection of the circle and the parabola in the first quadrant.

Substitute $y^2 = 4x$ into the circle equation $x^2 + y^2 = 8x$:

Rearrange the equation:

Factor out $x$:

This gives $x = 0$ or $x = 4$.

When $x = 0$, $y^2 = 4(0) \implies y = 0$. So, $(0, 0)$ is an intersection point.

When $x = 4$, $y^2 = 4(4) = 16 \implies y = \pm 4$. Since we are considering the first quadrant, we take $y = 4$. So, $(4, 4)$ is an intersection point.

The region is bounded above by the parabola $y = 2\sqrt{x}$ and below by the x-axis, from $x=0$ up to some point, and then bounded above by the circle and below by the x-axis from that point up to $x=4$. The problem states "between the circle $x^2 + y^2 = 8x$ and the parabola $y^2 = 4x$". This implies the area is enclosed by these two curves.

We need to integrate with respect to $x$. The upper boundary is defined by the parabola ($y = 2\sqrt{x}$) up to their intersection point in the first quadrant, and then by the circle. However, the wording "lying above the x-axis and included between the circle ... and the parabola ..." suggests the area is bounded by the *lower* of the two curves and the x-axis, up to the intersection, and then by the *upper* of the two curves.

Let's re-evaluate the phrasing. "Included between" usually refers to the area where one curve is above the other.

The region lies above the x-axis. The boundaries are the parabola $y = 2\sqrt{x}$ and the circle's upper half. The intersection points are $(0,0)$ and $(4,4)$.

We need to compare the values of $y$ for the parabola and the circle for $x$ between $0$ and $4$.

From the circle equation $(x - 4)^2 + y^2 = 16$, for the upper semi-circle, $y = \sqrt{16 - (x-4)^2}$.

Let's check which curve is above the other for $x \in (0, 4)$. We found the intersection points, so one curve must be above the other in between.

At $x = 2$:

Parabola $y = 2\sqrt{2} \approx 2.828$.

Circle $y = \sqrt{16 - (2-4)^2} = \sqrt{16 - (-2)^2} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3} \approx 3.464$.

Since $3.464 > 2.828$, the circle is above the parabola in the interval $(0, 4)$.

The area is the integral of the difference between the upper curve (circle) and the lower curve (parabola) from $x = 0$ to $x = 4$.

This integral can be split into two parts:

Part 1: $\int_{0}^{4} \sqrt{16 - (x-4)^2} \, dx$. This represents the area of a quarter circle of radius 4, since the circle is $(x-4)^2 + y^2 = 16$, centered at $(4,0)$, and we are integrating from $x=0$ to $x=4$ for the upper half. This corresponds to the area of the quarter circle to the left of the vertical line $x=4$ if it were centered at $(0,0)$. However, the center is at $(4,0)$. The integration range from $0$ to $4$ covers the left half of the circle's area that is above the x-axis. This is indeed a semi-circle with radius 4, or rather, the upper half of the circle's area, which corresponds to a quarter circle in terms of the integration range. The integral $\int_{0}^{4} \sqrt{16 - (x-4)^2} \, dx$ calculates the area of the upper semi-circle of radius 4, but restricted to the interval [0, 4]. Geometrically, this represents the area of a quarter circle of radius 4. The area of a full circle is $\pi r^2 = \pi (4^2) = 16\pi$. So, the area of a quarter circle is $4\pi$.

Part 2: $\int_{0}^{4} 2\sqrt{x} \, dx$. This is the area under the parabola $y = 2\sqrt{x}$ from $x = 0$ to $x = 4$.

Let's calculate Part 2 first:

Now for Part 1, the integral $\int_{0}^{4} \sqrt{16 - (x-4)^2} \, dx$ represents the area under the upper semi-circle of $(x-4)^2 + y^2 = 16$ from $x=0$ to $x=4$. This geometrical shape is indeed a quarter circle with radius 4. The area of a quarter circle of radius 4 is $\frac{1}{4} \pi r^2 = \frac{1}{4} \pi (4^2) = 4\pi$.

So, the total area is:

The area lying above the x-axis and included between the circle $x^2 + y^2 = 8x$ and the parabola $y^2 = 4x$ is $4\pi - \frac{32}{3}$ square units.

Given:

The curve is $y = \sqrt{x}$.

The bounding lines are $x = 2$, $x = 4$, and the x-axis ($y = 0$).

To Find:

The area of the region bounded by the given curve and lines.

Solution:

The area of the region bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by the integral:

In this case, $a = 2$, $b = 4$, and $y = \sqrt{x}$. Since $\sqrt{x} \ge 0$ for $x \in [2, 4]$, the curve is above the x-axis, and the area is directly given by the integral.

So, the area is:

Rewrite $\sqrt{x}$ as $x^{1/2}$:

Now, integrate $x^{1/2}$ with respect to $x$:

Now, evaluate the definite integral from 2 to 4:

Substitute the upper and lower limits:

Calculate the terms:

Substitute these values back into the area equation:

The area of the region bounded by the curve $y = \sqrt{x}$, the lines $x=2$ and $x=4$, and the x-axis is $\frac{16 - 4\sqrt{2}}{3}$ square units.

Given:

Four lines that define the boundaries of the region: $y = x$, $y = 2x$, $x = 1$, and $x = 2$.

To Find:

The area of the region enclosed by these lines.

Solution:

The region is bounded by two lines $y = x$ and $y = 2x$ between the vertical lines $x = 1$ and $x = 2$.

To find the area between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$, where $f(x) \ge g(x)$ on $[a, b]$, we use the formula:

In this case, the interval is from $a = 1$ to $b = 2$.

We need to determine which function is greater in this interval. Let's check the values of $y$ for the two lines at a point within the interval, say $x = 1.5$.

For $y = x$: $y = 1.5$.

For $y = 2x$: $y = 2 \times 1.5 = 3$.

Since $3 > 1.5$, the line $y = 2x$ is above the line $y = x$ in the interval $[1, 2]$.

So, $f(x) = 2x$ and $g(x) = x$.

The area is:

Now, integrate $x$ with respect to $x$:

Evaluate the definite integral from 1 to 2:

Substitute the upper and lower limits:

Simplify:

The area enclosed by the lines $y = x$, $y = 2x$ and $x = 1$, $x = 2$ is $\frac{3}{2}$ square units.

Given:

A parabola: $x^2 = y$, which can be written as $y = x^2$.

A line: $y = x+2$.

To Find:

The area of the region bounded by the parabola and the line.

Solution:

First, find the points of intersection of the parabola $y = x^2$ and the line $y = x+2$.

Set the equations equal to each other:

Rearrange the equation:

Factor the quadratic equation:

This gives the x-coordinates of the intersection points: $x = 2$ and $x = -1$.

Now, find the corresponding y-coordinates:

If $x = 2$, then $y = 2+2 = 4$. The point is $(2, 4)$.

If $x = -1$, then $y = -1+2 = 1$. The point is $(-1, 1)$.

To find the area between the curves, we integrate the difference between the upper curve and the lower curve with respect to $x$, from the smaller x-value to the larger x-value.

In the interval $[-1, 2]$, let's check which function is greater.

Test a value, for instance $x = 0$:

For $y = x^2$: $y = 0^2 = 0$.

For $y = x+2$: $y = 0+2 = 2$.

Since $2 > 0$, the line $y = x+2$ is above the parabola $y = x^2$ in the interval $[-1, 2]$.

The area is given by:

Here, $a = -1$, $b = 2$, $y_{upper} = x+2$, and $y_{lower} = x^2$.

So, the area is:

Now, integrate the expression $(x+2-x^2)$:

Evaluate the definite integral from -1 to 2:

Substitute the upper and lower limits:

Simplify:

The area of the region bounded by the parabola $x^2 = y$ and the line $y = x+2$ is $\frac{9}{2}$ square units.

Given:

Circle 1: $x^2 + y^2 = 4$. This is a circle centered at $(0,0)$ with radius $r=2$.

Circle 2: $(x-2)^2 + y^2 = 4$. This is a circle centered at $(2,0)$ with radius $r=2$.

To Find:

The area of the region enclosed between these two circles.

Solution:

First, find the points of intersection of the two circles.

From the first equation, $y^2 = 4 - x^2$.

Substitute this into the second equation:

Expand $(x-2)^2$:

Simplify:

Now, find the corresponding $y$ values by substituting $x = 1$ into either circle equation.

Using $x^2 + y^2 = 4$:

The intersection points are $(1, \sqrt{3})$ and $(1, -\sqrt{3})$.

The area enclosed by the two circles can be calculated by considering the area of a segment of each circle. The region of intersection is symmetrical about the x-axis.

Let's focus on the upper half of the region. This area is bounded by the upper arc of the first circle $y = \sqrt{4 - x^2}$ and the upper arc of the second circle $y = \sqrt{4 - (x-2)^2}$. The integration limits will be from the leftmost point of the intersection to the rightmost point. The intersection occurs at $x=1$. The first circle extends from $x=-2$ to $x=2$. The second circle extends from $x=0$ to $x=4$. The common area starts from $x=0$ (left edge of the second circle) and ends at $x=2$ (right edge of the first circle). However, the enclosed region is formed by the intersection, so we are interested in the area between $x=0$ and $x=2$ where one circle's arc is above the other.

The region of intersection is formed by two segments, one from each circle. Each segment is defined by the chord connecting $(1, \sqrt{3})$ and $(1, -\sqrt{3})$.

Consider the first circle $x^2 + y^2 = 4$. The chord is at $x=1$. The area of the segment is the area of the sector minus the area of the triangle.

For the first circle, center $(0,0)$, radius $r=2$. The chord is at $x=1$. The points are $(1, \sqrt{3})$ and $(1, -\sqrt{3})$.

Let $\theta$ be the angle such that $\cos \theta = \frac{1}{2}$ (adjacent/hypotenuse = 1/2). So, $\theta = \frac{\pi}{3}$. The full angle subtended by the chord is $2\theta = \frac{2\pi}{3}$.

Area of the sector of the first circle = $\frac{1}{2}r^2 (2\theta) = \frac{1}{2}(2^2)\left(\frac{2\pi}{3}\right) = \frac{1}{2}(4)\left(\frac{2\pi}{3}\right) = \frac{4\pi}{3}$.

Area of the triangle formed by $(0,0)$, $(1, \sqrt{3})$, and $(1, -\sqrt{3})$: base $2\sqrt{3}$, height $1$. Area $= \frac{1}{2} \times 2\sqrt{3} \times 1 = \sqrt{3}$.

Area of the segment of the first circle (to the right of $x=1$) = Area of sector - Area of triangle $= \frac{4\pi}{3} - \sqrt{3}$.

Consider the second circle $(x-2)^2 + y^2 = 4$. Center $(2,0)$, radius $r=2$. The chord is at $x=1$. The points are $(1, \sqrt{3})$ and $(1, -\sqrt{3})$.

The distance from the center $(2,0)$ to the chord $x=1$ is $|1-2| = |-1| = 1$. So the geometry is the same as for the first circle relative to its center.

The angle calculation relative to the center $(2,0)$ is the same. The chord is 1 unit away from the center. The radius is 2. So, $\cos \phi = \frac{1}{2}$, meaning the angle from the line connecting the center to the chord is $\frac{\pi}{3}$. The total angle is $\frac{2\pi}{3}$.

Area of the sector of the second circle = $\frac{1}{2}r^2 (2\phi) = \frac{1}{2}(2^2)\left(\frac{2\pi}{3}\right) = \frac{4\pi}{3}$.

Area of the triangle formed by $(2,0)$, $(1, \sqrt{3})$, and $(1, -\sqrt{3})$: base $2\sqrt{3}$, height $|1-2| = 1$. Area $= \frac{1}{2} \times 2\sqrt{3} \times 1 = \sqrt{3}$.

Area of the segment of the second circle (to the left of $x=1$) = Area of sector - Area of triangle $= \frac{4\pi}{3} - \sqrt{3}$.

The area enclosed between the circles is the sum of these two segments.

The area of the region enclosed between the circles $x^2 + y^2 = 4$ and $(x-2)^2 + y^2 = 4$ is $\frac{8\pi}{3} - 2\sqrt{3}$ square units.

Given:

Two curves: $y = x^2 + 2$ (a parabola) and $y = x$ (a line).

The bounding vertical lines are $x = 0$ and $x = 3$.

To Find:

The area of the region bounded by these curves and lines.

Solution:

We need to find the area between the curve $y = x^2 + 2$ and the line $y = x$ over the interval $[0, 3]$.

First, let's check if the curves intersect within this interval by setting their equations equal:

Rearranging the equation:

To find the roots of this quadratic equation, we can use the discriminant, $\Delta = b^2 - 4ac$.

Here, $a = 1$, $b = -1$, $c = 2$.

Since the discriminant is negative ($\Delta < 0$), there are no real roots, meaning the parabola and the line do not intersect.

Now, we need to determine which curve is above the other in the interval $[0, 3]$.

Let's pick a test value, say $x = 1$.

For $y = x^2 + 2$: $y = 1^2 + 2 = 3$.

For $y = x$: $y = 1$.

Since $3 > 1$, the parabola $y = x^2 + 2$ is above the line $y = x$ in the interval $[0, 3]$.

The area between the curves is given by:

Here, $a = 0$, $b = 3$, $y_{upper} = x^2 + 2$, and $y_{lower} = x$.

So, the area is:

Rearrange the integrand:

Now, integrate the expression $(x^2 - x + 2)$:

Evaluate the definite integral from 0 to 3:

Substitute the upper and lower limits:

Simplify:

The area of the region bounded by the curves $y = x^2 + 2$, $y = x$, $x = 0$ and $x = 3$ is $\frac{21}{2}$ square units.

Given:

The vertices of the triangle are A(1, 0), B(2, 2), and C(3, 1).

To Find:

The area of the triangle using integration.

Solution:

To find the area using integration, we need to determine the equations of the lines forming the sides of the triangle.

Line AB: Passing through (1, 0) and (2, 2).

Slope $m_{AB} = \frac{2 - 0}{2 - 1} = \frac{2}{1} = 2$.

Equation of line AB using point-slope form ($y - y_1 = m(x - x_1)$):

Line BC: Passing through (2, 2) and (3, 1).

Slope $m_{BC} = \frac{1 - 2}{3 - 2} = \frac{-1}{1} = -1$.

Equation of line BC:

Line AC: Passing through (1, 0) and (3, 1).

Slope $m_{AC} = \frac{1 - 0}{3 - 1} = \frac{1}{2}$.

Equation of line AC:

To find the area using integration with respect to x, we need to consider the region in vertical strips. The x-coordinates range from 1 to 3. The upper boundary of the triangle changes at $x=2$.

From $x = 1$ to $x = 2$, the upper boundary is line AB ($y = 2x - 2$) and the lower boundary is line AC ($y = \frac{1}{2}x - \frac{1}{2}$).

From $x = 2$ to $x = 3$, the upper boundary is line BC ($y = -x + 4$) and the lower boundary is line AC ($y = \frac{1}{2}x - \frac{1}{2}$).

Area $= \int_{1}^{2} (y_{AB} - y_{AC}) \, dx + \int_{2}^{3} (y_{BC} - y_{AC}) \, dx$

First integral:

Integrate:

Evaluate:

Second integral:

Integrate:

Evaluate:

Total Area = $\frac{3}{4} + \frac{3}{4} = \frac{6}{4} = \frac{3}{2}$.

Alternative Method using Geometry:

The triangle has vertices A(1, 0), B(2, 2), C(3, 1).

We can find the area using the determinant formula:

Area $= \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

Both methods give the same result.

The area of the region bounded by the triangle is $\frac{3}{2}$ square units.

Given:

Parabola: $y^2 = 4ax$

Circle: $x^2 + y^2 = 4ax$, which is $(x-2a)^2 + y^2 = (2a)^2$ (center $(2a,0)$, radius $2a$)

To Find:

Common area.

Solution:

The common region is bounded by the parabola $x = \frac{y^2}{4a}$ and the left arc of the circle $x = 2a - \sqrt{4a^2 - y^2}$. The intersection points are $(0,0)$ and implicitly at $y=2a$ where $x_p=a$ and $x_c=2a$. The common area is found by integrating with respect to $y$ from $0$ to $2a$.

Area $= \int_0^{2a} (x_{circle} - x_{parabola}) dy$

Area $= \int_0^{2a} \left( (2a - \sqrt{4a^2 - y^2}) - \frac{y^2}{4a} \right) dy$

Area $= \int_0^{2a} 2a \, dy - \int_0^{2a} \sqrt{4a^2 - y^2} \, dy - \int_0^{2a} \frac{y^2}{4a} \, dy$

1. $\int_0^{2a} 2a \, dy = [2ay]_0^{2a} = 4a^2$.

2. $\int_0^{2a} \sqrt{4a^2 - y^2} \, dy$ is the area of a quarter circle of radius $2a$. Area $= \frac{1}{4}\pi (2a)^2 = \pi a^2$.

3. $\int_0^{2a} \frac{y^2}{4a} \, dy = \left[\frac{y^3}{12a}\right]_0^{2a} = \frac{(2a)^3}{12a} = \frac{8a^3}{12a} = \frac{2a^2}{3}$.

Area $= 4a^2 - \pi a^2 - \frac{2a^2}{3} = a^2 \left(4 - \pi - \frac{2}{3}\right) = a^2 \left(\frac{12 - 3\pi - 2}{3}\right) = a^2 \left(\frac{10 - 3\pi}{3}\right)$.

The area of the common region is $a^2 \left(\frac{10}{3} - \pi\right)$ square units.