Chapter 8 Application Of Integrals

Solution:

We need to find the area under the curve $y = \sin x$ from $x=0$ to $x=\pi$.

The area A is given by the definite integral of the function over the specified interval.

The area A is calculated as:

$A = \int_{0}^{\pi} \sin x \, dx$

The integral of $\sin x$ with respect to $x$ is $-\cos x$.

$A = [-\cos x]_{0}^{\pi}$

Now, we evaluate the definite integral by substituting the upper and lower limits:

$A = (-\cos \pi) - (-\cos 0)$

We know that $\cos \pi = -1$ and $\cos 0 = 1$. Substitute these values into the expression:

$A = (-(-1)) - (-1)$

$A = (1) - (-1)$

$A = 1 + 1$

$A = 2$

Thus, the area of the curve $y = \sin x$ between $x=0$ and $x=\pi$ is 2 square units.

Solution:

The region is bounded by the curve $ay^2 = x^3$, the y-axis (which is the line $x=0$), and the horizontal lines $y=a$ and $y=2a$.

From the equation of the curve, we can express $x$ in terms of $y$:

$x^3 = ay^2$

$x = (ay^2)^{1/3}$

$x = a^{1/3} y^{2/3}$

Since the region is bounded by the y-axis and lines parallel to the x-axis, it is more convenient to find the area by integrating with respect to $y$. The area $A$ is given by the definite integral of $x$ with respect to $y$ from $y=a$ to $y=2a$.

$A = \int\limits_{a}^{2a} x \, dy$

Substitute the expression for $x$ we found:

$A = \int\limits_{a}^{2a} a^{1/3} y^{2/3} \, dy$

We can factor out the constant $a^{1/3}$ from the integral:

$A = a^{1/3} \int\limits_{a}^{2a} y^{2/3} \, dy$

Now, we evaluate the integral of $y^{2/3}$. Using the power rule for integration $\int y^n \, dy = \frac{y^{n+1}}{n+1}$, where $n = 2/3$:

$\int y^{2/3} \, dy = \frac{y^{2/3 + 1}}{2/3 + 1} = \frac{y^{5/3}}{5/3} = \frac{3}{5} y^{5/3}$

So, the definite integral is:

$A = a^{1/3} \left[ \frac{3}{5} y^{5/3} \right]_{a}^{2a}$

Next, we apply the limits of integration by substituting the upper limit ($2a$) and subtracting the result of substituting the lower limit ($a$):

$A = a^{1/3} \left( \frac{3}{5} (2a)^{5/3} - \frac{3}{5} (a)^{5/3} \right)$

$A = a^{1/3} \frac{3}{5} \left( (2a)^{5/3} - a^{5/3} \right)$

We simplify the terms inside the parentheses using the property $(mn)^p = m^p n^p$ and $(m^p)^q = m^{pq}$:

$(2a)^{5/3} = 2^{5/3} a^{5/3}$

Substitute this back into the expression for A:

$A = a^{1/3} \frac{3}{5} \left( 2^{5/3} a^{5/3} - a^{5/3} \right)$

Factor out $a^{5/3}$ from the terms inside the parentheses:

$A = a^{1/3} \frac{3}{5} a^{5/3} (2^{5/3} - 1)$

Combine the terms with $a$ using the property $a^m a^n = a^{m+n}$:

$a^{1/3} a^{5/3} = a^{1/3 + 5/3} = a^{6/3} = a^2$

So, the area A is:

$A = \frac{3}{5} a^2 (2^{5/3} - 1)$

We can also write $2^{5/3}$ as $\sqrt[3]{2^5} = \sqrt[3]{32}$.

$A = \frac{3}{5} a^2 (\sqrt[3]{32} - 1)$

The area of the specified region is $\frac{3}{5} a^2 (2^{5/3} - 1)$ square units.

Solution:

The equations of the given curves are:

Parabola: $y^2 = 2x$

Straight line: $x - y = 4$

To find the area of the region bounded by these curves, we first find their points of intersection. We can express $x$ from the equation of the line:

$x = y + 4$

Substitute this expression for $x$ into the equation of the parabola:

$y^2 = 2(y + 4)$

$y^2 = 2y + 8$

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

$y^2 - 2y - 8 = 0$

We can solve this quadratic equation by factoring:

$(y - 4)(y + 2) = 0$

The solutions for $y$ are $y = 4$ and $y = -2$. These are the y-coordinates of the points of intersection.

Now, find the corresponding x-coordinates using the line equation $x = y + 4$:

If $y = 4$, then $x = 4 + 4 = 8$. The intersection point is $(8, 4)$.

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

The points of intersection are $(2, -2)$ and $(8, 4)$.

To find the area bounded by the curves, it is convenient to integrate with respect to $y$. From the given equations, we express $x$ in terms of $y$ for both curves:

Parabola: $x = \frac{y^2}{2}$

Line: $x = y + 4$

We need to determine which curve is to the right in the region between $y=-2$ and $y=4$. For any value of $y$ in this interval, say $y=0$, the line is at $x=0+4=4$ and the parabola is at $x=0^2/2=0$. Since $4 > 0$, the line is to the right of the parabola.

The area $A$ is given by the integral of (right curve $x$) - (left curve $x$) with respect to $y$ from the lower y-limit to the upper y-limit:

$A = \int\limits_{-2}^{4} \left( (y + 4) - \frac{y^2}{2} \right) \, dy$

Now, we evaluate the definite integral:

$A = \left[ \frac{y^2}{2} + 4y - \frac{1}{2} \cdot \frac{y^3}{3} \right]_{-2}^{4}$

$A = \left[ \frac{y^2}{2} + 4y - \frac{y^3}{6} \right]_{-2}^{4}$

Apply the limits of integration:

$A = \left( \frac{(4)^2}{2} + 4(4) - \frac{(4)^3}{6} \right) - \left( \frac{(-2)^2}{2} + 4(-2) - \frac{(-2)^3}{6} \right)$

$A = \left( \frac{16}{2} + 16 - \frac{64}{6} \right) - \left( \frac{4}{2} - 8 - \frac{-8}{6} \right)$

$A = \left( 8 + 16 - \frac{32}{3} \right) - \left( 2 - 8 + \frac{4}{3} \right)$

$A = \left( 24 - \frac{32}{3} \right) - \left( -6 + \frac{4}{3} \right)$

Continue simplifying:

$A = 24 - \frac{32}{3} + 6 - \frac{4}{3}$

$A = (24 + 6) + \left( -\frac{32}{3} - \frac{4}{3} \right)$

$A = 30 - \frac{36}{3}$

$A = 30 - 12$

$A = 18$

The area of the region bounded by the parabola $y^2 = 2x$ and the line $x - y = 4$ is 18 square units.

Solution:

The equations of the given parabolas are:

(1) $y^2 = 6x$

(2) $x^2 = 6y$

To find the area of the region bounded by these curves, we first find their points of intersection. From equation (2), we can write $y = \frac{x^2}{6}$.

Substitute this expression for $y$ into equation (1):

$\left(\frac{x^2}{6}\right)^2 = 6x$

$\frac{x^4}{36} = 6x$

Rearrange the equation to solve for $x$:

$x^4 = 36 \times 6x$

$x^4 = 216x$

$x^4 - 216x = 0$

$x(x^3 - 216) = 0$

This gives us two possibilities for $x$:

$x = 0$ or $x^3 - 216 = 0$

If $x^3 - 216 = 0$, then $x^3 = 216$. Taking the cube root, we get $x = 6$.

The x-coordinates of the points of intersection are $x=0$ and $x=6$.

Now, find the corresponding y-coordinates using either equation. Using $y = \frac{x^2}{6}$:

If $x = 0$, then $y = \frac{0^2}{6} = 0$. The intersection point is $(0, 0)$.

If $x = 6$, then $y = \frac{6^2}{6} = \frac{36}{6} = 6$. The intersection point is $(6, 6)$.

The points of intersection are $(0, 0)$ and $(6, 6)$.

The bounded region is between $x=0$ and $x=6$. We need to determine which parabola is the upper curve and which is the lower curve in this interval. From equation (1), $y = \sqrt{6x}$ (since $y^2=6x$ and in the first quadrant where $x>0$, $y$ is positive). From equation (2), $y = \frac{x^2}{6}$.

For $0 < x < 6$, let's pick a value, say $x=1$.

For $y^2=6x$, $y = \sqrt{6(1)} = \sqrt{6} \approx 2.45$.

For $x^2=6y$, $y = \frac{1^2}{6} = \frac{1}{6} \approx 0.17$.

Since $\sqrt{6} > \frac{1}{6}$ for $x=1$, the curve $y=\sqrt{6x}$ (from $y^2=6x$) is above the curve $y=\frac{x^2}{6}$ (from $x^2=6y$) in the interval $(0, 6)$.

The area $A$ is given by the integral of the difference between the upper curve and the lower curve from $x=0$ to $x=6$:

$A = \int\limits_{0}^{6} \left( \sqrt{6x} - \frac{x^2}{6} \right) \, dx$

$A = \int\limits_{0}^{6} \left( \sqrt{6} x^{1/2} - \frac{1}{6} x^2 \right) \, dx$

Now, we evaluate the definite integral:

$A = \left[ \sqrt{6} \cdot \frac{x^{1/2 + 1}}{1/2 + 1} - \frac{1}{6} \cdot \frac{x^{2+1}}{2+1} \right]_{0}^{6}$

$A = \left[ \sqrt{6} \cdot \frac{x^{3/2}}{3/2} - \frac{1}{6} \cdot \frac{x^3}{3} \right]_{0}^{6}$

$A = \left[ \sqrt{6} \cdot \frac{2}{3} x^{3/2} - \frac{1}{18} x^3 \right]_{0}^{6}$

$A = \left[ \frac{2\sqrt{6}}{3} x^{3/2} - \frac{1}{18} x^3 \right]_{0}^{6}$

Apply the limits of integration:

$A = \left( \frac{2\sqrt{6}}{3} (6)^{3/2} - \frac{1}{18} (6)^3 \right) - \left( \frac{2\sqrt{6}}{3} (0)^{3/2} - \frac{1}{18} (0)^3 \right)$

$A = \left( \frac{2\sqrt{6}}{3} (6\sqrt{6}) - \frac{216}{18} \right) - (0 - 0)$

$A = \left( \frac{2 \times 6 \times 6}{3} - 12 \right)$

$A = \left( \frac{72}{3} - 12 \right)$

$A = 24 - 12$

$A = 12$

The area of the region bounded by the parabolas $y^2 = 6x$ and $x^2 = 6y$ is 12 square units.

Solution:

The given parametric equations of the curve are:

These are the parametric equations of an ellipse centered at the origin. To see this, we can rearrange the equations:

From (1), $\frac{x}{3} = \cos t$

From (2), $\frac{y}{2} = \sin t$

Squaring both equations and adding them, we get:

$\left(\frac{x}{3}\right)^2 + \left(\frac{y}{2}\right)^2 = \cos^2 t + \sin^2 t$

$\frac{x^2}{9} + \frac{y^2}{4} = 1$

This is the equation of an ellipse with semi-major axis $a=3$ and semi-minor axis $b=2$.

The area enclosed by a parametric curve $x = x(t)$, $y = y(t)$ as $t$ varies from $t_1$ to $t_2$ is given by the integral formula:

$A = \frac{1}{2} \int\limits_{t_1}^{t_2} (x(t) y'(t) - y(t) x'(t)) \, dt$

Alternatively, for a simple closed curve traversed counterclockwise, the area can be found using:

$A = \int\limits_{t_1}^{t_2} x(t) y'(t) \, dt$ or $A = -\int\limits_{t_1}^{t_2} y(t) x'(t) \, dt$

For the given ellipse parameterized by $x = 3 \cos t$ and $y = 2 \sin t$, a single traversal of the closed curve occurs as $t$ goes from $0$ to $2\pi$. This traversal is counterclockwise.

We calculate the derivatives of $x(t)$ and $y(t)$ with respect to $t$:

$x'(t) = \frac{d}{dt}(3 \cos t) = -3 \sin t$

$y'(t) = \frac{d}{dt}(2 \sin t) = 2 \cos t$

Let's use the formula $A = \int\limits_{0}^{2\pi} x(t) y'(t) \, dt$:

$A = \int\limits_{0}^{2\pi} (3 \cos t)(2 \cos t) \, dt$

$A = \int\limits_{0}^{2\pi} 6 \cos^2 t \, dt$

Use the trigonometric identity $\cos^2 t = \frac{1 + \cos(2t)}{2}$:

$A = \int\limits_{0}^{2\pi} 6 \left( \frac{1 + \cos(2t)}{2} \right) \, dt$

$A = \int\limits_{0}^{2\pi} 3 (1 + \cos(2t)) \, dt$

$A = 3 \int\limits_{0}^{2\pi} (1 + \cos(2t)) \, dt$

Evaluate the integral:

$A = 3 \left[ t + \frac{\sin(2t)}{2} \right]_{0}^{2\pi}$

$A = 3 \left[ \left( 2\pi + \frac{\sin(2 \times 2\pi)}{2} \right) - \left( 0 + \frac{\sin(2 \times 0)}{2} \right) \right]$

$A = 3 \left[ \left( 2\pi + \frac{\sin(4\pi)}{2} \right) - \left( 0 + \frac{\sin(0)}{2} \right) \right]$

Since $\sin(4\pi) = 0$ and $\sin(0) = 0$:

$A = 3 \left[ (2\pi + 0) - (0 + 0) \right]$

$A = 3 (2\pi)$

$A = 6\pi$

Alternate Solution:

The given parametric equations $x = 3 \cos t$, $y = 2 \sin t$ represent an ellipse with semi-major axis $a=3$ and semi-minor axis $b=2$.

The area of an ellipse with semi-major axis $a$ and semi-minor axis $b$ is given by the formula $A = \pi ab$.

Substituting $a=3$ and $b=2$ into the formula:

$A = \pi (3)(2)$

$A = 6\pi$

The area enclosed by the curve $x = 3 \cos t$, $y = 2 \sin t$ is $6\pi$ square units.

Solution:

The equations of the given curves are:

Parabola: $y = \frac{3x^2}{4}$

Line: $3x - 2y + 12 = 0$

To find the area of the region bounded by these curves, we first find their points of intersection. From the equation of the line, we can express $y$ in terms of $x$:

$2y = 3x + 12$

$y = \frac{3x + 12}{2} = \frac{3}{2}x + 6$

Now, substitute this expression for $y$ into the equation of the parabola:

$\frac{3x^2}{4} = \frac{3}{2}x + 6$

Multiply both sides by 4 to eliminate the denominators:

$3x^2 = 4 \left( \frac{3}{2}x + 6 \right)$

$3x^2 = 6x + 24$

Rearrange the equation to form a quadratic equation in terms of $x$:

$3x^2 - 6x - 24 = 0$

Divide the entire equation by 3:

$x^2 - 2x - 8 = 0$

Factor the quadratic equation to find the x-coordinates of the intersection points:

$(x - 4)(x + 2) = 0$

The solutions for $x$ are $x = 4$ and $x = -2$. These are the limits of integration.

The bounded region is between $x=-2$ and $x=4$. We need to determine which curve is the upper curve ($y_{upper}$) and which is the lower curve ($y_{lower}$) in this interval.

Let's compare the y-values for a point within the interval, say $x=0$:

For the line: $y = \frac{3}{2}(0) + 6 = 6$

For the parabola: $y = \frac{3(0)^2}{4} = 0$

Since $6 > 0$, the line $y = \frac{3}{2}x + 6$ is above the parabola $y = \frac{3x^2}{4}$ in the interval $(-2, 4)$.

The area $A$ of the region between the curves is given by the definite integral of the difference between the upper and lower curves from $x=-2$ to $x=4$:

$A = \int\limits_{-2}^{4} \left( \left(\frac{3}{2}x + 6\right) - \left(\frac{3x^2}{4}\right) \right) \, dx$

$A = \int\limits_{-2}^{4} \left( \frac{3}{2}x + 6 - \frac{3}{4}x^2 \right) \, dx$

Now, evaluate the indefinite integral:

$\int \left( \frac{3}{2}x + 6 - \frac{3}{4}x^2 \right) \, dx = \frac{3}{2} \cdot \frac{x^2}{2} + 6x - \frac{3}{4} \cdot \frac{x^3}{3} + C$

$= \frac{3}{4}x^2 + 6x - \frac{1}{4}x^3 + C$

Apply the limits of integration ($x=4$ and $x=-2$):

$A = \left[ \frac{3}{4}x^2 + 6x - \frac{1}{4}x^3 \right]_{-2}^{4}$

$A = \left( \frac{3}{4}(4)^2 + 6(4) - \frac{1}{4}(4)^3 \right) - \left( \frac{3}{4}(-2)^2 + 6(-2) - \frac{1}{4}(-2)^3 \right)$

$A = \left( \frac{3}{4}(16) + 24 - \frac{1}{4}(64) \right) - \left( \frac{3}{4}(4) - 12 - \frac{1}{4}(-8) \right)$

$A = \left( 12 + 24 - 16 \right) - \left( 3 - 12 + 2 \right)$

$A = \left( 36 - 16 \right) - \left( -9 + 2 \right)$

$A = 20 - (-7)$

$A = 20 + 7$

$A = 27$

The area of the region included between the parabola $y = \frac{3x^2}{4}$ and the line $3x - 2y + 12 = 0$ is 27 square units.

Solution:

The given parametric equations of the curve are:

$x = at^2$

$y = 2at$

We need to find the area of the region bounded by this curve and the vertical lines (ordinates) corresponding to the parameter values $t = 1$ and $t = 2$.

The area under a parametric curve $y=y(t)$, $x=x(t)$ between the ordinates corresponding to $t=t_1$ and $t=t_2$ is given by the integral formula:

$A = \int\limits_{t_1}^{t_2} y(t) \frac{dx}{dt} \, dt$

First, we find the derivative of $x$ with respect to $t$:

$\frac{dx}{dt} = \frac{d}{dt}(at^2) = a \cdot 2t = 2at$

The given limits for the parameter $t$ are $t_1 = 1$ and $t_2 = 2$.

Substitute $y(t) = 2at$ and $\frac{dx}{dt} = 2at$ into the area formula:

$A = \int\limits_{1}^{2} (2at)(2at) \, dt$

$A = \int\limits_{1}^{2} 4a^2 t^2 \, dt$

Now, evaluate the definite integral:

$A = 4a^2 \int\limits_{1}^{2} t^2 \, dt$

$A = 4a^2 \left[ \frac{t^3}{3} \right]_{1}^{2}$

Apply the limits of integration:

$A = 4a^2 \left( \frac{(2)^3}{3} - \frac{(1)^3}{3} \right)$

$A = 4a^2 \left( \frac{8}{3} - \frac{1}{3} \right)$

$A = 4a^2 \left( \frac{8 - 1}{3} \right)$

$A = 4a^2 \left( \frac{7}{3} \right)$

$A = \frac{28a^2}{3}$

The area of the region bounded by the given curve and the ordinates corresponding to $t=1$ and $t=2$ is $\frac{28a^2}{3}$ square units.

Solution:

The equations of the given curves are:

Parabola: $y^2 = ax$

Circle: $x^2 + y^2 = 2ax$

We are interested in the region above the x-axis, so $y \ge 0$. For the parabola, $y = \sqrt{ax}$. For the circle, $y = \sqrt{2ax - x^2}$.

To find the points of intersection, substitute $y^2 = ax$ into the circle equation:

$x^2 + ax = 2ax$

$x^2 - ax = 0$

$x(x - a) = 0$

This gives $x = 0$ or $x = a$.

If $x = 0$, $y^2 = a(0) = 0 \implies y = 0$. Intersection point: $(0, 0)$.

If $x = a$, $y^2 = a(a) = a^2 \implies y = \pm a$. Since we are considering the region above the x-axis, $y = a$. Intersection point: $(a, a)$.

The curves intersect at $(0, 0)$ and $(a, a)$. The region is bounded between $x=0$ and $x=a$.

We rewrite the circle equation by completing the square to identify its center and radius:

$x^2 - 2ax + y^2 = 0$

$(x^2 - 2ax + a^2) + y^2 = a^2$

$(x - a)^2 + y^2 = a^2$

This is a circle centered at $(a, 0)$ with radius $a$.

In the region between $x=0$ and $x=a$, the upper boundary is the circle and the lower boundary is the parabola. The area $A$ is given by the integral of the difference between the y-values of the circle and the parabola from $x=0$ to $x=a$:

$A = \int\limits_{0}^{a} (\sqrt{2ax - x^2} - \sqrt{ax}) \, dx$

$A = \int\limits_{0}^{a} \sqrt{2ax - x^2} \, dx - \int\limits_{0}^{a} \sqrt{ax} \, dx$

Evaluate the first integral $\int\limits_{0}^{a} \sqrt{2ax - x^2} \, dx$. Complete the square under the radical:

$2ax - x^2 = a^2 - (x - a)^2$

So, $\int\limits_{0}^{a} \sqrt{a^2 - (x - a)^2} \, dx$. Let $u = x - a$, $du = dx$. When $x=0$, $u=-a$. When $x=a$, $u=0$.

$\int\limits_{-a}^{0} \sqrt{a^2 - u^2} \, du$

This integral represents the area of a quarter circle of radius $a$ in the upper half plane (from $u=-a$ to $u=0$).

$\left[ \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{u}{a}\right) \right]_{-a}^{0}$

$= \left( 0 + \frac{a^2}{2} \sin^{-1}(0) \right) - \left( \frac{-a}{2}\sqrt{a^2 - a^2} + \frac{a^2}{2} \sin^{-1}(-1) \right)$

$= (0 + 0) - (0 + \frac{a^2}{2} (-\frac{\pi}{2}))$

$= \frac{\pi a^2}{4}$

Evaluate the second integral $\int\limits_{0}^{a} \sqrt{ax} \, dx$:

$\int\limits_{0}^{a} \sqrt{a} x^{1/2} \, dx = \sqrt{a} \int\limits_{0}^{a} x^{1/2} \, dx$

$= \sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{a}$

$= \sqrt{a} \left( \frac{2}{3} a^{3/2} - 0 \right)$

$= a^{1/2} \cdot \frac{2}{3} a^{3/2} = \frac{2}{3} a^{(1/2 + 3/2)} = \frac{2}{3} a^2$

Subtract the second integral from the first to find the area A:

$A = \frac{\pi a^2}{4} - \frac{2a^2}{3}$

$A = a^2 \left( \frac{\pi}{4} - \frac{2}{3} \right)$

$A = a^2 \left( \frac{3\pi - 8}{12} \right)$

The area of the region is $\frac{a^2}{12}(3\pi - 8)$ square units.

Solution:

The equation of the circle is $x^2 + y^2 = a^2$, which is a circle centered at the origin $(0,0)$ with radius $a$.

The equation of the line is $x = \frac{a}{2}$.

The minor segment is the region bounded by the circle and the line $x = \frac{a}{2}$ to the right of the line (since $a/2 < a$).

We need to find the points of intersection of the circle and the line. Substitute $x = \frac{a}{2}$ into the circle equation:

$\left(\frac{a}{2}\right)^2 + y^2 = a^2$

$\frac{a^2}{4} + y^2 = a^2$

$y^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}$

$y = \pm \sqrt{\frac{3a^2}{4}} = \pm \frac{\sqrt{3}a}{2}$

The points of intersection are $\left(\frac{a}{2}, \frac{\sqrt{3}a}{2}\right)$ and $\left(\frac{a}{2}, -\frac{\sqrt{3}a}{2}\right)$.

The area of the minor segment can be found by integrating the difference between the upper and lower parts of the circle with respect to $x$ from $x = \frac{a}{2}$ to $x = a$.

The upper half of the circle is $y = \sqrt{a^2 - x^2}$ and the lower half is $y = -\sqrt{a^2 - x^2}$. The region is symmetric about the x-axis, so we can calculate the area of the upper half of the segment (from $y=0$ to $y=\sqrt{a^2-x^2}$) and multiply it by 2.

The area A is given by:

$A = 2 \int\limits_{a/2}^{a} \sqrt{a^2 - x^2} \, dx$

We use the standard integration formula: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$.

Apply the limits of integration:

$A = 2 \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \right]_{a/2}^{a}$

Evaluate at the upper limit ($x=a$):

$\frac{a}{2}\sqrt{a^2 - a^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{a}{a}\right) = \frac{a}{2}\sqrt{0} + \frac{a^2}{2}\sin^{-1}(1) = 0 + \frac{a^2}{2} \cdot \frac{\pi}{2} = \frac{\pi a^2}{4}$

Evaluate at the lower limit ($x=a/2$):

$\frac{a/2}{2}\sqrt{a^2 - (a/2)^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{a/2}{a}\right)$

$= \frac{a}{4}\sqrt{a^2 - \frac{a^2}{4}} + \frac{a^2}{2}\sin^{-1}\left(\frac{1}{2}\right)$

$= \frac{a}{4}\sqrt{\frac{3a^2}{4}} + \frac{a^2}{2} \cdot \frac{\pi}{6}$

$= \frac{a}{4} \cdot \frac{\sqrt{3}a}{2} + \frac{\pi a^2}{12}$

$= \frac{\sqrt{3}a^2}{8} + \frac{\pi a^2}{12}$

Now, subtract the value at the lower limit from the value at the upper limit and multiply by 2:

$A = 2 \left[ \left(\frac{\pi a^2}{4}\right) - \left(\frac{\sqrt{3}a^2}{8} + \frac{\pi a^2}{12}\right) \right]$

$A = 2 \left[ \frac{\pi a^2}{4} - \frac{\sqrt{3}a^2}{8} - \frac{\pi a^2}{12} \right]$

$A = 2 \left[ \pi a^2 \left(\frac{1}{4} - \frac{1}{12}\right) - \frac{\sqrt{3}a^2}{8} \right]$

$A = 2 \left[ \pi a^2 \left(\frac{3-1}{12}\right) - \frac{\sqrt{3}a^2}{8} \right]$

$A = 2 \left[ \pi a^2 \left(\frac{2}{12}\right) - \frac{\sqrt{3}a^2}{8} \right]$

$A = 2 \left[ \frac{\pi a^2}{6} - \frac{\sqrt{3}a^2}{8} \right]$

$A = \frac{2\pi a^2}{6} - \frac{2\sqrt{3}a^2}{8}$

$A = \frac{\pi a^2}{3} - \frac{\sqrt{3}a^2}{4}$

$A = a^2 \left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right)$

The area of the minor segment is $a^2 \left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right)$ square units.

Solution:

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

This is the equation of a circle centered at the origin $(0,0)$ with radius $r$. Comparing the given equation with the standard equation of a circle $x^2 + y^2 = r^2$, we have $r^2 = 2$.

So, the radius of the circle is $r = \sqrt{2}$.

The area enclosed by a circle with radius $r$ is given by the formula $A = \pi r^2$.

$A = \pi (\sqrt{2})^2$

$A = \pi (2)$

$A = 2\pi$

Alternatively, we can find the area using integration. The area of the circle can be calculated as $A = \int\limits_{-r}^{r} 2\sqrt{r^2 - x^2} \, dx$.

For this circle, $r = \sqrt{2}$, so $r^2 = 2$. The area is $A = \int\limits_{-\sqrt{2}}^{\sqrt{2}} 2\sqrt{2 - x^2} \, dx$.

$A = 2 \int\limits_{-\sqrt{2}}^{\sqrt{2}} \sqrt{2 - x^2} \, dx$

Using the standard integral formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$, with $a = \sqrt{2}$:

$A = 2 \left[ \frac{x}{2}\sqrt{2 - x^2} + \frac{2}{2}\sin^{-1}\left(\frac{x}{\sqrt{2}}\right) \right]_{-\sqrt{2}}^{\sqrt{2}}$

$A = 2 \left[ \frac{x}{2}\sqrt{2 - x^2} + \sin^{-1}\left(\frac{x}{\sqrt{2}}\right) \right]_{-\sqrt{2}}^{\sqrt{2}}$

Apply the limits of integration:

$A = 2 \left( \frac{\sqrt{2}}{2}\sqrt{2 - (\sqrt{2})^2} + \sin^{-1}\left(\frac{\sqrt{2}}{\sqrt{2}}\right) \right) - 2 \left( \frac{-\sqrt{2}}{2}\sqrt{2 - (-\sqrt{2})^2} + \sin^{-1}\left(\frac{-\sqrt{2}}{\sqrt{2}}\right) \right)$

$A = 2 \left( \frac{\sqrt{2}}{2}\sqrt{0} + \sin^{-1}(1) \right) - 2 \left( \frac{-\sqrt{2}}{2}\sqrt{0} + \sin^{-1}(-1) \right)$

$A = 2 \left( 0 + \frac{\pi}{2} \right) - 2 \left( 0 - \frac{\pi}{2} \right)$

$A = \pi - (-\pi)$

$A = 2\pi$

The area enclosed by the circle is $2\pi$ square units.

Comparing this with the given options, the correct answer is (D).

The final answer is (D) 2$\pi$ sq units.

Solution:

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

From this equation, we can express $y^2$ and then $y$ in terms of $x$:

$\frac{y^2}{b^2} = 1 - \frac{x^2}{a^2}$

$y^2 = b^2 \left( 1 - \frac{x^2}{a^2} \right) = \frac{b^2}{a^2} (a^2 - x^2)$

$y = \pm \frac{b}{a} \sqrt{a^2 - x^2}$

The area of the ellipse can be calculated by integrating the difference between the upper and lower halves of the ellipse from $x=-a$ to $x=a$. Due to symmetry, we can calculate the area of the upper half and multiply it by 2, or calculate the area of the portion in the first quadrant (from $x=0$ to $x=a$) and multiply it by 4.

Let's calculate the area in the first quadrant and multiply by 4:

$A = 4 \int\limits_{0}^{a} y \, dx$

$A = 4 \int\limits_{0}^{a} \frac{b}{a} \sqrt{a^2 - x^2} \, dx$

$A = \frac{4b}{a} \int\limits_{0}^{a} \sqrt{a^2 - x^2} \, dx$

Use the standard integration formula: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$.

Apply the limits of integration:

$A = \frac{4b}{a} \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \right]_{0}^{a}$

Evaluate at the upper limit ($x=a$):

$\frac{a}{2}\sqrt{a^2 - a^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{a}{a}\right) = \frac{a}{2}\sqrt{0} + \frac{a^2}{2}\sin^{-1}(1) = 0 + \frac{a^2}{2} \cdot \frac{\pi}{2} = \frac{\pi a^2}{4}$

Evaluate at the lower limit ($x=0$):

$\frac{0}{2}\sqrt{a^2 - 0^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{0}{a}\right) = 0 + \frac{a^2}{2}\sin^{-1}(0) = 0 + 0 = 0$

Subtract the value at the lower limit from the value at the upper limit and multiply by $\frac{4b}{a}$:

$A = \frac{4b}{a} \left( \frac{\pi a^2}{4} - 0 \right)$

$A = \frac{4b}{a} \cdot \frac{\pi a^2}{4}$

$A = \pi ab$

The area enclosed by the ellipse is $\pi ab$ square units.

Comparing this with the given options, the correct answer is (B).

The final answer is (B) $\pi$ab.

Solution:

The equations of the given curves are:

Parabola: $y = x^2$

Line: $y = 16$

To find the area of the region bounded by these curves, we first find their points of intersection. Set the y-values equal:

$x^2 = 16$

Taking the square root of both sides:

$x = \pm \sqrt{16}$

$x = \pm 4$

The x-coordinates of the points of intersection are $x = -4$ and $x = 4$. The intersection points are $(-4, 16)$ and $(4, 16)$.

The bounded region is between $x=-4$ and $x=4$. In this region, the line $y=16$ is above the parabola $y=x^2$.

The area $A$ of the region between the curves is given by the definite integral of the difference between the upper curve ($y=16$) and the lower curve ($y=x^2$) from $x=-4$ to $x=4$:

$A = \int\limits_{-4}^{4} (16 - x^2) \, dx$

The integrand $(16 - x^2)$ is an even function, and the limits of integration are symmetric about 0. Therefore, we can write the integral as:

$A = 2 \int\limits_{0}^{4} (16 - x^2) \, dx$

Now, evaluate the indefinite integral:

$\int (16 - x^2) \, dx = 16x - \frac{x^3}{3} + C$

Apply the limits of integration from 0 to 4 and multiply by 2:

$A = 2 \left[ 16x - \frac{x^3}{3} \right]_{0}^{4}$

$A = 2 \left( \left( 16(4) - \frac{(4)^3}{3} \right) - \left( 16(0) - \frac{(0)^3}{3} \right) \right)$

$A = 2 \left( \left( 64 - \frac{64}{3} \right) - (0 - 0) \right)$

$A = 2 \left( 64 - \frac{64}{3} \right)$

Combine the terms inside the parentheses:

$64 - \frac{64}{3} = \frac{64 \times 3}{3} - \frac{64}{3} = \frac{192 - 64}{3} = \frac{128}{3}$

Substitute this back into the expression for A:

$A = 2 \left( \frac{128}{3} \right)$

$A = \frac{256}{3}$

The area of the region bounded by the curve $y = x^2$ and the line $y = 16$ is $\frac{256}{3}$ square units.

Comparing this with the given options, the correct answer is (B).

The final answer is (B) $\frac{256}{3}$.

Solution:

The region is bounded by the curve $x = y^2$, the y-axis (which is the line $x=0$), and the horizontal lines $y = 3$ and $y = 4$.

Since the region is bounded by the y-axis and lines parallel to the x-axis, it is convenient to find the area by integrating with respect to $y$. The limits of integration for $y$ are given as $3$ and $4$.

In the region from $y=3$ to $y=4$, the curve $x = y^2$ is to the right of the y-axis ($x=0$).

The area $A$ is given by the definite integral of the right curve ($x=y^2$) minus the left curve ($x=0$) with respect to $y$ from $y=3$ to $y=4$.

$A = \int\limits_{3}^{4} (y^2 - 0) \, dy$

$A = \int\limits_{3}^{4} y^2 \, dy$

Now, we evaluate the definite integral:

$A = \left[ \frac{y^3}{3} \right]_{3}^{4}$

Apply the limits of integration:

$A = \left( \frac{(4)^3}{3} \right) - \left( \frac{(3)^3}{3} \right)$

$A = \frac{64}{3} - \frac{27}{3}$

$A = \frac{64 - 27}{3}$

$A = \frac{37}{3}$

The area of the region is $\frac{37}{3}$ square units.

The area of the region bounded by the curve x = y², y-axis and the line y = 3 and y = 4 is $\frac{37}{3}$ sq units.

Solution:

The region is bounded by the curve $y = x^2 + x$, the x-axis ($y=0$), and the vertical lines $x = 2$ and $x = 5$.

We need to find the area under the curve $y = x^2 + x$ from $x=2$ to $x=5$. First, let's check the sign of $y$ in the interval $[2, 5]$. For $x$ in this interval, $x > 0$ and $x+1 > 0$, so $y = x(x+1) > 0$. The curve lies above the x-axis in the given interval.

The area A is given by the definite integral of the function $y$ with respect to $x$ from $x=2$ to $x=5$:

$A = \int\limits_{2}^{5} (x^2 + x) \, dx$

Now, we evaluate the definite integral:

$A = \left[ \frac{x^{2+1}}{2+1} + \frac{x^{1+1}}{1+1} \right]_{2}^{5}$

$A = \left[ \frac{x^3}{3} + \frac{x^2}{2} \right]_{2}^{5}$

Apply the limits of integration by substituting the upper limit ($x=5$) and subtracting the result of substituting the lower limit ($x=2$):

$A = \left( \frac{(5)^3}{3} + \frac{(5)^2}{2} \right) - \left( \frac{(2)^3}{3} + \frac{(2)^2}{2} \right)$

$A = \left( \frac{125}{3} + \frac{25}{2} \right) - \left( \frac{8}{3} + \frac{4}{2} \right)$

$A = \left( \frac{125}{3} + \frac{25}{2} \right) - \left( \frac{8}{3} + 2 \right)$

Combine the terms with common denominators:

$A = \frac{125}{3} + \frac{25}{2} - \frac{8}{3} - 2$

$A = \left( \frac{125}{3} - \frac{8}{3} \right) + \left( \frac{25}{2} - 2 \right)$

$A = \frac{117}{3} + \left( \frac{25}{2} - \frac{4}{2} \right)$

$A = 39 + \frac{21}{2}$

Combine the terms into a single fraction:

$A = \frac{39 \times 2}{2} + \frac{21}{2}$

$A = \frac{78}{2} + \frac{21}{2}$

$A = \frac{78 + 21}{2}$

$A = \frac{99}{2}$

The area of the region bounded by the curve $y = x^2 + x$, the x-axis and the line $x = 2$ and $x = 5$ is equal to $\frac{99}{2}$.

The area of the region bounded by the curve y = x² + x, x-axis and the line x = 2 and x = 5 is equal to $\frac{99}{2}$ sq units.

Given:

The curves are $y^2 = 9x$ and $y = 3x$.

To Find:

The area of the region bounded by the given curves.

Solution:

The given curves are:

To find the points of intersection, substitute equation (2) into equation (1):

This gives two possible values for x:

Now, find the corresponding y-values using $y = 3x$:

When $x = 0$, $y = 3(0) = 0$. Point of intersection is $(0, 0)$.

When $x = 1$, $y = 3(1) = 3$. Point of intersection is $(1, 3)$.

The region is bounded by the curves between $x = 0$ and $x = 1$.

For $x \in (0, 1)$, we need to determine which curve is above the other. From $y^2 = 9x$, we get $y = \pm 3\sqrt{x}$. Since the intersection points are in the first quadrant and the line $y=3x$ is in the first quadrant for $x>0$, we consider the upper half of the parabola, $y = 3\sqrt{x}$.

Let's test a value, say $x = 0.5$:

For the parabola: $y = 3\sqrt{0.5} \approx 3 \times 0.707 = 2.121$

For the line: $y = 3(0.5) = 1.5$

Since $2.121 > 1.5$, the curve $y = 3\sqrt{x}$ is above $y = 3x$ in the interval $(0, 1)$.

The area A of the region bounded by the curves from $x=0$ to $x=1$ is given by the integral:

Evaluate the integral:

Apply the limits of integration:

The area of the region bounded by the curves is $\frac{1}{2}$ square units.

Given:

The curves are $y^2 = 2px$ and $x^2 = 2py$, where $p > 0$.

To Find:

The area of the region bounded by the given parabolas.

Solution:

The given curves are:

To find the points of intersection, we solve the system of equations. From equation (2), we can express y in terms of x:

Substitute equation (3) into equation (1):

This equation gives the x-coordinates of the intersection points:

or

Now, find the corresponding y-coordinates using $y = \frac{x^2}{2p}$:

When $x = 0$, $y = \frac{(0)^2}{2p} = 0$. The intersection point is $(0, 0)$.

When $x = 2p$, $y = \frac{(2p)^2}{2p} = \frac{4p^2}{2p} = 2p$. The intersection point is $(2p, 2p)$.

The region bounded by the two parabolas is in the first quadrant, from $x = 0$ to $x = 2p$.

To find the area, we need to determine which curve is above the other in the interval $(0, 2p)$.

From $y^2 = 2px$, we get $y = \sqrt{2px}$ (since we are in the first quadrant, $y \ge 0$).

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

Consider a value of x between $0$ and $2p$, for example, $x=p$.

For $y = \sqrt{2px}$: $y = \sqrt{2p \cdot p} = \sqrt{2p^2} = p\sqrt{2}$.

For $y = \frac{x^2}{2p}$: $y = \frac{p^2}{2p} = \frac{p}{2}$.

Since $p\sqrt{2} > \frac{p}{2}$ (as $\sqrt{2} > 1/2$), the curve $y = \sqrt{2px}$ is above $y = \frac{x^2}{2p}$ in the interval $(0, 2p)$.

The area A of the bounded region is given by the definite integral:

Now, evaluate the integral:

Apply the limits of integration:

The area of the region bounded by the parabolas is $\frac{4p^2}{3}$ square units.

Given:

The curves are $y = x^3$, $y = x + 6$, and the line $x = 0$.

To Find:

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

Solution:

The given equations are:

To find the points of intersection of the curves (1) and (2), set the expressions for y equal to each other:

We can find the integer roots by testing divisors of 6 ($\pm 1, \pm 2, \pm 3, \pm 6$).

Let $P(x) = x^3 - x - 6$.

$P(1) = 1^3 - 1 - 6 = 1 - 1 - 6 = -6$

$P(2) = 2^3 - 2 - 6 = 8 - 2 - 6 = 0$

Since $P(2) = 0$, $x = 2$ is a root. Thus, $(x - 2)$ is a factor of $x^3 - x - 6$.

We can perform polynomial division to find the other factors:

So, $x^3 - x - 6 = (x - 2)(x^2 + 2x + 3) = 0$.

The quadratic factor $x^2 + 2x + 3$ has a discriminant $\Delta = b^2 - 4ac = (2)^2 - 4(1)(3) = 4 - 12 = -8$. Since $\Delta < 0$, the quadratic has no real roots.

Thus, the only real intersection point of $y = x^3$ and $y = x + 6$ occurs at $x = 2$.

When $x = 2$, $y = 2^3 = 8$. The intersection point is $(2, 8)$.

The region is bounded by the curve $y = x^3$, the line $y = x + 6$, and the y-axis ($x=0$). The region extends from $x=0$ to the intersection point at $x=2$.

In the interval $(0, 2)$, we need to determine which curve is above the other. Let's pick $x = 1$:

For $y = x^3$, $y(1) = 1^3 = 1$.

For $y = x + 6$, $y(1) = 1 + 6 = 7$.

Since $7 > 1$, the line $y = x + 6$ is above the curve $y = x^3$ in the interval $[0, 2]$.

The area A of the bounded region is given by the integral of the difference between the upper curve ($y = x + 6$) and the lower curve ($y = x^3$) from $x = 0$ to $x = 2$.

Evaluate the integral:

Apply the limits of integration:

The area of the region bounded by the curves is 10 square units.

Given:

The curves are $y^2 = 4x$ and $x^2 = 4y$.

To Find:

The area of the region bounded by the given parabolas.

Solution:

The given curves are parabolas. The first parabola $y^2 = 4x$ opens to the right, and the second parabola $x^2 = 4y$ opens upwards.

To find the points of intersection, we solve the system of equations. From equation (2), we can express y in terms of x:

Substitute equation (3) into equation (1):

This equation gives the x-coordinates of the intersection points:

or

Now, find the corresponding y-coordinates using $y = \frac{x^2}{4}$:

When $x = 0$, $y = \frac{(0)^2}{4} = 0$. The intersection point is $(0, 0)$.

When $x = 4$, $y = \frac{(4)^2}{4} = \frac{16}{4} = 4$. The intersection point is $(4, 4)$.

The region bounded by the two parabolas is in the first quadrant, between $x = 0$ and $x = 4$.

To find the area, we need to determine which curve is above the other in the interval $(0, 4)$.

From $y^2 = 4x$, the upper half of the parabola is $y = \sqrt{4x} = 2\sqrt{x}$ (since $y \ge 0$ in the first quadrant).

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

Consider a value of x between $0$ and $4$, for example, $x=1$.

For $y = 2\sqrt{x}$: $y(1) = 2\sqrt{1} = 2$.

For $y = \frac{x^2}{4}$: $y(1) = \frac{1^2}{4} = \frac{1}{4}$.

Since $2 > \frac{1}{4}$, the curve $y = 2\sqrt{x}$ is above $y = \frac{x^2}{4}$ in the interval $(0, 4)$.

The area A of the bounded region is given by the definite integral of the difference between the upper curve and the lower curve from $x = 0$ to $x = 4$.

Evaluate the integral:

Apply the limits of integration:

Simplify the second term: $\frac{64}{12} = \frac{16 \times 4}{3 \times 4} = \frac{16}{3}$.

The area of the region bounded by the parabolas is $\frac{16}{3}$ square units.

Given:

The curves are $y^2 = 9x$ and $y = x$.

To Find:

The area of the region included between the given curves.

Solution:

The given curves are:

To find the points of intersection of the curves, substitute equation (2) into equation (1):

This gives two possible values for x:

or

Now, find the corresponding y-values using $y = x$ (from equation 2):

When $x = 0$, $y = 0$. The intersection point is $(0, 0)$.

When $x = 9$, $y = 9$. The intersection point is $(9, 9)$.

The region included between the curves is bounded from $x = 0$ to $x = 9$. Both intersection points are in the first quadrant, and for $x \in (0, 9)$, the y-values given by $y=x$ and $y^2=9x$ (where $y=\sqrt{9x}=3\sqrt{x}$ for $y \ge 0$) are positive. We consider the part of the parabola $y^2=9x$ where $y \ge 0$, which is $y = 3\sqrt{x}$.

To determine which curve is above the other in the interval $(0, 9)$, let's pick a value, say $x = 4$:

For the parabola $y = 3\sqrt{x}$: $y(4) = 3\sqrt{4} = 3 \times 2 = 6$.

For the line $y = x$: $y(4) = 4$.

Since $6 > 4$, the curve $y = 3\sqrt{x}$ is above the line $y = x$ in the interval $(0, 9)$.

The area A of the region bounded by the curves is given by the definite integral of the difference between the upper curve ($y = 3\sqrt{x}$) and the lower curve ($y = x$) from $x = 0$ to $x = 9$.

Evaluate the integral (3):

Apply the limits of integration from 0 to 9:

The area of the region included between the curves is $\frac{27}{2}$ square units.

Given:

The curves are the parabola $x^2 = y$ and the line $y = x + 2$.

To Find:

The area of the region enclosed by the given curves.

Solution:

The given curves are:

To find the points of intersection of the curves, set the expressions for y equal to each other:

Rearrange the equation to form a quadratic equation:

Factor the quadratic equation:

This gives the x-coordinates of the intersection points:

or

The region enclosed by the curves is bounded between these two x-values, i.e., from $x = -1$ to $x = 2$.

To determine which curve is above the other in the interval $[-1, 2]$, let's pick a test point, say $x = 0$:

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

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

Since $2 > 0$, the line $y = x + 2$ is above the parabola $y = x^2$ for $x \in (-1, 2)$.

The area A of the enclosed region is given by the definite integral of the difference between the upper curve ($y = x + 2$) and the lower curve ($y = x^2$) from $x = -1$ to $x = 2$.

Evaluate the integral:

Apply the limits of integration:

Simplify the fraction:

The area of the region enclosed by the parabola and the line is $\frac{11}{2}$ square units.

Given:

The curve is $y^2 = 8x$ and the line is $x = 2$.

To Find:

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

Solution:

The given equations are:

Equation (1), $y^2 = 8x$, represents a parabola opening to the right with its vertex at the origin $(0,0)$.

Equation (2), $x = 2$, represents a vertical line parallel to the y-axis.

To find the points of intersection, substitute $x=2$ into equation (1):

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

The region bounded by the parabola $y^2 = 8x$ and the line $x=2$ extends from $x=0$ (the vertex of the parabola) to $x=2$.

For a given x in the interval $[0, 2]$, the y-values on the parabola are $y = \pm \sqrt{8x} = \pm \sqrt{4 \times 2 \times x} = \pm 2\sqrt{2}\sqrt{x}$.

The upper boundary of the region is $y = 2\sqrt{2}\sqrt{x}$, and the lower boundary is $y = -2\sqrt{2}\sqrt{x}$. The region is symmetric about the x-axis.

The area A of the bounded region can be calculated by integrating the difference between the upper and lower curves with respect to x from $x=0$ to $x=2$.

Evaluate the integral:

Apply the limits of integration from 0 to 2:

The area of the region bounded by the line $x=2$ and the parabola $y^2=8x$ is $\frac{32}{3}$ square units.

Alternate Solution (Integrating with respect to y):

The parabola is $x = \frac{y^2}{8}$ and the line is $x = 2$. The intersection points are $(2, -4)$ and $(2, 4)$. The region is bounded between $y=-4$ and $y=4$.

For a given y in the interval $[-4, 4]$, the right boundary is the line $x=2$ and the left boundary is the parabola $x = \frac{y^2}{8}$.

The area A is given by:

Due to symmetry, this is equal to:

Evaluate the integral:

Both methods yield the same result.

Given:

The region defined by the equation $y = \sqrt{4−x^2}$ and the x-axis.

To Sketch and Find:

To sketch the region $\left\{ (x, y) \;:\; y = \sqrt{4−x^2} \right\}$ and x-axis, and find the area of this region using integration.

Solution:

The given equation is $y = \sqrt{4−x^2}$.

Squaring both sides, we get $y^2 = 4 - x^2$.

Rearranging the terms, we have $x^2 + y^2 = 4$.

This is the equation of a circle centered at the origin $(0, 0)$ with radius $r = \sqrt{4} = 2$.

Since $y = \sqrt{4−x^2}$, the value of $y$ must be non-negative, i.e., $y \geq 0$.

Therefore, the equation $y = \sqrt{4−x^2}$ represents the upper semi-circle of the circle $x^2 + y^2 = 4$.

The region is bounded by this upper semi-circle and the x-axis ($y=0$).

The intersection points of the semi-circle with the x-axis are found by setting $y=0$ in the equation $y = \sqrt{4−x^2}$:

$0 = \sqrt{4-x^2}$

$0 = 4 - x^2$

$x^2 = 4$

$x = \pm 2$

The intersection points are $(-2, 0)$ and $(2, 0)$.

The region to be sketched is the area enclosed by the upper semi-circle $y = \sqrt{4-x^2}$ and the x-axis from $x=-2$ to $x=2$. This region is the upper semi-circle of radius 2.

To find the area of this region using integration, we integrate the function $y = \sqrt{4−x^2}$ with respect to $x$ from the lower limit $x=-2$ to the upper limit $x=2$.

The area $A$ is given by:

$A = \int_{-2}^{2} \sqrt{4-x^2} \, dx$

This integral is of the form $\int \sqrt{a^2 - x^2} \, dx$ where $a=2$. The standard formula for this integral is $\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$.

Using this formula, we evaluate the definite integral:

$A = \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{-2}^{2}$

$A = \left[ \frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) \right]_{-2}^{2}$

Now, we evaluate the expression at the limits of integration:

$A = \left( \frac{2}{2}\sqrt{4-2^2} + 2\sin^{-1}\left(\frac{2}{2}\right) \right) - \left( \frac{-2}{2}\sqrt{4-(-2)^2} + 2\sin^{-1}\left(\frac{-2}{2}\right) \right)$

$A = \left( 1\sqrt{4-4} + 2\sin^{-1}(1) \right) - \left( -1\sqrt{4-4} + 2\sin^{-1}(-1) \right)$

$A = \left( 1\sqrt{0} + 2\sin^{-1}(1) \right) - \left( -1\sqrt{0} + 2\sin^{-1}(-1) \right)$

We know that $\sin^{-1}(1) = \frac{\pi}{2}$ and $\sin^{-1}(-1) = -\frac{\pi}{2}$.

$A = \left( 0 + 2 \times \frac{\pi}{2} \right) - \left( 0 + 2 \times \left(-\frac{\pi}{2}\right) \right)$

$A = (\pi) - (-\pi)$

$A = \pi + \pi$

$A = 2\pi$

The area of the region is $2\pi$ square units.

As a check, the region is a semi-circle of radius $r=2$. The area of a semi-circle is given by $\frac{1}{2}\pi r^2$.

Area $= \frac{1}{2}\pi (2)^2 = \frac{1}{2}\pi (4) = 2\pi$. This confirms the result obtained by integration.

The sketch of the region is the upper half of a circle centered at $(0,0)$ with radius 2, bounded by the x-axis from $x=-2$ to $x=2$. The area of this region is $2\pi$ square units.

Given:

The curve $y = 2 \sqrt{x}$ and the lines $x = 0$ and $x = 1$.

To Find:

The area under the curve $y = 2 \sqrt{x}$ included between the lines $x = 0$ and $x = 1$.

Solution:

The area under the 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, the function is $y = 2 \sqrt{x}$, and the limits of integration are from $x = 0$ to $x = 1$.

The area $A$ is given by:

$A = \int_{0}^{1} 2 \sqrt{x} \, dx$

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

$A = \int_{0}^{1} 2 x^{1/2} \, dx$

Now, we integrate the function. The integral of $x^n$ is $\frac{x^{n+1}}{n+1}$ (for $n \neq -1$).

$\int x^{1/2} \, dx = \frac{x^{1/2 + 1}}{1/2 + 1} = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2}$

So, $\int 2 x^{1/2} \, dx = 2 \times \frac{2}{3} x^{3/2} = \frac{4}{3} x^{3/2}$.

Now, we evaluate the definite integral using the limits $0$ and $1$:

$A = \left[ \frac{4}{3} x^{3/2} \right]_{0}^{1}$

$A = \frac{4}{3} (1)^{3/2} - \frac{4}{3} (0)^{3/2}$

$A = \frac{4}{3} (1) - \frac{4}{3} (0)$

$A = \frac{4}{3} - 0$

$A = \frac{4}{3}$

The area under the curve $y = 2 \sqrt{x}$ between $x=0$ and $x=1$ is $\frac{4}{3}$ square units.

The area under the curve is $\frac{4}{3}$ square units.

Given:

The region is bounded by the line $2y = 5x + 7$, the x-axis, and the lines $x = 2$ and $x = 8$.

To Find:

The area of the region bounded by the given line, the x-axis, and the specified vertical lines using integration.

Solution:

The equation of the given line is $2y = 5x + 7$.

We can express $y$ in terms of $x$:

$y = \frac{5x + 7}{2}$

$y = \frac{5}{2}x + \frac{7}{2}$

The region is bounded by this line, the x-axis ($y=0$), and the vertical lines $x = 2$ and $x = 8$.

The area $A$ 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 definite integral $\int_{a}^{b} f(x) \, dx$, provided $f(x) \geq 0$ in the interval $[a, b]$.

In this case, $f(x) = \frac{5}{2}x + \frac{7}{2}$, the lower limit of integration is $a=2$, and the upper limit is $b=8$.

Let's check if $y \geq 0$ for $x$ in the interval $[2, 8]$.

For $x=2$, $y = \frac{5}{2}(2) + \frac{7}{2} = 5 + \frac{7}{2} = \frac{10+7}{2} = \frac{17}{2}$.

For $x=8$, $y = \frac{5}{2}(8) + \frac{7}{2} = 20 + \frac{7}{2} = \frac{40+7}{2} = \frac{47}{2}$.

Since the function $y = \frac{5}{2}x + \frac{7}{2}$ is a linear function with a positive slope ($\frac{5}{2}$) and positive y-intercept ($\frac{7}{2}$), and the value of $y$ is positive at both endpoints $x=2$ and $x=8$, the function is positive throughout the interval $[2, 8]$. Therefore, the area can be calculated by the integral directly.

The area $A$ is given by:

$A = \int_{2}^{8} \left( \frac{5}{2}x + \frac{7}{2} \right) \, dx$

We can integrate term by term:

$A = \left[ \int \frac{5}{2}x \, dx + \int \frac{7}{2} \, dx \right]_{2}^{8}$

$A = \left[ \frac{5}{2} \int x \, dx + \frac{7}{2} \int 1 \, dx \right]_{2}^{8}$

$A = \left[ \frac{5}{2} \left( \frac{x^2}{2} \right) + \frac{7}{2} (x) \right]_{2}^{8}$

$A = \left[ \frac{5x^2}{4} + \frac{7x}{2} \right]_{2}^{8}$

Now, we evaluate the expression at the upper and lower limits:

$A = \left( \frac{5(8)^2}{4} + \frac{7(8)}{2} \right) - \left( \frac{5(2)^2}{4} + \frac{7(2)}{2} \right)$

$A = \left( \frac{5(64)}{4} + \frac{56}{2} \right) - \left( \frac{5(4)}{4} + \frac{14}{2} \right)$

$A = \left( \frac{320}{4} + 28 \right) - \left( \frac{20}{4} + 7 \right)$

$A = \left( 80 + 28 \right) - \left( 5 + 7 \right)$

$A = 108 - 12$

$A = 96$

The area of the region is 96 square units.

The area of the region bounded by the line $2y = 5x + 7$, x-axis, and the lines $x = 2$ and $x = 8$ is 96 square units.

Given:

The curve $y = \sqrt{x−1}$ and the interval $[1, 5]$, which defines the region between the lines $x = 1$ and $x = 5$ and the x-axis.

To Sketch and Find:

To draw a rough sketch of the curve $y = \sqrt{x−1}$ in the interval $[1, 5]$ and find the area under the curve between the lines $x = 1$ and $x = 5$ using integration.

Solution:

The given equation of the curve is $y = \sqrt{x−1}$.

The function is defined for $x-1 \geq 0$, which means $x \geq 1$. The given interval is $[1, 5]$, which falls within the domain of the function.

To sketch the curve, let's consider the relationship $y^2 = x-1$, which is $x = y^2 + 1$. This is the equation of a parabola opening to the right with its vertex at $(1, 0)$.

Since $y = \sqrt{x-1}$, $y$ must be greater than or equal to 0. Thus, the graph of $y = \sqrt{x-1}$ is the upper half of the parabola $x = y^2 + 1$, starting from the vertex $(1, 0)$.

For the interval $[1, 5]$:

When $x = 1$, $y = \sqrt{1-1} = \sqrt{0} = 0$. The point is $(1, 0)$.

When $x = 5$, $y = \sqrt{5-1} = \sqrt{4} = 2$. The point is $(5, 2)$.

The rough sketch is the segment of the upper half of the parabola $y = \sqrt{x-1}$ starting from $(1,0)$ and ending at $(5,2)$. The region whose area is to be found is bounded by this curve, the x-axis ($y=0$), and the vertical lines $x=1$ and $x=5$. Since $y = \sqrt{x-1} \geq 0$ for $x \geq 1$, the curve lies above the x-axis in the given interval.

The area $A$ under the curve $y = f(x)$ from $x=a$ to $x=b$ is given by the integral $\int_{a}^{b} f(x) \, dx$.

Here, $f(x) = \sqrt{x-1}$, the lower limit is $a=1$, and the upper limit is $b=5$.

The area is:

$A = \int_{1}^{5} \sqrt{x-1} \, dx$

To evaluate this integral, we can use a substitution. Let $u = x-1$.

Then, $du = dx$.

When $x = 1$, $u = 1 - 1 = 0$.

When $x = 5$, $u = 5 - 1 = 4$.

The integral becomes:

$A = \int_{0}^{4} \sqrt{u} \, du$

$A = \int_{0}^{4} u^{1/2} \, du$

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

$\int u^{1/2} \, du = \frac{u^{1/2+1}}{1/2+1} = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}$

Evaluate the definite integral with the limits from 0 to 4:

$A = \left[ \frac{2}{3} u^{3/2} \right]_{0}^{4}$

$A = \frac{2}{3} (4)^{3/2} - \frac{2}{3} (0)^{3/2}$

$A = \frac{2}{3} (\sqrt{4})^3 - \frac{2}{3} (0)$

$A = \frac{2}{3} (2)^3 - 0$

$A = \frac{2}{3} \times 8$

$A = \frac{16}{3}$

The area under the curve $y = \sqrt{x-1}$ between $x=1$ and $x=5$ is $\frac{16}{3}$ square units.

The area under the curve is $\frac{16}{3}$ square units.

Given:

The curve $y = \sqrt{a^2−x^2}$ and the lines $x = 0$ and $x = a$.

To Determine:

The area under the curve $y = \sqrt{a^2−x^2}$ included between the lines $x = 0$ and $x = a$ using integration.

Solution:

The given equation is $y = \sqrt{a^2−x^2}$.

Squaring both sides, we get $y^2 = a^2 - x^2$.

Rearranging the terms, we have $x^2 + y^2 = a^2$.

This is the equation of a circle centered at the origin $(0, 0)$ with radius $a$.

Since $y = \sqrt{a^2−x^2}$, the value of $y$ must be non-negative, i.e., $y \geq 0$.

Therefore, the equation $y = \sqrt{a^2−x^2}$ represents the upper semi-circle of the circle $x^2 + y^2 = a^2$.

The region is bounded by this upper semi-circle, the x-axis ($y=0$), and the vertical lines $x = 0$ and $x = a$.

Since $a > 0$, the lines $x=0$ and $x=a$ correspond to the y-axis and a vertical line in the first quadrant (or on the boundary). The interval $[0, a]$ lies within the domain of the function $\sqrt{a^2-x^2}$ (which requires $a^2-x^2 \geq 0$, or $-a \leq x \leq a$).

The region whose area is to be found is the part of the upper semi-circle in the first quadrant, bounded by $x=0$ and $x=a$. Since $y = \sqrt{a^2-x^2} \geq 0$ for $x \in [0, a]$, the curve lies above the x-axis in this interval.

The area $A$ under the 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, $f(x) = \sqrt{a^2-x^2}$, the lower limit of integration is $x = 0$, and the upper limit is $x = a$.

The area $A$ is given by:

$A = \int_{0}^{a} \sqrt{a^2-x^2} \, dx$

This is a standard integral. The formula is:

$\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$

Using the definite integral, we evaluate the expression at the limits:

$A = \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \right]_{0}^{a}$

Evaluate at the upper limit ($x=a$):

$\left( \frac{a}{2}\sqrt{a^2 - a^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{a}{a}\right) \right) = \left( \frac{a}{2}\sqrt{0} + \frac{a^2}{2}\sin^{-1}(1) \right) = \left( 0 + \frac{a^2}{2} \times \frac{\pi}{2} \right) = \frac{\pi a^2}{4}$

Evaluate at the lower limit ($x=0$):

$\left( \frac{0}{2}\sqrt{a^2 - 0^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{0}{a}\right) \right) = \left( 0\sqrt{a^2} + \frac{a^2}{2}\sin^{-1}(0) \right) = \left( 0 + \frac{a^2}{2} \times 0 \right) = 0$

Subtract the lower limit value from the upper limit value:

$A = \frac{\pi a^2}{4} - 0$

$A = \frac{\pi a^2}{4}$

The area of the region is $\frac{\pi a^2}{4}$ square units.

This result corresponds to the area of a quarter circle of radius $a$, which is expected as the region is the part of the circle $x^2 + y^2 = a^2$ in the first quadrant ($x \ge 0$, $y \ge 0$).

The area under the curve $y = \sqrt{a^2−x^2}$ included between the lines $x = 0$ and $x = a$ is $\frac{\pi a^2}{4}$ square units.

Given:

The region is bounded by the curves $y = \sqrt{x}$ and $y = x$.

To Find:

The area of the region bounded by the curves $y = \sqrt{x}$ and $y = x$ using integration.

Solution:

To find the area of the region bounded by the two curves, we first need to find their points of intersection.

Set the equations equal to each other:

$ \sqrt{x} = x $

Square both sides:

$ (\sqrt{x})^2 = x^2 $

$ x = x^2 $

Rearrange the equation:

$ x^2 - x = 0 $

Factor out $x$:

$ x(x - 1) = 0 $

This gives the intersection points at $x = 0$ and $x = 1$.

The corresponding y-values are:

For $x = 0$, $y = \sqrt{0} = 0$ (or $y = 0$). Point is $(0, 0)$.

For $x = 1$, $y = \sqrt{1} = 1$ (or $y = 1$). Point is $(1, 1)$.

The region is bounded between $x = 0$ and $x = 1$.

Next, we need to determine which curve is above the other in the interval $(0, 1)$. Let's pick a test point, for example, $x = 0.5$.

For $y = \sqrt{x}$, $y = \sqrt{0.5} = \frac{1}{\sqrt{2}} \approx 0.707$

For $y = x$, $y = 0.5$

Since $0.707 > 0.5$, the curve $y = \sqrt{x}$ is above the curve $y = x$ in the interval $(0, 1)$.

The area $A$ of the region between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x=b$, where $f(x) \geq g(x)$ on $[a, b]$, is given by $\int_{a}^{b} (f(x) - g(x)) \, dx$.

Here, $f(x) = \sqrt{x}$, $g(x) = x$, $a = 0$, and $b = 1$.

The area is:

$ A = \int_{0}^{1} (\sqrt{x} - x) \, dx $

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

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

Integrate term by term:

$ A = \left[ \frac{x^{1/2+1}}{1/2+1} - \frac{x^{1+1}}{1+1} \right]_{0}^{1} $

$ A = \left[ \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_{0}^{1} $

$ A = \left[ \frac{2}{3} x^{3/2} - \frac{1}{2} x^2 \right]_{0}^{1} $

Evaluate the expression at the upper and lower limits:

$ A = \left( \frac{2}{3} (1)^{3/2} - \frac{1}{2} (1)^2 \right) - \left( \frac{2}{3} (0)^{3/2} - \frac{1}{2} (0)^2 \right) $

$ A = \left( \frac{2}{3} (1) - \frac{1}{2} (1) \right) - (0 - 0) $

$ A = \left( \frac{2}{3} - \frac{1}{2} \right) - 0 $

Find a common denominator for the fractions:

$ A = \frac{4}{6} - \frac{3}{6} $

$ A = \frac{4 - 3}{6} $

$ A = \frac{1}{6} $

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

The area of the region is $\frac{1}{6}$ square units.

Given:

The region is bounded by the curve $y = -x^2$ and the straight line $x + y + 2 = 0$.

To Find:

The area enclosed by the curve $y = -x^2$ and the straight line $x + y + 2 = 0$ using integration.

Solution:

We have the equations of the two curves:

Curve 1: $y = -x^2$

Curve 2: $x + y + 2 = 0$

From Curve 2, we can express $y$ in terms of $x$:

$y = -x - 2$

To find the points of intersection of the two curves, we set their y-values equal:

$-x^2 = -x - 2$

Rearrange the equation to form a quadratic equation:

$x^2 - x - 2 = 0$

Factor the quadratic equation:

$(x - 2)(x + 1) = 0$

This gives the x-coordinates of the intersection points:

$x - 2 = 0 \implies x = 2$

$x + 1 = 0 \implies x = -1$

The corresponding y-coordinates can be found by substituting these x-values into either equation. Using $y = -x - 2$:

For $x = -1$: $y = -(-1) - 2 = 1 - 2 = -1$. Intersection point: $(-1, -1)$.

For $x = 2$: $y = -(2) - 2 = -2 - 2 = -4$. Intersection point: $(2, -4)$.

The region whose area is to be found is bounded by these two curves between $x = -1$ and $x = 2$.

To find the area between the curves, we need to determine which curve is above the other in the interval $(-1, 2)$. The curves are $y_{parabola} = -x^2$ (a downward-opening parabola) and $y_{line} = -x - 2$ (a straight line).

Let's pick a test point in the interval, say $x=0$.

For the parabola: $y_{parabola}(0) = -(0)^2 = 0$.

For the line: $y_{line}(0) = -(0) - 2 = -2$.

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

The area $A$ enclosed by the curves is given by the integral of the difference between the upper curve and the lower curve from the lower x-limit to the upper x-limit:

$A = \int_{-1}^{2} (y_{upper} - y_{lower}) \, dx$

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

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

Now, we evaluate the definite integral:

$A = \left[ -\frac{x^{2+1}}{2+1} + \frac{x^{1+1}}{1+1} + 2x \right]_{-1}^{2}$

$A = \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^{2}$

Evaluate the expression at the upper limit ($x=2$) and the lower limit ($x=-1$):

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

$A = \left( -\frac{8}{3} + \frac{4}{2} + 4 \right) - \left( -\frac{-1}{3} + \frac{1}{2} - 2 \right)$

$A = \left( -\frac{8}{3} + 2 + 4 \right) - \left( \frac{1}{3} + \frac{1}{2} - 2 \right)$

$A = \left( -\frac{8}{3} + 6 \right) - \left( \frac{1}{3} + \frac{1}{2} - 2 \right)$

$A = -\frac{8}{3} + 6 - \frac{1}{3} - \frac{1}{2} + 2$

$A = \left(-\frac{8}{3} - \frac{1}{3}\right) + (6 + 2) - \frac{1}{2}$

$A = -\frac{9}{3} + 8 - \frac{1}{2}$

$A = -3 + 8 - \frac{1}{2}$

$A = 5 - \frac{1}{2}$

$A = \frac{10}{2} - \frac{1}{2}$

$A = \frac{9}{2}$

The area enclosed by the curve $y = -x^2$ and the straight line $x + y + 2 = 0$ is $\frac{9}{2}$ square units.

The area of the enclosed region is $\frac{9}{2}$ square units.

Given:

The region is bounded by the curve $y = \sqrt{x}$, the curve $x = 2y + 3$, and the x-axis ($y=0$), in the first quadrant.

To Find:

The area of the region bounded by the given curves and the x-axis in the first quadrant using integration.

Solution:

We are given the curves:

and the boundary $y = 0$ (x-axis).

First, let's find the points of intersection of these curves.

Intersection of curve (1) and $y = 0$:

Set $y = 0$ in equation (1): $0 = \sqrt{x} \implies x = 0$. The point is $(0, 0)$.

Intersection of curve (2) and $y = 0$:

Set $y = 0$ in equation (2): $x = 2(0) + 3 \implies x = 3$. The point is $(3, 0)$.

Intersection of curve (1) and curve (2):

Substitute $y = \sqrt{x}$ from equation (1) into equation (2):

$x = 2\sqrt{x} + 3$

$x - 3 = 2\sqrt{x}$

Square both sides:

$(x - 3)^2 = (2\sqrt{x})^2$

$x^2 - 6x + 9 = 4x$

Factor the quadratic equation (3):

$(x - 1)(x - 9) = 0$

Possible x-values are $x = 1$ and $x = 9$.

We must check these solutions in the equation $x - 3 = 2\sqrt{x}$ (obtained before squaring) as squaring can introduce extraneous roots:

For $x = 1$: $1 - 3 = 2\sqrt{1} \implies -2 = 2$. This is false. Thus, $x=1$ is an extraneous solution.

For $x = 9$: $9 - 3 = 2\sqrt{9} \implies 6 = 2(3) \implies 6 = 6$. This is true. Thus, $x=9$ is a valid solution.

So, the only valid x-coordinate for the intersection in the first quadrant is $x = 9$.

The corresponding y-coordinate can be found using equation (1): $y = \sqrt{9} = 3$. The point is $(9, 3)$.

The intersection points relevant to the region in the first quadrant bounded by the three curves are $(0, 0)$, $(3, 0)$, and $(9, 3)$.

The region is bounded by $y=0$ below, and by the curves $y=\sqrt{x}$ (or $x=y^2$) and $x=2y+3$.

Observing the sketch of these curves and the region in the first quadrant, the region is best described by integrating with respect to $y$. The relevant range of y is from $0$ to $3$ (the y-coordinate of the intersection point $(9,3)$).

For a given $y$ in the range $[0, 3]$, the region extends horizontally from the curve $x = y^2$ on the left to the line $x = 2y + 3$ on the right.

The area $A$ of the region can be calculated as the integral of the difference between the right boundary ($x_{right} = 2y+3$) and the left boundary ($x_{left} = y^2$) with respect to $y$, from $y=0$ to $y=3$.

Now, we evaluate the definite integral (4):

$A = \int_{0}^{3} (2y + 3 - y^2) \, dy$

Integrate term by term:

$A = \left[ \frac{2y^{2}}{2} + 3y - \frac{y^{3}}{3} \right]_{0}^{3}$

$A = \left[ y^2 + 3y - \frac{y^3}{3} \right]_{0}^{3}$

Evaluate the expression at the upper limit ($y=3$) and the lower limit ($y=0$):

$A = \left( (3)^2 + 3(3) - \frac{(3)^3}{3} \right) - \left( (0)^2 + 3(0) - \frac{(0)^3}{3} \right)$

$A = \left( 9 + 9 - \frac{27}{3} \right) - (0 + 0 - 0)$

$A = \left( 18 - 9 \right) - 0$

$A = 9$

The area of the region bounded by the curve $y = \sqrt{x}$, the line $x = 2y + 3$, and the x-axis in the first quadrant is 9 square units.

The area of the region is 9 square units.

Given:

The region is bounded by the curves $y^2 = 2x$ and $x^2 + y^2 = 4x$.

To Find:

The area of the region bounded by the curve $y^2 = 2x$ and $x^2 + y^2 = 4x$ using integration.

Solution:

We are given two curves:

Equation (1), $y^2 = 2x$, represents a parabola opening to the right with its vertex at the origin $(0,0)$.

Equation (2), $x^2 + y^2 = 4x$, can be rewritten by completing the square for the x terms:

$x^2 - 4x + y^2 = 0$

$(x^2 - 4x + 4) - 4 + y^2 = 0$

Equation (3) represents a circle centered at $(2, 0)$ with a radius of $r = \sqrt{4} = 2$.

To find the area of the region bounded by these two curves, we first find their points of intersection by solving the equations simultaneously.

Substitute $y^2 = 2x$ from equation (1) into equation (2):

$x^2 + 2x = 4x$

$x^2 - 2x = 0$

Factor out $x$:

$x(x - 2) = 0$

This gives the x-coordinates of the intersection points: $x = 0$ and $x = 2$.

Now, find the corresponding y-coordinates using equation (1), $y^2 = 2x$:

For $x = 0$: $y^2 = 2(0) = 0 \implies y = 0$. The intersection point is $(0, 0)$.

For $x = 2$: $y^2 = 2(2) = 4 \implies y = \pm 2$. The intersection points are $(2, 2)$ and $(2, -2)$.

The two curves intersect at three points: $(0, 0)$, $(2, 2)$, and $(2, -2)$. These points define the boundaries of the enclosed region.

To find the area of the enclosed region, we can integrate with respect to $y$. The range of $y$ values for the enclosed region is from $y = -2$ to $y = 2$.

From equation (1), the parabola is $x = \frac{y^2}{2}$. This curve forms the left boundary of the enclosed region for $y$ from $-2$ to $2$ (except at $y=0$).

From equation (3), the circle is $(x - 2)^2 + y^2 = 4$. Solving for $x$, we get $x - 2 = \pm \sqrt{4 - y^2}$, so $x = 2 \pm \sqrt{4 - y^2}$. The left half of the circle, $x = 2 - \sqrt{4 - y^2}$, forms the right boundary of the enclosed region between the intersection points $(2, \pm 2)$ and $(0, 0)$.

For a given $y$ in the interval $[-2, 2]$, the right boundary is $x_R = \frac{y^2}{2}$ (from the parabola) and the left boundary is $x_L = 2 - \sqrt{4 - y^2}$ (from the circle). Let's verify this by picking a point, e.g., $y=1$. $x_{parabola}(1) = 1^2/2 = 0.5$. $x_{circle\_left}(1) = 2 - \sqrt{4-1^2} = 2 - \sqrt{3} \approx 0.268$. Since $0.5 > 0.268$, the parabola is to the right of the left half of the circle in the region between $y=-2$ and $y=2$. Thus, $x_R = \frac{y^2}{2}$ and $x_L = 2 - \sqrt{4 - y^2}$.

The area $A$ of the region enclosed by the two curves is given by the integral of the difference between the right boundary curve and the left boundary curve with respect to $y$, from $y = -2$ to $y = 2$:

Substitute the expressions for $x_R$ and $x_L$ into the integral (4):

$A = \int_{-2}^{2} \left( \frac{y^2}{2} - 2 + \sqrt{4 - y^2} \right) \, dy$

We can split this into three separate integrals:

$A = \int_{-2}^{2} \frac{y^2}{2} \, dy - \int_{-2}^{2} 2 \, dy + \int_{-2}^{2} \sqrt{4 - y^2} \, dy$

Let's evaluate each integral:

1. $\int_{-2}^{2} \frac{y^2}{2} \, dy = \frac{1}{2} \int_{-2}^{2} y^2 \, dy$

Since $y^2$ is an even function, $\int_{-2}^{2} y^2 \, dy = 2 \int_{0}^{2} y^2 \, dy$.

$= \frac{1}{2} \left[ \frac{y^3}{3} \right]_{-2}^{2} = \frac{1}{6} [ (2)^3 - (-2)^3 ] = \frac{1}{6} [ 8 - (-8) ] = \frac{1}{6} [16] = \frac{16}{6} = \frac{8}{3}$

2. $\int_{-2}^{2} 2 \, dy = [2y]_{-2}^{2} = 2(2) - 2(-2) = 4 - (-4) = 8$

3. $\int_{-2}^{2} \sqrt{4 - y^2} \, dy$. This integral represents the area of a semi-circle of radius $a=2$ bounded by the y-axis (where $x = \sqrt{a^2 - y^2}$). The integration limits $y=-2$ to $y=2$ cover the full range of y for the semi-circle. The value of this integral is the area of the entire semi-circle with radius 2, which is $\frac{1}{2}\pi a^2 = \frac{1}{2}\pi (2)^2 = 2\pi$.

Alternatively, using the standard formula $\int \sqrt{a^2 - y^2} \, dy = \frac{y}{2}\sqrt{a^2 - y^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{y}{a}\right)$: $\int_{-2}^{2} \sqrt{4 - y^2} \, dy = \left[ \frac{y}{2}\sqrt{4 - y^2} + \frac{4}{2}\sin^{-1}\left(\frac{y}{2}\right) \right]_{-2}^{2}$ $= \left( \frac{2}{2}\sqrt{4 - 2^2} + 2\sin^{-1}\left(\frac{2}{2}\right) \right) - \left( \frac{-2}{2}\sqrt{4 - (-2)^2} + 2\sin^{-1}\left(\frac{-2}{2}\right) \right)$ $= \left( 1\sqrt{0} + 2\sin^{-1}(1) \right) - \left( -1\sqrt{0} + 2\sin^{-1}(-1) \right)$ $= \left( 0 + 2 \times \frac{\pi}{2} \right) - \left( 0 + 2 \times \left(-\frac{\pi}{2}\right) \right) = \pi - (-\pi) = 2\pi$

Now, combine the results of the three integrals:

$A = \frac{8}{3} - 8 + 2\pi$

$A = \frac{8}{3} - \frac{24}{3} + 2\pi$

$A = -\frac{16}{3} + 2\pi$

$A = 2\pi - \frac{16}{3}$

The area of the region bounded by the curve $y^2 = 2x$ and $x^2 + y^2 = 4x$ is $2\pi - \frac{16}{3}$ square units.

The area of the enclosed region is $2\pi - \frac{16}{3}$ square units.

Given:

The curve $y = \sin x$ and the interval from $x = 0$ to $x = 2\pi$ along the x-axis.

To Find:

The area bounded by the curve $y = \sin x$ and the x-axis between $x = 0$ and $x = 2\pi$ using integration.

Solution:

The area bounded by a curve $y = f(x)$ and the x-axis between $x=a$ and $x=b$ is given by the absolute value of the definite integral $\int_{a}^{b} f(x) \, dx$. However, if the curve crosses the x-axis within the interval $[a, b]$, we need to calculate the area of each subregion separately where the function has a constant sign and then sum the absolute values of these areas.

For the curve $y = \sin x$ in the interval $[0, 2\pi]$, the sine function is positive in $[0, \pi]$ and negative in $[\pi, 2\pi]$. The curve crosses the x-axis at $x = 0$, $x = \pi$, and $x = 2\pi$.

Therefore, the total area is the sum of the area in $[0, \pi]$ and the absolute value of the area in $[\pi, 2\pi]$.

Area from $x=0$ to $x=\pi$: $A_1 = \int_{0}^{\pi} \sin x \, dx$. Since $\sin x \geq 0$ in this interval, the area is the value of the integral.

Area from $x=\pi$ to $x=2\pi$: $A_2 = \int_{\pi}^{2\pi} \sin x \, dx$. Since $\sin x \leq 0$ in this interval, the area is the absolute value of the integral, $|A_2|$.

The total area $A$ is $A_1 + |A_2|$.

First, let's evaluate the indefinite integral of $\sin x$:

$\int \sin x \, dx = -\cos x + C$

Now, evaluate the definite integrals:

For $A_1$:

$A_1 = \int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi}$

$A_1 = (-\cos \pi) - (-\cos 0)$

$A_1 = (-(-1)) - (-1)$

$A_1 = 1 + 1$

For $A_2$:

$A_2 = \int_{\pi}^{2\pi} \sin x \, dx = [-\cos x]_{\pi}^{2\pi}$

$A_2 = (-\cos 2\pi) - (-\cos \pi)$

$A_2 = (-1) - (-(-1))$

$A_2 = -1 - 1$

The area from $\pi$ to $2\pi$ is the absolute value of $A_2$:

$|A_2| = |-2| = 2$

The total area $A$ is the sum of the areas of the two regions:

$A = A_1 + |A_2|$

$A = 2 + 2$

$A = 4$

The area bounded by the curve $y = \sin x$ and the x-axis between $x = 0$ and $x = 2\pi$ is 4 square units.

The area of the bounded region is 4 square units.

Given:

The vertices of the triangle are A(–1, 1), B(0, 5), and C(3, 2).

To Find:

The area of the region bounded by the triangle with the given vertices using integration.

Solution:

We first find the equations of the lines forming the sides of the triangle.

Equation of line AB (passing through A(–1, 1) and B(0, 5)):

Slope $m_{AB} = \frac{5 - 1}{0 - (-1)} = \frac{4}{1} = 4$.

Using the point-slope form with point B(0, 5):

$y - 5 = 4(x - 0)$

Equation of line BC (passing through B(0, 5) and C(3, 2)):

Slope $m_{BC} = \frac{2 - 5}{3 - 0} = \frac{-3}{3} = -1$.

Using the point-slope form with point B(0, 5):

$y - 5 = -1(x - 0)$

Equation of line AC (passing through A(–1, 1) and C(3, 2)):

Slope $m_{AC} = \frac{2 - 1}{3 - (-1)} = \frac{1}{4}$.

Using the point-slope form with point A(–1, 1):

$y - 1 = \frac{1}{4}(x - (-1))$

$y - 1 = \frac{1}{4}(x + 1)$

$4(y - 1) = x + 1$

$4y - 4 = x + 1$

$4y = x + 5$

To find the area of the triangle using integration with respect to x, we can consider the area under the upper boundary curves (AB from x=-1 to x=0 and BC from x=0 to x=3) and subtract the area under the lower boundary curve (AC from x=-1 to x=3).

The x-coordinates of the vertices are -1, 0, and 3. These will be our integration limits.

Area under AB from x = -1 to x = 0:

$A_{AB} = \int_{-1}^{0} (4x + 5) \, dx$

$A_{AB} = \left[ 2x^2 + 5x \right]_{-1}^{0}$

$A_{AB} = (2(0)^2 + 5(0)) - (2(-1)^2 + 5(-1))$

$A_{AB} = (0 + 0) - (2(1) - 5)$

$A_{AB} = 0 - (2 - 5) = -(-3) = 3$

Area under BC from x = 0 to x = 3:

$A_{BC} = \int_{0}^{3} (-x + 5) \, dx$

$A_{BC} = \left[ -\frac{x^2}{2} + 5x \right]_{0}^{3}$

$A_{BC} = \left( -\frac{(3)^2}{2} + 5(3) \right) - \left( -\frac{(0)^2}{2} + 5(0) \right)$

$A_{BC} = \left( -\frac{9}{2} + 15 \right) - (0 + 0)$

$A_{BC} = -\frac{9}{2} + \frac{30}{2} = \frac{21}{2}$

Area under AC from x = -1 to x = 3:

$A_{AC} = \int_{-1}^{3} \left( \frac{1}{4}x + \frac{5}{4} \right) \, dx$

$A_{AC} = \left[ \frac{1}{4} \frac{x^2}{2} + \frac{5}{4} x \right]_{-1}^{3}$

$A_{AC} = \left[ \frac{x^2}{8} + \frac{5x}{4} \right]_{-1}^{3}$

$A_{AC} = \left( \frac{(3)^2}{8} + \frac{5(3)}{4} \right) - \left( \frac{(-1)^2}{8} + \frac{5(-1)}{4} \right)$

$A_{AC} = \left( \frac{9}{8} + \frac{15}{4} \right) - \left( \frac{1}{8} - \frac{5}{4} \right)$

$A_{AC} = \left( \frac{9}{8} + \frac{30}{8} \right) - \left( \frac{1}{8} - \frac{10}{8} \right)$

$A_{AC} = \frac{39}{8} - \left( -\frac{9}{8} \right) = \frac{39}{8} + \frac{9}{8} = \frac{48}{8} = 6$

The area of the triangle is the sum of the areas under the upper boundaries minus the area under the lower boundary:

$A = A_{AB} + A_{BC} - A_{AC}$

$A = 3 + \frac{21}{2} - 6$

$A = (3 - 6) + \frac{21}{2}$

$A = -3 + \frac{21}{2}$

$A = -\frac{6}{2} + \frac{21}{2}$

$A = \frac{21 - 6}{2} = \frac{15}{2}$

The area of the triangle is $\frac{15}{2}$ square units.

The area of the region bounded by the triangle is $\frac{15}{2}$ square units.

Given:

The region defined by the inequalities $y^2 \leq 6ax$ and $x^2 + y^2 \leq 16a^2$. We assume $a > 0$ for the problem to describe a bounded region in the first and fourth quadrants.

To Sketch and Find:

To draw a rough sketch of the region $\left\{ (x, y) \;:\; y^2 ≤ 6ax \;and\; x^2 + y^2 ≤ 16a^2 \right\}$ and find its area using integration.

Solution:

The given inequalities are:

$y^2 \leq 6ax$

$x^2 + y^2 \leq 16a^2$

The boundary curves are the parabola $y^2 = 6ax$ and the circle $x^2 + y^2 = 16a^2$.

The equation $y^2 = 6ax$ (with $a>0$) represents a parabola opening to the right with vertex at $(0,0)$. The inequality $y^2 \leq 6ax$ represents the region on or to the right of the parabola, which implies $x \geq \frac{y^2}{6a}$.

The equation $x^2 + y^2 = 16a^2$ represents a circle centered at the origin $(0,0)$ with radius $4a$. The inequality $x^2 + y^2 \leq 16a^2$ represents the region on or inside the circle, which implies $x^2 \leq 16a^2 - y^2$, or $-\sqrt{16a^2 - y^2} \leq x \leq \sqrt{16a^2 - y^2}$.

The required region is the intersection of these two regions, i.e., the set of points $(x,y)$ satisfying both inequalities.

To find the points of intersection of the boundary curves, we solve the equations simultaneously:

Substitute equation (i) into equation (ii):

$x^2 + 6ax = 16a^2$

$x^2 + 6ax - 16a^2 = 0$

Factoring the quadratic equation:

$(x + 8a)(x - 2a) = 0$

The possible x-coordinates of intersection are $x = -8a$ and $x = 2a$.

For the parabola $y^2 = 6ax$ with $a>0$, we must have $x \geq 0$. Thus, the intersection point with $x = -8a$ is not relevant as it does not lie on the parabola $y^2=6ax$. The relevant intersection x-coordinate is $x=2a$.

Substitute $x = 2a$ into equation (i) to find the corresponding y-coordinates:

$y^2 = 6a(2a) = 12a^2$

$y = \pm \sqrt{12a^2} = \pm 2a\sqrt{3}$

The intersection points are $(0,0)$, $(2a, 2a\sqrt{3})$, and $(2a, -2a\sqrt{3})$. The point $(0,0)$ is obtained from $x=0$ in $y^2=6ax$ and $x^2+y^2=16a^2$.

Rough Sketch:

The sketch involves drawing the parabola $y^2=6ax$ (opening right from the origin) and the circle $x^2+y^2=(4a)^2$ (centered at the origin with radius $4a$). The two curves intersect at the origin and at $(2a, \pm 2a\sqrt{3})$. The region $y^2 \leq 6ax$ is the area to the right of the parabola $x = y^2/(6a)$. The region $x^2 + y^2 \leq 16a^2$ is the area inside the circle $x = \sqrt{16a^2 - y^2}$ and $x = -\sqrt{16a^2 - y^2}$. The intersection region is the part of the circle that is to the right of the parabola. This region is bounded on the left by the parabola $x = \frac{y^2}{6a}$ and on the right by the right half of the circle $x = \sqrt{16a^2 - y^2}$. The region extends vertically from the lowest intersection point $y = -2a\sqrt{3}$ to the highest intersection point $y = 2a\sqrt{3}$.

Area Calculation using Integration:

We calculate the area by integrating with respect to $y$. For a given $y$ between $-2a\sqrt{3}$ and $2a\sqrt{3}$, $x$ ranges from the parabola $x_L = \frac{y^2}{6a}$ to the circle $x_R = \sqrt{16a^2 - y^2}$.

The area $A$ is given by:

$A = \int_{-2a\sqrt{3}}^{2a\sqrt{3}} (x_R - x_L) \, dy$

$A = \int_{-2a\sqrt{3}}^{2a\sqrt{3}} \left( \sqrt{16a^2 - y^2} - \frac{y^2}{6a} \right) \, dy$

Due to the symmetry of the region about the x-axis, we can integrate from $0$ to $2a\sqrt{3}$ and multiply the result by 2:

$A = 2 \int_{0}^{2a\sqrt{3}} \sqrt{16a^2 - y^2} \, dy - 2 \int_{0}^{2a\sqrt{3}} \frac{y^2}{6a} \, dy$

We evaluate the two integrals separately.

Integral 1: $2 \int_{0}^{2a\sqrt{3}} \sqrt{16a^2 - y^2} \, dy$. This is of the form $\int \sqrt{A^2 - y^2} \, dy$ where $A=4a$. The standard integral is $\frac{y}{2}\sqrt{A^2 - y^2} + \frac{A^2}{2}\sin^{-1}\left(\frac{y}{A}\right)$.

$2 \left[ \frac{y}{2}\sqrt{(4a)^2 - y^2} + \frac{(4a)^2}{2}\sin^{-1}\left(\frac{y}{4a}\right) \right]_{0}^{2a\sqrt{3}}$

$= 2 \left[ \frac{y}{2}\sqrt{16a^2 - y^2} + 8a^2\sin^{-1}\left(\frac{y}{4a}\right) \right]_{0}^{2a\sqrt{3}}$

Evaluate at the upper limit ($y = 2a\sqrt{3}$):

$2 \left( \frac{2a\sqrt{3}}{2}\sqrt{16a^2 - (2a\sqrt{3})^2} + 8a^2\sin^{-1}\left(\frac{2a\sqrt{3}}{4a}\right) \right)$

$= 2 \left( a\sqrt{3}\sqrt{16a^2 - 12a^2} + 8a^2\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) \right)$

$= 2 \left( a\sqrt{3}\sqrt{4a^2} + 8a^2 \left(\frac{\pi}{3}\right) \right)$

$= 2 \left( a\sqrt{3}(2a) + \frac{8\pi a^2}{3} \right)$

$= 2 \left( 2a^2\sqrt{3} + \frac{8\pi a^2}{3} \right) = 4a^2\sqrt{3} + \frac{16\pi a^2}{3}$

Evaluate at the lower limit ($y = 0$):

$2 \left( \frac{0}{2}\sqrt{16a^2 - 0^2} + 8a^2\sin^{-1}\left(\frac{0}{4a}\right) \right) = 2 (0 + 0) = 0$

So, the value of the first integral term is $4a^2\sqrt{3} + \frac{16\pi a^2}{3}$.

Integral 2: $-2 \int_{0}^{2a\sqrt{3}} \frac{y^2}{6a} \, dy$

$= -\frac{2}{6a} \int_{0}^{2a\sqrt{3}} y^2 \, dy = -\frac{1}{3a} \left[ \frac{y^3}{3} \right]_{0}^{2a\sqrt{3}}$

$= -\frac{1}{9a} [ (2a\sqrt{3})^3 - (0)^3 ]$

$= -\frac{1}{9a} [ (2^3 a^3 (\sqrt{3})^3) ] = -\frac{1}{9a} [ 8a^3 \cdot 3\sqrt{3} ]$

$= -\frac{24a^3\sqrt{3}}{9a} = -\frac{8a^2\sqrt{3}}{3}$

Add the results of the two integral terms to get the total area $A$:

$A = \left( 4a^2\sqrt{3} + \frac{16\pi a^2}{3} \right) + \left( -\frac{8a^2\sqrt{3}}{3} \right)$

$A = 4a^2\sqrt{3} - \frac{8a^2\sqrt{3}}{3} + \frac{16\pi a^2}{3}$

$A = \frac{12a^2\sqrt{3} - 8a^2\sqrt{3}}{3} + \frac{16\pi a^2}{3}$

$A = \frac{4a^2\sqrt{3}}{3} + \frac{16\pi a^2}{3}$

$A = \frac{a^2}{3} (4\sqrt{3} + 16\pi)$

$A = \frac{4a^2}{3} (\sqrt{3} + 4\pi)$

The area of the region bounded by the inequalities $y^2 \leq 6ax$ and $x^2 + y^2 \leq 16a^2$ is $\frac{4a^2}{3} (4\pi + \sqrt{3})$ square units.

The area of the region is $\frac{4a^2}{3} (4\pi + \sqrt{3})$ square units.

Given:

The region is bounded by the lines $x + 2y = 2$, $y – x = 1$, and $2x + y = 7$.

To Compute:

The area of the region bounded by the given lines using integration.

Solution:

The given lines are:

The region bounded by these three lines is a triangle. To find its area, we first find the vertices of the triangle by finding the intersection points of the lines.

Intersection of $L_1$ and $L_2$:

From $L_2$, $y = x + 1$. Substitute this into $L_1$:

$x + 2(x + 1) = 2$

$x + 2x + 2 = 2$

$3x = 0 \implies x = 0$

Substitute $x=0$ into $y = x+1$: $y = 0 + 1 = 1$.

Vertex A: $(0, 1)$.

Intersection of $L_1$ and $L_3$:

From $L_3$, $y = 7 - 2x$. Substitute this into $L_1$:

$x + 2(7 - 2x) = 2$

$x + 14 - 4x = 2$

$-3x = -12 \implies x = 4$

Substitute $x=4$ into $y = 7 - 2x$: $y = 7 - 2(4) = 7 - 8 = -1$.

Vertex B: $(4, -1)$.

Intersection of $L_2$ and $L_3$:

From $L_2$, $y = x + 1$. Substitute this into $L_3$:

$2x + (x + 1) = 7$

$3x + 1 = 7$

$3x = 6 \implies x = 2$

Substitute $x=2$ into $y = x + 1$: $y = 2 + 1 = 3$.

Vertex C: $(2, 3)$.

The vertices of the triangle are A(0, 1), B(4, -1), and C(2, 3).

To find the area using integration with respect to $x$, we need to identify the upper and lower boundary lines in the relevant intervals of $x$. The x-coordinates of the vertices are 0, 2, and 4. We split the integral at $x=2$ (the x-coordinate of vertex C).

Rewrite the lines in terms of $y$:

$L_1: y = 1 - \frac{1}{2}x$

$L_2: y = x + 1$

$L_3: y = 7 - 2x$

The lower boundary of the triangle is the line connecting A(0,1) and B(4,-1), which is $L_1: y = 1 - \frac{1}{2}x$. This line is below the other two lines in the region bounded by the triangle.

The upper boundary is formed by two line segments:

- From A(0,1) to C(2,3), which is the line connecting these points: $y = x+1$. This is $L_2$. This segment is the upper boundary for $x \in [0, 2]$.

- From C(2,3) to B(4,-1), which is the line connecting these points: $y = 7-2x$. This is $L_3$. This segment is the upper boundary for $x \in [2, 4]$.

The area $A$ of the triangle is the sum of the areas between the upper boundary and the lower boundary in the intervals $[0, 2]$ and $[2, 4]$.

Substitute the expressions for $y_{L1}$, $y_{L2}$, and $y_{L3}$ into equation (1):

Simplify the integrands:

For the first integral: $(x + 1) - (1 - \frac{1}{2}x) = x + 1 - 1 + \frac{1}{2}x = \frac{3}{2}x$

For the second integral: $(7 - 2x) - (1 - \frac{1}{2}x) = 7 - 2x - 1 + \frac{1}{2}x = 6 - \frac{3}{2}x$

Now, equation (2) becomes:

Evaluate the definite integrals:

First integral: $\int_{0}^{2} \frac{3}{2}x \, dx = \left[ \frac{3}{2} \frac{x^2}{2} \right]_{0}^{2} = \left[ \frac{3}{4}x^2 \right]_{0}^{2}$

$= \frac{3}{4}(2)^2 - \frac{3}{4}(0)^2 = \frac{3}{4}(4) - 0 = 3$

Second integral: $\int_{2}^{4} (6 - \frac{3}{2}x) \, dx = \left[ 6x - \frac{3}{2} \frac{x^2}{2} \right]_{2}^{4} = \left[ 6x - \frac{3}{4}x^2 \right]_{2}^{4}$

$= \left( 6(4) - \frac{3}{4}(4)^2 \right) - \left( 6(2) - \frac{3}{4}(2)^2 \right)$

$= \left( 24 - \frac{3}{4}(16) \right) - \left( 12 - \frac{3}{4}(4) \right)$

$= (24 - 12) - (12 - 3)$

$= 12 - 9 = 3$

Add the results of the two integrals:

$A = 3 + 3 = 6$

The area of the region bounded by the three lines is 6 square units.

The area of the bounded region is 6 square units.

Given:

The region is bounded by the lines $y = 4x + 5$, $y = 5 – x$, and $4y = x + 5$.

To Compute:

The area of the region bounded by the given lines using integration.

Solution:

The given lines are:

The region bounded by these three lines is a triangle. To find its area, we first find the vertices of the triangle by finding the intersection points of the lines.

Intersection of $L_1$ and $L_2$:

$4x + 5 = 5 - x$

$5x = 0 \implies x = 0$

Substitute $x=0$ into $y = 5-x$: $y = 5 - 0 = 5$.

Vertex A: $(0, 5)$.

Intersection of $L_1$ and $L_3$:

$4x + 5 = \frac{1}{4}x + \frac{5}{4}$

Multiply by 4: $16x + 20 = x + 5$

$15x = -15 \implies x = -1$

Substitute $x=-1$ into $y = 4x+5$: $y = 4(-1) + 5 = -4 + 5 = 1$.

Vertex B: $(-1, 1)$.

Intersection of $L_2$ and $L_3$:

$5 - x = \frac{1}{4}x + \frac{5}{4}$

Multiply by 4: $20 - 4x = x + 5$

$15 = 5x \implies x = 3$

Substitute $x=3$ into $y = 5-x$: $y = 5 - 3 = 2$.

Vertex C: $(3, 2)$.

The vertices of the triangle are A(0, 5), B(-1, 1), and C(3, 2).

To find the area using integration with respect to $x$, we need to identify the upper and lower boundary lines in the relevant intervals of $x$. The x-coordinates of the vertices are -1, 0, and 3. We split the integral at $x=0$ (the x-coordinate of vertex A) and $x=3$ (the x-coordinate of vertex C). The x-coordinate of B is -1, which is the starting point.

The lower boundary of the triangle is the line connecting B(-1,1) and C(3,2), which is $L_3: y = \frac{1}{4}x + \frac{5}{4}$. This line is below the other two lines in the region bounded by the triangle.

The upper boundary is formed by two line segments:

- From B(-1,1) to A(0,5), which is $L_1: y = 4x + 5$. This segment is the upper boundary for $x \in [-1, 0]$.

- From A(0,5) to C(3,2), which is $L_2: y = 5 - x$. This segment is the upper boundary for $x \in [0, 3]$.

The area $A$ of the triangle is the sum of the areas between the upper boundary and the lower boundary in the intervals $[-1, 0]$ and $[0, 3]$.

$A = \int_{-1}^{0} (y_{L1} - y_{L3}) \, dx + \int_{0}^{3} (y_{L2} - y_{L3}) \, dx$

$A = \int_{-1}^{0} \left( (4x + 5) - \left(\frac{1}{4}x + \frac{5}{4}\right) \right) \, dx + \int_{0}^{3} \left( (5 - x) - \left(\frac{1}{4}x + \frac{5}{4}\right) \right) \, dx$

Simplify the integrands:

For the first integral: $(4x + 5) - (\frac{1}{4}x + \frac{5}{4}) = 4x + 5 - \frac{1}{4}x - \frac{5}{4} = \left(4 - \frac{1}{4}\right)x + \left(5 - \frac{5}{4}\right) = \frac{15}{4}x + \frac{20-5}{4} = \frac{15}{4}x + \frac{15}{4}$

For the second integral: $(5 - x) - (\frac{1}{4}x + \frac{5}{4}) = 5 - x - \frac{1}{4}x - \frac{5}{4} = \left(-1 - \frac{1}{4}\right)x + \left(5 - \frac{5}{4}\right) = -\frac{5}{4}x + \frac{15}{4}$

Now, the area integral becomes:

$A = \int_{-1}^{0} \left(\frac{15}{4}x + \frac{15}{4}\right) \, dx + \int_{0}^{3} \left(-\frac{5}{4}x + \frac{15}{4}\right) \, dx$

Evaluate the definite integrals:

First integral: $\int_{-1}^{0} \left(\frac{15}{4}x + \frac{15}{4}\right) \, dx = \left[ \frac{15}{4} \frac{x^2}{2} + \frac{15}{4}x \right]_{-1}^{0} = \left[ \frac{15}{8}x^2 + \frac{15}{4}x \right]_{-1}^{0}$

$= \left( \frac{15}{8}(0)^2 + \frac{15}{4}(0) \right) - \left( \frac{15}{8}(-1)^2 + \frac{15}{4}(-1) \right)$

$= (0) - \left( \frac{15}{8} - \frac{15}{4} \right) = - \left( \frac{15}{8} - \frac{30}{8} \right) = - \left( -\frac{15}{8} \right) = \frac{15}{8}$

Second integral: $\int_{0}^{3} \left(-\frac{5}{4}x + \frac{15}{4}\right) \, dx = \left[ -\frac{5}{4} \frac{x^2}{2} + \frac{15}{4}x \right]_{0}^{3} = \left[ -\frac{5}{8}x^2 + \frac{15}{4}x \right]_{0}^{3}$

$= \left( -\frac{5}{8}(3)^2 + \frac{15}{4}(3) \right) - \left( -\frac{5}{8}(0)^2 + \frac{15}{4}(0) \right)$

$= \left( -\frac{5 \times 9}{8} + \frac{45}{4} \right) - (0)$

$= -\frac{45}{8} + \frac{90}{8} = \frac{90 - 45}{8} = \frac{45}{8}$

Add the results of the two integrals to find the total area:

$A = \frac{15}{8} + \frac{45}{8} = \frac{15 + 45}{8} = \frac{60}{8}$

Simplify the fraction:

$A = \frac{\cancel{60}^{15}}{\cancel{8}_{2}} = \frac{15}{2}$

The area of the region bounded by the three lines is $\frac{15}{2}$ square units.

The area of the bounded region is $\frac{15}{2}$ square units.

Given:

The curve $y = 2 \cos x$ and the interval from $x = 0$ to $x = 2\pi$ along the x-axis.

To Find:

The area bounded by the curve $y = 2 \cos x$ and the x-axis from $x = 0$ to $x = 2\pi$ using integration.

Solution:

To find the area bounded by the curve $y = f(x)$ and the x-axis over an interval $[a, b]$, we need to consider where the function is positive and negative in that interval. The total area is the sum of the absolute values of the integrals over the subintervals where the sign of $f(x)$ is constant.

The given curve is $y = 2 \cos x$ and the interval is $[0, 2\pi]$.

The curve $y = 2 \cos x$ is positive when $\cos x > 0$ and negative when $\cos x < 0$.

In the interval $[0, 2\pi]$, $\cos x$ is positive for $x \in [0, \frac{\pi}{2})$ and $x \in (\frac{3\pi}{2}, 2\pi]$. $\cos x$ is negative for $x \in (\frac{\pi}{2}, \frac{3\pi}{2})$. $\cos x$ is zero at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.

The total area $A$ is the sum of the areas of the regions above and below the x-axis:

$A = \int_{0}^{2\pi} |2 \cos x| \, dx$

We split the integral based on the sign of $\cos x$:

$A = \int_{0}^{\pi/2} 2 \cos x \, dx + \int_{\pi/2}^{3\pi/2} |2 \cos x| \, dx + \int_{3\pi/2}^{2\pi} 2 \cos x \, dx$

Since $\cos x \geq 0$ on $[0, \pi/2]$ and $[3\pi/2, 2\pi]$, and $\cos x \leq 0$ on $[\pi/2, 3\pi/2]$:

$A = \int_{0}^{\pi/2} 2 \cos x \, dx + \int_{\pi/2}^{3\pi/2} (-2 \cos x) \, dx + \int_{3\pi/2}^{2\pi} 2 \cos x \, dx$

First, find the indefinite integral of $2 \cos x$:

$\int 2 \cos x \, dx = 2 \sin x + C$

Now, evaluate the definite integrals:

Integral 1: $\int_{0}^{\pi/2} 2 \cos x \, dx = [2 \sin x]_{0}^{\pi/2}$

$= 2 \sin(\frac{\pi}{2}) - 2 \sin(0) = 2(1) - 2(0) = 2$

Integral 2: $\int_{\pi/2}^{3\pi/2} (-2 \cos x) \, dx = [-2 \sin x]_{\pi/2}^{3\pi/2}$

$= (-2 \sin(\frac{3\pi}{2})) - (-2 \sin(\frac{\pi}{2}))$

$= (-2(-1)) - (-2(1)) = 2 - (-2) = 2 + 2 = 4$

Integral 3: $\int_{3\pi/2}^{2\pi} 2 \cos x \, dx = [2 \sin x]_{3\pi/2}^{2\pi}$

$= 2 \sin(2\pi) - 2 \sin(\frac{3\pi}{2})$

$= 2(0) - 2(-1) = 0 - (-2) = 2$

Summing the areas from the three intervals:

$A = 2 + 4 + 2 = 8$

The area bounded by the curve $y = 2 \cos x$ and the x-axis from $x = 0$ to $x = 2\pi$ is 8 square units.

The area of the bounded region is 8 square units.

Given:

The region is bounded by the curve $y = 1 + |x + 1|$, the lines $x = –3$, $x = 3$, and the x-axis ($y = 0$).

To Sketch and Find:

To draw a rough sketch of the region bounded by the given curve and lines, and find the area of this region using integration.

Solution:

The given curve is $y = 1 + |x + 1|$.

The absolute value function $|x+1|$ is defined as:

$|x+1| = \begin{cases} x+1 & \text{if } x+1 \geq 0 \text{ (i.e., } x \geq -1) \\ -(x+1) = -x-1 & \text{if } x+1 < 0 \text{ (i.e., } x < -1) \end{cases}$

So, the curve $y = 1 + |x + 1|$ can be written as a piecewise function:

$y = \begin{cases} 1 + (x+1) = x + 2 & \text{if } x \geq -1 \\ 1 + (-x-1) = -x & \text{if } x < -1 \end{cases}$

The region is bounded by this curve, the vertical lines $x = -3$ and $x = 3$, and the x-axis ($y = 0$).

Rough Sketch:

The curve $y = 1 + |x+1|$ has its vertex at $x = -1$. At $x=-1$, $y = 1 + |-1+1| = 1$. The vertex is at $(-1, 1)$.

For $x < -1$, the curve is the line $y = -x$. For example, at $x=-3$, $y = -(-3) = 3$. Point $(-3, 3)$.

For $x \geq -1$, the curve is the line $y = x+2$. For example, at $x=3$, $y = 3+2 = 5$. Point $(3, 5)$. At $x=0$, $y = 0+2=2$. Point $(0, 2)$.

The curve is a V-shape opening upwards, with its lowest point at $(-1, 1)$. Since the minimum y-value is 1, the curve is always above the x-axis ($y=0$).

The region whose area we need to find is bounded below by the x-axis ($y=0$) and above by the curve $y = 1 + |x+1|$, between the vertical lines $x = -3$ and $x = 3$.

Area Calculation using Integration:

The area $A$ of the region is given by the definite integral of the upper boundary curve minus the lower boundary curve, from $x = -3$ to $x = 3$. Since the lower boundary is $y=0$ and the upper boundary $y=1+|x+1|$ is always positive, the area is:

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

We need to split the integral at $x = -1$ because the definition of $|x+1|$ changes at this point:

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

Using the piecewise definition of the curve:

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

Now, we evaluate the definite integrals:

First integral: $\int_{-3}^{-1} (-x) \, dx$

$\left[ -\frac{x^2}{2} \right]_{-3}^{-1} = \left( -\frac{(-1)^2}{2} \right) - \left( -\frac{(-3)^2}{2} \right)$

$= \left( -\frac{1}{2} \right) - \left( -\frac{9}{2} \right) = -\frac{1}{2} + \frac{9}{2} = \frac{8}{2} = 4$

Second integral: $\int_{-1}^{3} (x + 2) \, dx$

$\left[ \frac{x^2}{2} + 2x \right]_{-1}^{3} = \left( \frac{(3)^2}{2} + 2(3) \right) - \left( \frac{(-1)^2}{2} + 2(-1) \right)$

$= \left( \frac{9}{2} + 6 \right) - \left( \frac{1}{2} - 2 \right)$

$= \left( \frac{9}{2} + \frac{12}{2} \right) - \left( \frac{1}{2} - \frac{4}{2} \right)$

$= \frac{21}{2} - \left( -\frac{3}{2} \right) = \frac{21}{2} + \frac{3}{2} = \frac{24}{2} = 12$

The total area is the sum of the areas from the two intervals:

$A = 4 + 12 = 16$

The area of the region bounded by the curve $y = 1 + |x + 1|$, $x = –3$, $x = 3$, and $y = 0$ is 16 square units.

The area of the bounded region is 16 square units.

The region is bounded by the y-axis ($x=0$), the curve $y = \cos x$ and the curve $y = \sin x$. We are considering this region in the interval $0 \le x \le \frac{\pi}{2}$.

To find the area bounded by two curves, we need to find their intersection points. Let's find the intersection of $y = \cos x$ and $y = \sin x$ in the given interval $[0, \frac{\pi}{2}]$.

Setting the functions equal:

$\cos x = \sin x$

Dividing by $\cos x$ (assuming $\cos x \ne 0$):

$\tan x = 1$

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

The region bounded by the y-axis ($x=0$), $y=\cos x$, and $y=\sin x$ is the area enclosed by these curves. As $x$ increases from $0$, $y=\cos x$ starts at $(0,1)$ and $y=\sin x$ starts at $(0,0)$. They intersect at $x=\frac{\pi}{4}$. For $0 \le x \le \frac{\pi}{4}$, we know that $\cos x \ge \sin x$. Thus, the area bounded by the y-axis, $y=\cos x$ and $y=\sin x$ is the area between the curves $y=\cos x$ and $y=\sin x$ from $x=0$ to $x=\frac{\pi}{4}$.

The area, $A$, is given by the integral:

$A = \int_{0}^{\pi/4} (\cos x - \sin x) dx$

Now, we evaluate the integral:

$A = [\sin x - (-\cos x)]_{0}^{\pi/4}$

$A = [\sin x + \cos x]_{0}^{\pi/4}$

$A = (\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4})) - (\sin(0) + \cos(0))$

$A = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)$

$A = \frac{2}{\sqrt{2}} - 1$

$A = \sqrt{2} - 1$

The area of the region is $(\sqrt{2} - 1)$ square units.

This matches option (C).

The final answer is $\boxed{(\sqrt{2} – 1) \text{ sq units}}$.

The given curve is $x^2 = 4y$, which can be written as $y = \frac{x^2}{4}$. This is a parabola opening upwards with its vertex at the origin.

The given straight line is $x = 4y - 2$, which can be written as $4y = x + 2$, or $y = \frac{x+2}{4}$.

To find the area of the region bounded by the parabola and the line, we first need to find the points of intersection of these two curves.

We set the $y$ values equal:

$\frac{x^2}{4} = \frac{x+2}{4}$

Multiplying both sides by 4:

$x^2 = x + 2$

Rearranging the equation:

$x^2 - x - 2 = 0$

Factoring the quadratic equation:

$(x - 2)(x + 1) = 0$

The solutions for $x$ are $x = 2$ and $x = -1$. These are the x-coordinates of the intersection points, which define the limits of integration.

To determine which function is above the other in the interval $[-1, 2]$, we can test a point, for example $x=0$.

For the parabola $y = \frac{x^2}{4}$, at $x=0$, $y = \frac{0^2}{4} = 0$.

For the line $y = \frac{x+2}{4}$, at $x=0$, $y = \frac{0+2}{4} = \frac{2}{4} = \frac{1}{2}$.

Since $\frac{1}{2} > 0$, the line $y = \frac{x+2}{4}$ is above the parabola $y = \frac{x^2}{4}$ in the interval $[-1, 2]$.

The area $A$ of the region bounded by the two curves is given by the definite integral of the difference between the upper curve and the lower curve from $x=-1$ to $x=2$:

$A = \int_{-1}^{2} \left(\frac{x+2}{4} - \frac{x^2}{4}\right) dx$

$A = \frac{1}{4} \int_{-1}^{2} (x + 2 - x^2) dx$

Now, we evaluate the integral:

$A = \frac{1}{4} \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}$

Evaluate at the upper limit $x=2$:

$\left(\frac{2^2}{2} + 2(2) - \frac{2^3}{3}\right) = \left(\frac{4}{2} + 4 - \frac{8}{3}\right) = \left(2 + 4 - \frac{8}{3}\right) = \left(6 - \frac{8}{3}\right) = \left(\frac{18}{3} - \frac{8}{3}\right) = \frac{10}{3}$

Evaluate at the lower limit $x=-1$:

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

To combine the terms at the lower limit, find a common denominator (6):

$\frac{3}{6} - \frac{12}{6} + \frac{2}{6} = \frac{3 - 12 + 2}{6} = \frac{-7}{6}$

Subtract the value at the lower limit from the value at the upper limit:

$A = \frac{1}{4} \left[\frac{10}{3} - \left(\frac{-7}{6}\right)\right]$

$A = \frac{1}{4} \left[\frac{10}{3} + \frac{7}{6}\right]$

Find a common denominator (6) for the terms inside the bracket:

$A = \frac{1}{4} \left[\frac{20}{6} + \frac{7}{6}\right]$

$A = \frac{1}{4} \left[\frac{27}{6}\right]$

Multiply the fractions:

$A = \frac{27}{24}$

Simplify the fraction by dividing the numerator and denominator by 3:

$A = \frac{\cancel{27}^{9}}{\cancel{24}_{8}}$

$A = \frac{9}{8}$

The area of the region bounded by the curve $x^2 = 4y$ and the straight line $x = 4y - 2$ is $\frac{9}{8}$ square units.

This matches option (D).

The final answer is $\boxed{\frac{9}{8} \text{ sq units}}$.

The given curve is $y = \sqrt{16-x^2}$.

Squaring both sides, we get $y^2 = 16 - x^2$.

Rearranging the terms, we have $x^2 + y^2 = 16$.

This equation represents a circle centered at the origin $(0,0)$ with a radius $r^2 = 16$, so the radius is $r = 4$.

Since the original equation is $y = \sqrt{16-x^2}$, this implies that $y \ge 0$. Therefore, the curve represents the upper semicircle of the circle $x^2 + y^2 = 16$.

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

The x-axis intersects the circle when $y=0$, i.e., $x^2 + 0^2 = 16$, which gives $x^2 = 16$, so $x = \pm 4$. The semicircle extends from $x=-4$ to $x=4$.

The area of the region bounded by the curve $y = \sqrt{16-x^2}$ and the x-axis is the area of this upper semicircle with radius 4.

The area of a circle with radius $r$ is $\pi r^2$. The area of a semicircle is half the area of the full circle.

Area of semicircle $= \frac{1}{2} \pi r^2$

Substitute the radius $r=4$:

Area $= \frac{1}{2} \pi (4)^2 = \frac{1}{2} \pi (16) = 8\pi$

The area of the region bounded by the curve $y = \sqrt{16−x^2}$ and the x-axis is $8\pi$ square units.

The provided options are:

(A) $\sqrt{2}$ sq units

(B) $(\sqrt{2} + 1)$ sq units

(C) $(\sqrt{2} – 1)$ sq units

(D) $(2 \sqrt{2} –1)$ sq units

Comparing the calculated area $8\pi$ with the given options, none of the options match the correct area.

Based on the calculation, the area is $8\pi$ sq units.

The region is bounded by the x-axis ($y=0$), the line $y = x$, and the circle $x^2 + y^2 = 32$ in the first quadrant.

The equation of the circle is $x^2 + y^2 = 32$. This is a circle centered at the origin $(0,0)$ with radius $r = \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$.

We need to find the intersection points of the boundaries in the first quadrant.

The line $y=x$ and the circle $x^2 + y^2 = 32$: Substitute $y=x$ into the circle equation:

$x^2 + x^2 = 32$

$2x^2 = 32$

$x^2 = 16$

Since we are in the first quadrant, we take the positive root: $x = 4$.

When $x=4$, $y=x=4$. So, the intersection point in the first quadrant is $(4,4)$.

The line $y=x$ passes through the origin $(0,0)$.

The x-axis ($y=0$) intersects the circle $x^2 + y^2 = 32$ at $x^2 + 0^2 = 32$, so $x = \pm \sqrt{32} = \pm 4\sqrt{2}$. In the first quadrant, the intersection is $(4\sqrt{2}, 0)$.

The region is bounded by the line $y=x$ (a ray from the origin into the first quadrant), the x-axis (the ray $\theta = 0$), and the arc of the circle $x^2 + y^2 = 32$ in the first quadrant.

The line $y=x$ makes an angle $\theta$ with the positive x-axis such that $\tan \theta = \frac{y}{x} = \frac{x}{x} = 1$. In the first quadrant, this angle is $\theta = \frac{\pi}{4}$.

The x-axis corresponds to the angle $\theta = 0$.

The region bounded by the x-axis ($y=0$, or $\theta = 0$), the line $y=x$ (or $\theta = \frac{\pi}{4}$), and the circle $x^2 + y^2 = 32$ (with radius $r=4\sqrt{2}$) in the first quadrant is a sector of the circle.

The area of a sector of a circle with radius $r$ and central angle $\theta$ (in radians) is given by $\frac{1}{2}r^2\theta$.

In this case, the radius is $r = 4\sqrt{2}$, and the angle of the sector is the angle between the line $y=x$ and the x-axis, which is $\theta = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

Area $= \frac{1}{2} r^2 \theta$

Area $= \frac{1}{2} (4\sqrt{2})^2 \left(\frac{\pi}{4}\right)$

Area $= \frac{1}{2} (16 \times 2) \left(\frac{\pi}{4}\right)$

Area $= \frac{1}{2} (32) \left(\frac{\pi}{4}\right)$

Area $= 16 \left(\frac{\pi}{4}\right)$

Area $= 4\pi$

The area of the region is $4\pi$ square units.

This matches option (B).

The final answer is $\boxed{4\pi \text{ sq units}}$.

The region is bounded by the curve $y = \cos x$ and the x-axis ($y=0$) in the interval $[0, \pi]$.

To find the area, we need to consider the portions of the curve above and below the x-axis separately, or integrate the absolute value of the function.

We find where the curve intersects the x-axis in the given interval by setting $y = \cos x = 0$.

In the interval $[0, \pi]$, $\cos x = 0$ at $x = \frac{\pi}{2}$.

This divides the interval $[0, \pi]$ into two sub-intervals: $[0, \frac{\pi}{2}]$ and $[\frac{\pi}{2}, \pi]$.

In the interval $[0, \frac{\pi}{2}]$, $\cos x \ge 0$. The area in this sub-interval is given by $\int_{0}^{\pi/2} \cos x dx$.

In the interval $[\frac{\pi}{2}, \pi]$, $\cos x \le 0$. The area in this sub-interval is given by $\int_{\pi/2}^{\pi} |\cos x| dx = \int_{\pi/2}^{\pi} (-\cos x) dx$.

The total area $A$ is the sum of the areas in these two sub-intervals:

$A = \int_{0}^{\pi/2} \cos x dx + \int_{\pi/2}^{\pi} (-\cos x) dx$

Evaluate the first integral:

$\int_{0}^{\pi/2} \cos x dx = [\sin x]_{0}^{\pi/2} = \sin(\frac{\pi}{2}) - \sin(0) = 1 - 0 = 1$

Evaluate the second integral:

$\int_{\pi/2}^{\pi} (-\cos x) dx = [-\sin x]_{\pi/2}^{\pi} = (-\sin(\pi)) - (-\sin(\frac{\pi}{2})) = (0) - (-1) = 1$

The total area is the sum of these two areas:

$A = 1 + 1 = 2$

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

This matches option (A).

The final answer is $\boxed{2 \text{ sq units}}$.

The given curves are the parabola $y^2 = x$ and the straight line $2y = x$.

To find the area of the region bounded by these curves, we first find their points of intersection.

We can substitute $x = 2y$ from the line equation into the parabola equation:

$y^2 = 2y$

Rearranging the equation:

$y^2 - 2y = 0$

Factor out $y$:

$y(y - 2) = 0$

This gives the y-coordinates of the intersection points: $y = 0$ and $y = 2$.

Now find the corresponding x-coordinates using the line equation $x = 2y$:

For $y=0$, $x = 2(0) = 0$. The intersection point is $(0,0)$.

For $y=2$, $x = 2(2) = 4$. The intersection point is $(4,2)$.

The limits of integration with respect to $y$ are from $y=0$ to $y=2$.

We need to determine which curve is to the right and which is to the left in the region between $y=0$ and $y=2$.

We can rewrite the equations in terms of $x$ as a function of $y$:

Line: $x = 2y$

Parabola: $x = y^2$

For $y \in (0, 2)$, let's test a value, say $y=1$. For the line, $x = 2(1) = 2$. For the parabola, $x = (1)^2 = 1$. Since $2 > 1$, the line $x = 2y$ is to the right of the parabola $x = y^2$ in the interval $(0, 2)$.

The area $A$ of the region bounded by the curves is given by the integral of the difference between the right curve and the left curve with respect to $y$, from $y=0$ to $y=2$:

$A = \int_{0}^{2} (x_{\text{line}} - x_{\text{parabola}}) dy$

$A = \int_{0}^{2} (2y - y^2) dy$

Now, we evaluate the definite integral:

$A = \left[\frac{2y^2}{2} - \frac{y^3}{3}\right]_{0}^{2}$

$A = \left[y^2 - \frac{y^3}{3}\right]_{0}^{2}$

Evaluate at the upper limit $y=2$:

$(2)^2 - \frac{(2)^3}{3} = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}$

Evaluate at the lower limit $y=0$:

$(0)^2 - \frac{(0)^3}{3} = 0 - 0 = 0$

Subtract the value at the lower limit from the value at the upper limit:

$A = \frac{4}{3} - 0 = \frac{4}{3}$

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

This matches option (A).

The final answer is $\boxed{\frac{4}{3} \text{ sq units}}$.

The region is bounded by the curve $y = \sin x$, the vertical lines $x = 0$ and $x = \frac{\pi}{2}$, and the x-axis ($y=0$).

We are interested in the area under the curve $y = \sin x$ from $x = 0$ to $x = \frac{\pi}{2}$, above the x-axis.

In the interval $[0, \frac{\pi}{2}]$, the value of $\sin x$ is non-negative, i.e., $\sin x \ge 0$.

Therefore, the area $A$ of the region is given by the definite integral of $\sin x$ from $x = 0$ to $x = \frac{\pi}{2}$.

$A = \int_{0}^{\pi/2} \sin x dx$

Now, we evaluate the integral:

The antiderivative of $\sin x$ is $-\cos x$.

$A = [-\cos x]_{0}^{\pi/2}$

Evaluate the antiderivative at the upper limit $x = \frac{\pi}{2}$ and the lower limit $x = 0$ and subtract:

$A = (-\cos(\frac{\pi}{2})) - (-\cos(0))$

$A = (-\cos(\frac{\pi}{2})) + \cos(0)$

We know that $\cos(\frac{\pi}{2}) = 0$ and $\cos(0) = 1$.

$A = (-0) + 1$

$A = 0 + 1$

$A = 1$

The area of the region is 1 square unit.

This matches option (D).

The final answer is $\boxed{1 \text{ sq unit}}$.

The given equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{16} = 1$.

This is an ellipse centered at the origin $(0,0)$.

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

Comparing the given equation with the standard form, we have $a^2 = 25$ and $b^2 = 16$.

This means the semi-major axis length is $a = \sqrt{25} = 5$ and the semi-minor axis length is $b = \sqrt{16} = 4$.

Since $a > b$, the major axis is along the x-axis.

The formula for the area of an ellipse with semi-major axis $a$ and semi-minor axis $b$ is $A = \pi ab$.

In this case, $a = 5$ and $b = 4$.

Area $A = \pi (5)(4)$

$A = 20\pi$

The area of the region bounded by the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ is $20\pi$ square units.

This matches option (A).

The final answer is $\boxed{20\pi \text{ sq units}}$.

The given equation of the circle is $x^2 + y^2 = 1$.

This is a circle centered at the origin $(0,0)$.

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

Comparing the given equation with the standard form, we have $r^2 = 1$, so the radius is $r = \sqrt{1} = 1$.

The formula for the area of a circle with radius $r$ is $A = \pi r^2$.

In this case, the radius is $r = 1$.

Area $A = \pi (1)^2$

$A = \pi (1)$

$A = \pi$

The area of the region bounded by the circle $x^2 + y^2 = 1$ is $\pi$ square units.

This matches option (B).

The final answer is $\boxed{\pi \text{ sq units}}$.

The region is bounded by the curve $y = x+1$, the vertical lines $x = 2$ and $x = 3$, and the x-axis ($y=0$).

In the interval $[2, 3]$, the function $y = x+1$ is positive, since for $x=2$, $y=3$ and for $x=3$, $y=4$.

Therefore, the curve $y = x+1$ is above the x-axis in the interval $[2, 3]$.

The area $A$ 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 $\int_{a}^{b} f(x) dx$, provided $f(x) \ge 0$ for all $x$ in $[a, b]$.

Here, $f(x) = x+1$, $a=2$, and $b=3$. Since $x+1 > 0$ for $x \in [2, 3]$, the area is:

$A = \int_{2}^{3} (x+1) dx$

Now, we evaluate the definite integral:

The antiderivative of $(x+1)$ is $\frac{x^2}{2} + x$.

$A = \left[\frac{x^2}{2} + x\right]_{2}^{3}$

Evaluate the antiderivative at the upper limit $x=3$:

$\frac{(3)^2}{2} + 3 = \frac{9}{2} + 3 = \frac{9}{2} + \frac{6}{2} = \frac{15}{2}$

Evaluate the antiderivative at the lower limit $x=2$:

$\frac{(2)^2}{2} + 2 = \frac{4}{2} + 2 = 2 + 2 = 4$

Subtract the value at the lower limit from the value at the upper limit:

$A = \frac{15}{2} - 4$

$A = \frac{15}{2} - \frac{8}{2}$

$A = \frac{15 - 8}{2}$

$A = \frac{7}{2}$

The area of the region bounded by the curve $y = x + 1$ and the lines $x = 2$ and $x = 3$ is $\frac{7}{2}$ square units.

This matches option (A).

The final answer is $\boxed{\frac{7}{2} \text{ sq units}}$.

The region is bounded by the curve $x = 2y + 3$, the y-axis ($x=0$), and the horizontal lines $y = 1$ and $y = -1$.

We need to find the area of the region enclosed by these boundaries.

Since the boundaries are given in terms of y ($y=-1$, $y=1$) and the equation of the curve is given as $x$ in terms of $y$ ($x = 2y + 3$), it is convenient to integrate with respect to $y$.

The limits of integration for $y$ are from $-1$ to $1$.

For $y \in [-1, 1]$, let's check the position of the line $x = 2y+3$ relative to the y-axis ($x=0$).

If $y=-1$, $x = 2(-1) + 3 = -2 + 3 = 1$.

If $y=0$, $x = 2(0) + 3 = 3$.

If $y=1$, $x = 2(1) + 3 = 2 + 3 = 5$.

Since $x = 2y + 3 \ge 1$ for all $y \in [-1, 1]$, the line $x = 2y + 3$ is always to the right of the y-axis ($x=0$) in the interval $[-1, 1]$.

The area $A$ of the region bounded by the curve $x = f(y)$, the y-axis, and the lines $y=c$ and $y=d$ is given by the integral $\int_{c}^{d} |f(y)| dy$. Since $x = 2y+3 > 0$ for $y \in [-1, 1]$, the area is:

$A = \int_{-1}^{1} (2y + 3) dy$

Now, we evaluate the definite integral:

The antiderivative of $2y + 3$ with respect to $y$ is $\frac{2y^{2}}{2} + 3y = y^2 + 3y$.

$A = [y^2 + 3y]_{-1}^{1}$

Evaluate the antiderivative at the upper limit $y=1$:

$(1)^2 + 3(1) = 1 + 3 = 4$

Evaluate the antiderivative at the lower limit $y=-1$:

$(-1)^2 + 3(-1) = 1 - 3 = -2$

Subtract the value at the lower limit from the value at the upper limit:

$A = 4 - (-2)$

$A = 4 + 2$

$A = 6$

The area of the region bounded by the curve $x = 2y + 3$ and the y-axis between $y = -1$ and $y = 1$ is 6 square units.

This matches option (C).

The final answer is $\boxed{6 \text{ sq units}}$.