Chapter 8 Application of Integrals (Concepts)

Areas of Bounded Regions

One of the most important and direct applications of definite integrals is the calculation of the area of a region in the xy-plane that is bounded by one or more curves and straight lines. While elementary geometry provides formulas for areas of simple shapes like rectangles, triangles, and circles, calculus, specifically integration, gives us the tools to find areas of much more complex regions with curved boundaries. The definite integral is fundamentally linked to the concept of area, as its definition arises from summing up infinitesimal areas (like areas of thin rectangles).

Area Under a Curve and Above the X-axis

Let $y = f(x)$ be a continuous function on the closed interval $[a, b]$. If $f(x)$ is non-negative on this interval, meaning $f(x) \geq 0$ for all $x \in [a, b]$, then the definite integral $\int\limits_{a}^{b} f(x) dx$ represents the exact area of the region bounded by the curve $y = f(x)$, the x-axis ($y=0$), and the vertical lines $x=a$ and $x=b$.

Geometrically, we can think of this area as being formed by infinitely many vertical strips of infinitesimal width $dx$. Each strip has height $f(x)$ at a particular $x$ and area approximately $f(x) dx$. The integral sums these infinitesimal areas from $x=a$ to $x=b$.

Area Below the X-axis

If the function $y = f(x)$ is continuous and non-positive on $[a, b]$, meaning $f(x) \leq 0$ for all $x \in [a, b]$, then the curve lies below or on the x-axis. The definite integral $\int\limits_{a}^{b} f(x) dx$ will result in a negative value, which is the negative of the area of the region bounded by the curve, the x-axis, $x=a$, and $x=b$.

To find the actual geometric area, which is always non-negative, we take the absolute value of the integral. Since $f(x) \leq 0$, $|f(x)| = -f(x)$.

Area When the Curve Crosses the X-axis

If the function $y = f(x)$ is continuous on $[a, b]$ but takes both positive and negative values within this interval, the definite integral $\int\limits_{a}^{b} f(x) dx$ calculates the net signed area. This is the sum of the areas of the regions above the x-axis minus the sum of the areas of the regions below the x-axis.

To find the total geometric area bounded by the curve, the x-axis, $x=a$, and $x=b$, we need to find the points where the curve intersects the x-axis (i.e., where $f(x)=0$). Let these points be $c_1, c_2, \dots, c_k$ such that $a < c_1 < c_2 < \dots < c_k < b$. These points divide the interval $[a, b]$ into subintervals $[a, c_1], [c_1, c_2], \dots, [c_k, b]$. Within each subinterval, $f(x)$ has a constant sign (either $\geq 0$ or $\leq 0$).

The total area is the sum of the absolute values of the definite integrals over each subinterval:

For each subinterval, if $f(x) \geq 0$, $|f(x)| = f(x)$, and if $f(x) \leq 0$, $|f(x)| = -f(x)$. So we integrate $f(x)$ or $-f(x)$ accordingly over each subinterval.

Area Bounded by a Curve and the Y-axis

Sometimes it is easier or necessary to integrate with respect to $y$ to find the area of a region. This is typically the case when the region is bounded by a curve given by $x = g(y)$, the y-axis, and horizontal lines $y=c$ and $y=d$.

If the function $x = g(y)$ is continuous and non-negative on the closed interval $[c, d]$, meaning $g(y) \geq 0$ for all $y \in [c, d]$, the curve lies to the right of or on the y-axis. The area of the region bounded by $x = g(y)$, the y-axis ($x=0$), and the horizontal lines $y=c$ and $y=d$ is given by the definite integral of $g(y)$ from $c$ to $d$ with respect to $y$.

If $g(y) \leq 0$ on $[c, d]$ (curve is to the left of the y-axis), the area is the absolute value of the integral:

If $g(y)$ changes sign on $[c, d]$, find the points where $g(y)=0$, split the integral, and sum the absolute values of the integrals over the subintervals, integrating $|g(y)|$.

Area Bounded by Some Simple Curves

We can apply the area formulas to find the areas of regions bounded by common geometric shapes like circles and ellipses, derived from their equations.

Area of a Circle

Consider a circle centered at the origin with radius $a$, given by the equation $x^2 + y^2 = a^2$. We can find the area by integrating the upper semi-circle, $y = \sqrt{a^2 - x^2}$, from $x=-a$ to $x=a$, and multiplying the result by 2 (to account for the lower semi-circle). Since $y = \sqrt{a^2 - x^2} \geq 0$ on $[-a, a]$, the area of the upper semi-circle is $\int\limits_{-a}^{a} \sqrt{a^2 - x^2} dx$.

We use the standard integral formula (from section 7.10, formula 8):

Applying the limits from $-a$ to $a$:

This is the area of the semi-circle. Multiply by 2 for the full circle area:

The area of a circle with radius $a$ is $\mathbf{\pi a^2}$ square units.

Area of an Ellipse

Consider an ellipse centered at the origin with semi-major axis $a$ and semi-minor axis $b$, given by the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. We can solve for $y$: $\frac{y^2}{b^2} = 1 - \frac{x^2}{a^2} = \frac{a^2 - x^2}{a^2}$. $y^2 = \frac{b^2}{a^2}(a^2 - x^2)$. The equation for the upper half of the ellipse is $y = \sqrt{\frac{b^2}{a^2}(a^2 - x^2)} = \frac{b}{a} \sqrt{a^2 - x^2}$ for $-a \leq x \leq a$.

The area of the entire ellipse is twice the area of the upper half from $x=-a$ to $x=a$. Since the upper half is non-negative, we integrate directly.

Move the constant $\frac{b}{a}$ outside the integral:

From the circle derivation, we already found the value of the definite integral $\int\limits_{-a}^{a} \sqrt{a^2 - x^2} dx$, which is $\frac{\pi a^2}{2}$ (the area of a semi-circle of radius $a$).

Simplify the expression:

The area of an ellipse with semi-major axis $a$ and semi-minor axis $b$ is $\mathbf{\pi ab}$ square units.

General Steps for Finding Area of Bounded Regions

To find the area of a region bounded by curves and lines:

1. Sketch the Region: Always start by drawing a clear diagram of the region. This helps identify the boundaries, intersection points, and whether to integrate with respect to $x$ or $y$.

2. Choose the Variable of Integration: Decide whether integrating with respect to $x$ or $y$ is more convenient. If the boundaries are easily expressed as $y=f(x)$ and the limits are $x$-values, integrate with respect to $x$. If the boundaries are easily expressed as $x=g(y)$ and the limits are $y$-values, integrate with respect to $y$. Choose the variable that simplifies the setup and evaluation of the integral(s).

3. Find Intersection Points/Limits: Determine the points of intersection of the bounding curves/lines. These points define the limits of integration. If integrating with respect to $x$, find the x-coordinates of intersection. If integrating with respect to $y$, find the y-coordinates.

4. Set up the Integral(s): Write the formula for the area. If integrating w.r.t. $x$, the area is $\int\limits_{a}^{b} |f(x)| dx$. If integrating w.r.t. $y$, the area is $\int\limits_{c}^{d} |g(y)| dy$. Use the sketch to determine where $f(x) \ge 0$ or $f(x) \le 0$ (or $g(y) \ge 0$ or $g(y) \le 0$) and split the integral if the function changes sign within the limits.

5. Evaluate the Integral(s): Compute the definite integral(s) using the Fundamental Theorem of Calculus and appropriate integration techniques.

Answer:

Given:

The curve $y = x^2$.

The x-axis ($y=0$).

The vertical lines $x=0$ and $x=2$.

To Find:

The area of the region bounded by these curves and lines.

Solution:

1. Sketch the region:

The curve $y=x^2$ is a standard parabola. The lines $x=0$ (y-axis) and $x=2$ are vertical lines. The x-axis ($y=0$) is a horizontal line.

(Note: The image should depict the area under the parabola $y=x^2$ from $x=0$ to $x=2$, bounded below by the x-axis).

2. Choose the integration variable: The boundaries are given conveniently in terms of $x$ (curve $y=f(x)$, vertical lines $x=a, x=b$). Integrating with respect to $x$ is appropriate.

3. Determine the limits of integration: The region is bounded horizontally by $x=0$ and $x=2$. These are the limits for the integral with respect to $x$. Lower limit $a=0$, upper limit $b=2$.

4. Set up the integral: On the interval $[0, 2]$, the function $f(x) = x^2$ is non-negative ($x^2 \ge 0$). The lower boundary is the x-axis ($y=0$). Therefore, the area is given by the definite integral $\int\limits_{a}^{b} f(x) dx$.

Area A = $\int\limits_{0}^{2} x^2 dx$

... (A)

5. Evaluate the integral: Use the Fundamental Theorem of Calculus.

Find an antiderivative of $f(x) = x^2$. Using the power rule $\int x^n dx = \frac{x^{n+1}}{n+1}$, an antiderivative is $F(x) = \frac{x^{2+1}}{2+1} = \frac{x^3}{3}$.

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

Apply the limits of integration (Upper limit value - Lower limit value):

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

A = $\frac{8}{3} - 0$

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

... (B)

The area of the bounded region is $\mathbf{\frac{8}{3}}$ square units.

Answer:

Given:

The parabola $y^2 = 4x$.

The y-axis ($x=0$).

The horizontal line $y=2$.

To Find:

The area of the region bounded by these curves and lines.

Solution:

1. Sketch the region:

The equation $y^2 = 4x$ represents a parabola that opens to the right, symmetric about the x-axis. It passes through the origin $(0,0)$. The y-axis ($x=0$) is the left boundary. The line $y=2$ is a horizontal line. The parabola intersects the line $y=2$ when $(2)^2 = 4x \implies 4 = 4x \implies x=1$. So, the intersection point is $(1, 2)$. The region is bounded below by the x-axis where the parabola touches the y-axis (y=0).

(Note: The image should show the parabola $y^2=4x$, the y-axis, and the line $y=2$, highlighting the region in the first quadrant bounded by these three).

2. Choose the integration variable: The region is bounded horizontally by the parabola and the y-axis ($x=0$), and vertically by horizontal lines ($y=0$ implicitly at the vertex, and $y=2$). It is simpler to integrate with respect to $y$ as the right boundary is a single function of $y$ ($x = y^2/4$) and the left boundary is a single vertical line ($x=0$). If we integrated w.r.t. x, we would need to split the parabola into $y=\sqrt{4x}$ and $y=-\sqrt{4x}$ and deal with two integrals or use symmetry.

Express the parabola equation as $x = g(y)$: From $y^2 = 4x$, we get $x = \frac{y^2}{4}$. So, $g(y) = \frac{y^2}{4}$.

3. Determine the limits of integration: The region is bounded vertically from $y=0$ (the vertex on the y-axis) up to $y=2$. These are the limits for the integral with respect to $y$. Lower limit $c=0$, upper limit $d=2$.

4. Set up the integral: On the interval $[0, 2]$ for $y$, the function $g(y) = \frac{y^2}{4}$ is non-negative ($\frac{y^2}{4} \ge 0$). The left boundary is the y-axis ($x=0$). Therefore, the area is given by the definite integral $\int\limits_{c}^{d} g(y) dy$.

Area A = $\int\limits_{0}^{2} \frac{y^2}{4} dy$

... (A)

5. Evaluate the integral: Use the Fundamental Theorem of Calculus.

Move the constant $\frac{1}{4}$ outside the integral using linearity:

A = $\frac{1}{4} \int\limits_{0}^{2} y^2 dy$

Find an antiderivative of $y^2$: $F(y) = \frac{y^3}{3}$.

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

Apply the limits of integration:

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

A = $\frac{1}{4} \left(\frac{8}{3} - 0\right)$

A = $\frac{1}{4} \cdot \frac{8}{3} = \frac{\cancel{8}^{2}}{\cancel{12}_{3}} = \frac{2}{3}$

... (B)

The area of the bounded region is $\mathbf{\frac{2}{3}}$ square units.

Area Between Two Curves

Integral calculus can also be used to find the area of a region that is bounded by two intersecting curves, or by a curve and a line. The principle is an extension of finding the area under a single curve. Instead of finding the area between a curve and the x-axis (or y-axis), we find the area between an upper curve and a lower curve (or a right curve and a left curve).

Area Between Two Curves with Respect to X

Consider a region bounded by two curves $y = f(x)$ and $y = g(x)$ over an interval $[a, b]$ on the x-axis. To find the area of such a region, we assume that one function is consistently greater than or equal to the other throughout the interval.

Suppose $f(x)$ and $g(x)$ are continuous functions on $[a, b]$ and $f(x) \geq g(x)$ for all $x \in [a, b]$. This means the curve $y = f(x)$ is above or touches the curve $y = g(x)$ throughout the interval from $x=a$ to $x=b$.

The area of the region bounded by $y = f(x)$, $y = g(x)$, and the vertical lines $x=a$ and $x=b$ is given by the definite integral of the difference between the upper function and the lower function, from $a$ to $b$:

Here, $f(x)$ is the upper function and $g(x)$ is the lower function.

Justification: We can think of this area as the area under the upper curve $y=f(x)$ from $a$ to $b$, minus the area under the lower curve $y=g(x)$ from $a$ to $b$. Area under $y=f(x)$ is $\int\limits_{a}^{b} f(x) dx$. Area under $y=g(x)$ is $\int\limits_{a}^{b} g(x) dx$. Area between curves = (Area under $y=f(x)$) - (Area under $y=g(x)$) = $\int\limits_{a}^{b} f(x) dx - \int\limits_{a}^{b} g(x) dx$. By the linearity property of definite integrals (Property P3), this is equal to $\int\limits_{a}^{b} [f(x) - g(x)] dx$. This holds even if parts of the curves are below the x-axis.

If the relative position of the curves changes within the interval of interest, i.e., if $f(x)$ is sometimes above $g(x)$ and sometimes below $g(x)$, we need to find the points of intersection where $f(x) = g(x)$. These points divide the interval into subintervals. In each subinterval, one function is consistently above the other. The total area is the sum of the areas of these sub-regions, calculated by integrating the absolute difference $|f(x) - g(x)|$ over the entire interval or summing $\int (\text{upper} - \text{lower}) dx$ over each subinterval.

If $f(x) \geq g(x)$ on $[a, c]$ and $g(x) \geq f(x)$ on $[c, b]$ ($a < c < b$), then

Area Between Two Curves with Respect to Y

Sometimes it is more convenient to determine the area by integrating with respect to $y$. This is suitable when the boundaries of the region are more easily expressed as functions of $y$, i.e., $x = g(y)$ and $x = h(y)$, and the region is bounded by horizontal lines $y=c$ and $y=d$.

Suppose $g(y)$ and $h(y)$ are continuous functions on $[c, d]$ and $g(y) \geq h(y)$ for all $y \in [c, d]$. This means the curve $x = g(y)$ is to the right of or touches the curve $x = h(y)$ throughout the interval from $y=c$ to $y=d$.

The area of the region bounded by $x = g(y)$, $x = h(y)$, and the horizontal lines $y=c$ and $y=d$ is given by the definite integral of the difference between the right-hand function and the left-hand function, from $c$ to $d$, integrated with respect to $y$:

Here, $g(y)$ is the right function and $h(y)$ is the left function.

If the relative position of the curves changes (which curve is to the right changes) within the interval $[c, d]$, find the intersection points by setting $g(y) = h(y)$. These points divide the interval into subintervals. The total area is the sum of the areas of these sub-regions, calculated by integrating the absolute difference $|g(y) - h(y)|$ over $[c, d]$ or summing $\int (\text{right} - \text{left}) dy$ over each subinterval.

Procedure for Finding Area Between Two Curves

1. Sketch the Region: Draw a clear sketch of the region bounded by the given curves. This is the most crucial step. Identify the boundaries of the region.

2. Find Intersection Points: Determine the points where the curves intersect by setting their equations equal to each other and solving for the variable ( $x$ or $y$). These intersection points will often determine the limits of integration.

3. Choose the Integration Variable ($dx$ or $dy$): Decide whether to integrate with respect to $x$ or $y$.

• Integrate with respect to $x$ if the upper and lower boundaries are functions of $x$ ($y=f(x)$ and $y=g(x)$) and the limits are $x$-values. This approach is suitable if vertical strips span the region from one curve to the other without crossing intersection points internally.
• Integrate with respect to $y$ if the right and left boundaries are functions of $y$ ($x=g(y)$ and $x=h(y)$) and the limits are $y$-values. This approach is suitable if horizontal strips span the region without crossing intersection points internally.

4. Determine Upper/Lower or Right/Left Function: On the interval of integration, identify which function is the upper/lower boundary (for $dx$ integration) or the right/left boundary (for $dy$ integration) by looking at the sketch or testing a point in the interval.

5. Set up the Integral(s): Write the definite integral for the area as $\int (\text{upper function} - \text{lower function}) dx$ or $\int (\text{right function} - \text{left function}) dy$. Use the limits of integration found in step 2 (or as given by vertical/horizontal lines). If the upper/lower or right/left function changes within the total region, split the region at the intersection points and set up a sum of integrals.

6. Evaluate the Integral(s): Compute the definite integral(s). The result is the area of the bounded region.

Answer:

Given:

The parabolas $y^2 = x$ and $x^2 = y$.

To Find:

The area of the region enclosed by these two parabolas.

Solution:

1. Sketch the region:

$y^2 = x$ is a parabola opening to the right, symmetric about the x-axis. $x^2 = y$ is a parabola opening upwards, symmetric about the y-axis.

(Note: The image should show the two parabolas intersecting in the first quadrant, highlighting the enclosed region).

2. Find intersection points: To find where the curves intersect, we set the equations equal to each other. Substitute $y=x^2$ from the second equation into the first equation $y^2=x$:

$(x^2)^2 = x$

x$^4$ = x

x$^4$ - x = 0

Factor out $x$:

x$(x^3 - 1)$ = 0

Factor the difference of cubes ($x^3 - 1 = (x-1)(x^2+x+1)$):

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

The real solutions are $x=0$ and $x-1=0 \implies x=1$. The quadratic factor $x^2+x+1$ has discriminant $1^2 - 4(1)(1) = 1 - 4 = -3 < 0$, so it has no real roots.

The x-coordinates of the intersection points are 0 and 1.

Find the corresponding y-coordinates using $y=x^2$:

• When $x=0$, $y = (0)^2 = 0$. Intersection point: $(0, 0)$.
• When $x=1$, $y = (1)^2 = 1$. Intersection point: $(1, 1)$.

The region is bounded by the curves between $x=0$ and $x=1$ (or between $y=0$ and $y=1$).

3. Choose the integration variable: Integrating with respect to $x$ is suitable as the region is bounded by vertical lines at $x=0$ and $x=1$. We need to express the curves as $y=f(x)$ and $y=g(x)$. From $y^2 = x$, we get $y = \pm \sqrt{x}$. Since the region is in the first quadrant (from the sketch, bounded between $(0,0)$ and $(1,1)$), we consider the upper branch $y = \sqrt{x}$. From $x^2 = y$, we have $y = x^2$. So, the curves are $y = \sqrt{x}$ and $y = x^2$. The limits are from $x=0$ to $x=1$.

4. Determine upper/lower function: On the interval $[0, 1]$ (between the intersection points), test a value, say $x=0.5$. For $y = \sqrt{x}$, at $x=0.5$, $y = \sqrt{0.5} \approx 0.707$. For $y = x^2$, at $x=0.5$, $y = (0.5)^2 = 0.25$. Since $0.707 > 0.25$, the curve $y = \sqrt{x}$ is above $y = x^2$ on the interval $(0, 1)$. Thus, the upper function is $f(x) = \sqrt{x}$ and the lower function is $g(x) = x^2$ on $[0, 1]$.

5. Set up the integral: The area is given by $\int\limits_{a}^{b} [f(x) - g(x)] dx$ with $a=0$, $b=1$, $f(x)=\sqrt{x}$, and $g(x)=x^2$.

Area A = $\int\limits_{0}^{1} (\sqrt{x} - x^2) dx$

... (A)

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

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

6. Evaluate the integral: Use linearity and the power rule for integration.

A = $\int\limits_{0}^{1} x^{1/2} dx - \int\limits_{0}^{1} x^2 dx$

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

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

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

Apply the limits to each part:

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

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

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

A = $\frac{1}{3}$

... (B)

The area of the region bounded by the two parabolas is $\mathbf{\frac{1}{3}}$ square units.

---

Alternative Method (Integrating with respect to y):

We can also find the area by integrating with respect to $y$. We need to express the curves as functions of $y$: From $x^2 = y$, we get $x = \sqrt{y}$ (right branch in the first quadrant). From $y^2 = x$, we have $x = y^2$. The intersection points are $(0,0)$ and $(1,1)$. The region is bounded between $y=0$ and $y=1$. On the interval $[0, 1]$ for $y$, the curve $x = \sqrt{y}$ is to the right of the curve $x = y^2$. (Test $y=0.5$: $\sqrt{0.5} \approx 0.707$, $(0.5)^2 = 0.25$. Since $0.707 > 0.25$, $\sqrt{y}$ is to the right). Right function $g(y) = \sqrt{y} = y^{1/2}$, left function $h(y) = y^2$. Limits for $y$ are from $0$ to $1$. Area $= \int\limits_{0}^{1} (\sqrt{y} - y^2) dy$ $= \int\limits_{0}^{1} (y^{1/2} - y^2) dy = \left[\frac{y^{3/2}}{3/2} - \frac{y^3}{3}\right]_{0}^{1} = \left[\frac{2}{3} y^{3/2} - \frac{y^3}{3}\right]_{0}^{1}$ $= \left(\frac{2}{3}(1)^{3/2} - \frac{(1)^3}{3}\right) - \left(\frac{2}{3}(0)^{3/2} - \frac{(0)^3}{3}\right) = \left(\frac{2}{3} - \frac{1}{3}\right) - (0 - 0) = \frac{1}{3}$. Both methods give the same result.

Answer:

Given:

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

The y-axis ($x=0$).

The x-axis ($y=0$).

The region is specified to be in the first quadrant ($x \ge 0, y \ge 0$).

To Find:

The area of the region in the first quadrant bounded by the ellipse, the x-axis, and the y-axis.

Solution:

1. Sketch the region:

The ellipse is centered at $(0,0)$ with semi-major axis 3 along the x-axis and semi-minor axis 2 along the y-axis. The lines $x=0$ (y-axis) and $y=0$ (x-axis) are the boundaries along the axes. The region in the first quadrant bounded by the ellipse and the axes is one-fourth of the total area of the ellipse.

(Note: The image should show the quarter of the ellipse in the first quadrant, bounded by the axes).

2. Find intersection points: The ellipse intersects the x-axis ($y=0$) when $\frac{x^2}{9} + \frac{0^2}{4} = 1 \implies \frac{x^2}{9} = 1 \implies x^2 = 9 \implies x = \pm 3$. In the first quadrant, $x=3$. The ellipse intersects the y-axis ($x=0$) when $\frac{0^2}{9} + \frac{y^2}{4} = 1 \implies \frac{y^2}{4} = 1 \implies y^2 = 4 \implies y = \pm 2$. In the first quadrant, $y=2$. The intersection points are $(3, 0)$ on the x-axis and $(0, 2)$ on the y-axis. The origin $(0,0)$ is also a boundary point.

3. Choose the integration variable: We can integrate with respect to $x$ or $y$. Integrating with respect to $x$ from 0 to 3 seems straightforward as the upper boundary is a function of $x$ and the lower boundary is the x-axis ($y=0$). Express the upper boundary (ellipse in first quadrant) as a function of $x$. From $\frac{x^2}{9} + \frac{y^2}{4} = 1$, solve for $y$:

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

y$^2 = \frac{4}{9}(9 - x^2)$

In the first quadrant, $y \ge 0$, so $y = \sqrt{\frac{4}{9}(9 - x^2)} = \frac{2}{3} \sqrt{9 - x^2}$. So, the upper curve is $y = f(x) = \frac{2}{3} \sqrt{9 - x^2}$. The lower curve is $y=0$ (x-axis). The limits of integration for $x$ are from 0 to 3.

4. Set up the integral: The area is given by $\int\limits_{a}^{b} [f(x) - g(x)] dx$. Here $a=0$, $b=3$, $f(x) = \frac{2}{3} \sqrt{9 - x^2}$, and $g(x) = 0$.

Area A = $\int\limits_{0}^{3} \left(\frac{2}{3} \sqrt{9 - x^2} - 0\right) dx$

... (A)

A = $\frac{2}{3} \int\limits_{0}^{3} \sqrt{9 - x^2} dx$

Rewrite $9$ as $3^2$: $A = \frac{2}{3} \int\limits_{0}^{3} \sqrt{3^2 - x^2} dx$. This is of the form $\int \sqrt{a^2 - x^2} dx$ with $a=3$.

5. Evaluate the integral: Use the standard integral formula (from Section 7.10, formula 8): $\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left(\frac{x}{a}\right) + C$.

A = $\frac{2}{3} \left[ \frac{x}{2}\sqrt{3^2 - x^2} + \frac{3^2}{2} \sin^{-1} \left(\frac{x}{3}\right) \right]_{0}^{3}$

A = $\frac{2}{3} \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2} \sin^{-1} \left(\frac{x}{3}\right) \right]_{0}^{3}$

Apply the limits (Upper limit value - Lower limit value):

A = $\frac{2}{3} \left[ \left(\frac{3}{2}\sqrt{9 - (3)^2} + \frac{9}{2} \sin^{-1} \left(\frac{3}{3}\right)\right) - \left(\frac{0}{2}\sqrt{9 - (0)^2} + \frac{9}{2} \sin^{-1} \left(\frac{0}{3}\right)\right) \right]$

A = $\frac{2}{3} \left[ \left(\frac{3}{2}\sqrt{0} + \frac{9}{2} \sin^{-1}(1)\right) - \left(0 + \frac{9}{2} \sin^{-1}(0)\right) \right]$

Using the values $\sin^{-1}(1) = \frac{\pi}{2}$ and $\sin^{-1}(0) = 0$:

A = $\frac{2}{3} \left[ \left(0 + \frac{9}{2} \cdot \frac{\pi}{2}\right) - (0 + 0) \right]$

A = $\frac{2}{3} \left[ \frac{9\pi}{4} \right]$

Simplify the expression:

A = $\frac{\cancel{2}}{3} \cdot \frac{\cancel{9}^{3}\pi}{\cancel{4}^{2}} = \frac{3\pi}{2}$

... (B)

The area of the region in the first quadrant bounded by the ellipse and the axes is $\mathbf{\frac{3\pi}{2}}$ square units.

This is indeed one-fourth of the total area of the ellipse, which is $\pi ab = \pi (3)(2) = 6\pi$. $\frac{1}{4}(6\pi) = \frac{3\pi}{2}$.