6 Applications of Integration Chapter Introduction 6.2 Volume by Cross-Sectional Area; Disk and Washer Methods

6.1 Area Between Curves

We are often interested in knowing the area of a region. Forget momentarily that we addressed this already in Section 5.4 and approach it instead using the technique described in Key Idea 6.0.1.

^†^†margin: (a) (b) (c) Figure 6.1.1: Subdividing a region into vertical slices and approximating the areas with rectangles. Λ

Let $Q$ be the area of a region bounded by continuous functions $f$ and $g$ . If we break the region into many subregions, we have an obvious equation:

Total Area = sum of the areas of the subregions.

The issue to address next is how to systematically break a region into subregions. A graph will help. Consider Figure 6.1.1 (a) where a region between two curves is shaded. While there are many ways to break this into subregions, one particularly efficient way is to “slice” it vertically, as shown in Figure 6.1.1 (b), into $n$ equally spaced slices.

We now approximate the area of a slice. Again, we have many options, but using a rectangle seems simplest. Picking any $x$ -value $c_{i}$ in the $i^{\text{ th}}$ slice, we set the height of the rectangle to be $f(c_{i})-g(c_{i})$ , the difference of the corresponding $y$ -values. The width of the rectangle is a small difference in $x$ -values, which we represent with $\Delta x$ . Figure 6.1.1 (c) shows sample points $c_{i}$ chosen in each subinterval and appropriate rectangles drawn. Each slice has an area approximately equal to $\bigl{(}f(c_{i})-g(c_{i})\bigr{)}\Delta x$ ; hence, the total area is approximately the Riemann Sum

Q\approx\sum_{i=1}^{n}\bigl{(}f(c_{i})-g(c_{i})\bigr{)}\Delta x.

Taking the limit as $n\to\infty$ gives the exact area as $\int_{a}^{b}\bigl{(}f(x)-g(x)\bigr{)}\operatorname{d}\!x.$

Theorem 6.1.1 Area Between Curves

Let $f(x)$ and $g(x)$ be continuous functions defined on $[a,b]$ where $f(x)\geq g(x)$ for all $x$ in $[a,b]$ . The area of the region bounded by the curves $y=f(x)$ , $y=g(x)$ and the lines $x=a$ and $x=b$ is

\int_{a}^{b}\bigl{(}f(x)-g(x)\bigr{)}\operatorname{d}\!x.

Often, we do not know which function is greater (or they switch within the domain of integration). If so, we can say that the area is $\int_{a}^{b}\left\lvert f(x)-g(x)\right\rvert\operatorname{d}\!x$ , which may involve dividing the domain of integration into pieces.

Watch the video:
Finding Areas Between Curves from https://youtu.be/DRFyNHdVgUA

Example 6.1.1 Finding area enclosed by curves

Find the area of the region bounded by $f(x)=\sin x+2$ , $g(x)=\frac{1}{2}\cos(2x)-1$ , $x=0$ and $x=4\pi$ , as shown in Figure 6.1.2.

Solution^†^†margin: Figure 6.1.2: Graphing an enclosed region in Example 6.1.1. Λ The graph verifies that the upper boundary of the region is given by $f$ and the lower bound is given by $g$ . Therefore the area of the region is the value of the integral

	$\displaystyle\int_{0}^{4\pi}\bigl{(}f(x)-g(x)\bigr{)}\operatorname{d}\!x$	$\displaystyle=\int_{0}^{4\pi}\Bigl{(}\sin x+2-\bigl{(}\frac{1}{2}\cos(2x)-1% \bigr{)}\Bigr{)}\operatorname{d}\!x$
		$\displaystyle=-\cos x-\frac{1}{4}\sin(2x)+3x\Big{\|}_{0}^{4\pi}$
		$\displaystyle=12\pi\ \text{units}^{2}.$

Example 6.1.2 Finding area between curves

Find the area of the region enclosed by $y=x^{2}+x-5$ and $y=3x-2$ .

SolutionIt will help to sketch these two functions, as done in Figure 6.1.3. ^†^†margin: Figure 6.1.3: Sketching the region enclosed by $y=x^{2}+x-5$ and $y=3x-2$ in Example 6.1.2. Λ The region whose area we seek is completely bounded by these two functions; they seem to intersect at $x=-1$ and $x=3$ . To check, set $x^{2}+x-5=3x-2$ and solve for $x$ :

	$\displaystyle x^{2}+x-5$	$\displaystyle=3x-2$
	$\displaystyle(x^{2}+x-5)-(3x-2)$	$\displaystyle=0$
	$\displaystyle x^{2}-2x-3$	$\displaystyle=0$
	$\displaystyle(x-3)(x+1)$	$\displaystyle=0$
	$\displaystyle x$	$\displaystyle=-1,\ 3.$

Following Theorem 6.1.1, the area is

	$\displaystyle\int_{-1}^{3}\bigl{(}3x-2-(x^{2}+x-5)\bigr{)}\operatorname{d}\!x$	$\displaystyle=\int_{-1}^{3}(-x^{2}+2x+3)\operatorname{d}\!x$
		$\displaystyle=\left.\left(-\frac{1}{3}x^{3}+x^{2}+3x\right)\right\|_{-1}^{3}$
		$\displaystyle=-\frac{1}{3}(27)+9+9-\left(\frac{1}{3}+1-3\right)$
		$\displaystyle=10\frac{2}{3}=10.\overline{6}$

^†^†margin: Figure 6.1.4: Graphing a region enclosed by two functions in Example 6.1.3. Λ

Example 6.1.3 Finding total area enclosed by curves

Find the total area of the region enclosed by the functions $f(x)=-2x+5$ and $g(x)=x^{3}-7x^{2}+12x-3$ as shown in Figure 6.1.4.

SolutionA quick calculation shows that $f=g$ at $x=1,2$ and 4. One can proceed thoughtlessly by computing $\displaystyle\int_{1}^{4}\bigl{(}f(x)-g(x)\bigr{)}\operatorname{d}\!x$ , but this ignores the fact that on $[1,2]$ , $g(x)>f(x)$ . (In fact, the thoughtless integration returns $-9/4$ , hardly the expected value of an area.) Thus we compute the total area by breaking the interval $[1,4]$ into two subintervals, $[1,2]$ and $[2,4]$ and using the proper integrand in each.

	Total Area	$\displaystyle=\int_{1}^{2}\bigl{(}g(x)-f(x)\bigr{)}\operatorname{d}\!x+\int_{2% }^{4}\bigl{(}f(x)-g(x)\bigr{)}\operatorname{d}\!x$
		$\displaystyle=\int_{1}^{2}\bigl{(}x^{3}-7x^{2}+14x-8\bigr{)}\operatorname{d}\!% x+\int_{2}^{4}\bigl{(}-x^{3}+7x^{2}-14x+8\bigr{)}\operatorname{d}\!x$
		$\displaystyle=\frac{5}{12}+\frac{8}{3}=\frac{37}{12}\ \text{units}^{2}.$

The previous example makes note that we are expecting area to be positive. When first learning about the definite integral, we interpreted it as “signed area under the curve,” allowing for “negative area.” That doesn’t apply here; area is to be positive.

The previous example also demonstrates that we often have to break a given region into subregions before applying Theorem 6.1.1. The following example shows another situation where this is applicable, along with an alternate view of applying the Theorem.

^†^†margin: Figure 6.1.5: Graphing a region for Example 6.1.4. Λ

Example 6.1.4 Finding area: integrating with respect to $y$

Find the area of the region enclosed by the functions $y=\sqrt{x}+2$ , $y=-(x-1)^{2}+3$ and $y=2$ , as shown in Figure 6.1.5.

SolutionWe give two approaches to this problem. In the first approach, we notice that the region’s “top” is defined by two different curves. On $[0,1]$ , the top function is $y=\sqrt{x}+2$ ; on $[1,2]$ , the top function is $y=-(x-1)^{2}+3$ . Thus we compute the area as the sum of two integrals:

	Total Area	$\displaystyle=\int_{0}^{1}\Bigl{(}\bigl{(}\sqrt{x}+2\bigr{)}-2\Bigr{)}% \operatorname{d}\!x+\int_{1}^{2}\Bigl{(}\bigl{(}-(x-1)^{2}+3\bigr{)}-2\Bigr{)}% \operatorname{d}\!x$
		$\displaystyle=2/3+2/3=4/3.$

The second approach is clever and very useful in certain situations. We are used to viewing curves as functions of $x$ ; we input an $x$ -value and a $y$ -value is returned. Some curves can also be described as functions of $y$ : input a $y$ -value and an $x$ -value is returned. We can rewrite the equations describing the boundary by solving for $x$ :

	$\displaystyle y=\sqrt{x}+2$	$\displaystyle\quad\Rightarrow\quad x=(y-2)^{2}$
	$\displaystyle y=-(x-1)^{2}+3$	$\displaystyle\quad\Rightarrow\quad x=\sqrt{3-y}+1.$

^†^†margin: Figure 6.1.6: The region used in Example 6.1.4 with boundaries relabeled as functions of

y

. Λ

Figure 6.1.6 shows the region with the boundaries relabeled. A horizontal rectangle is also pictured. The width of the rectangle is a small change in $y$ : $\Delta y$ . The height of the rectangle is a difference in $x$ -values. The “top” $x$ -value is the largest value, i.e., the rightmost. The “bottom” $x$ -value is the smaller, i.e., the leftmost. Therefore the height of the rectangle is

\bigl{(}\sqrt{3-y}+1\bigr{)}-(y-2)^{2}.

The area is found by integrating the above function with respect to $y$ with the appropriate bounds. We determine these by considering the $y$ -values the region occupies. It is bounded below by $y=2$ , and bounded above by $y=3$ . That is, both the “top” and “bottom” functions exist on the $y$ interval $[2,3]$ . Thus

	Total Area	$\displaystyle=\int_{2}^{3}\bigl{(}\sqrt{3-y}+1-(y-2)^{2}\bigr{)}\operatorname{% d}\!y$
		$\displaystyle=\Bigl{(}-\frac{2}{3}(3-y)^{3/2}+y-\frac{1}{3}(y-2)^{3}\Bigr{)}% \Big{\|}_{2}^{3}$
		$\displaystyle=4/3.$

The important thing to notice is that by integrating with respect to $y$ instead of $x$ , we only had to do one integral and did not need to find the point at which to switch from one integration to another.

This calculus-based technique of finding area can be useful even with shapes that we normally think of as “easy.” Example 6.1.5 computes the area of a triangle. While the formula “ $\frac{1}{2}\times\text{base}\times\text{height}$ ” is well known, in arbitrary triangles it can be nontrivial to compute the height. Calculus makes the problem simple.

Example 6.1.5 Finding the area of a triangle

Compute the area of the regions bounded by the lines ^†^†margin: Figure 6.1.7: Graphing a triangular region in Example 6.1.5. Λ $y=3-x$ , $y=x+1$ and $y=5x-15$ , as shown in Figure 6.1.7.

SolutionRecognize that there are two “bottom” functions to this region, causing us to use two definite integrals.

	Total Area	$\displaystyle=\int_{1}^{3}\bigl{(}(x+1)-(3-x)\bigr{)}\operatorname{d}\!x+\int_% {3}^{4}\bigl{(}(x+1)-(5x-15)\bigr{)}\operatorname{d}\!x$
		$\displaystyle=4+2=6.$

We can also approach this by converting each function into a function of $y$ . This also requires 2 integrals, so there isn’t really any advantage to doing so. We do it here for demonstration purposes.

The “top” function is always $x=\frac{y}{5}+3$ while there are two “bottom” functions: $x=3-y$ and $x=y-1$ . Being mindful of the proper integration bounds, we have

	Total Area	$\displaystyle=\int_{0}^{2}\left(\left(\frac{y}{5}+3\right)-(3-y)\right)% \operatorname{d}\!y+\int_{2}^{5}\left(\left(\frac{y}{5}+3\right)-(y-1)\right)% \operatorname{d}\!y$
		$\displaystyle=\frac{12}{5}+\frac{18}{5}=6.$

Of course, the final answer is the same (and we see that integrating with respect to $x$ was probably easier, since it avoided fractions).

In the next section we apply Key Idea 6.0.1 to finding the volumes of certain solids.

Exercises 6.1

Terms and Concepts

1.

T/F: The area between curves is always positive.
2.

T/F: Calculus can be used to find the area of basic geometric shapes.
3.

In your own words, describe how to find the total area enclosed by $y=f(x)$ and $y=g(x)$ .
4.

Describe a situation where it is advantageous to find an area enclosed by curves through integration with respect to $y$ instead of $x$ .

Problems

In Exercises 5–10., find the area of the shaded region in the given graph.

In Exercises 11–24., find the area of the region bounded by the given curves.

11.

$f(x)=2x^{2}+5x-3$ , $g(x)=x^{2}+4x-1$
12.

$f(x)=x^{2}-3x+2$ , $g(x)=-3x+3$
13.

$f(x)=\sin x$ , $g(x)=2x/\pi$
14.

$f(x)=x^{3}-4x^{2}+x-1$ , $g(x)=-x^{2}+2x-4$
15.

$f(x)=x$ , $g(x)=\sqrt{x}$
16.

$f(x)=-x^{3}+5x^{2}+2x+1$ , $g(x)=3x^{2}+x+3$
17.

$x=2y^{2}$ , $x+y=1$
18.

$x=y^{2}-1$ , $x=1-y^{2}$
19.

$4x+y^{2}=12$ , $x=y$
20.

$x=y^{2}-4y$ , $x=2y-y^{2}$
21.

$y=2x$ , $y=5x$ , $x=3$ .
22.

$y=-x+1$ , $y=3x+6$ , $x=2$ $x=-1$ .
23.

$y=x^{2}-2x+5$ , $y=5x-5$ .
24.

$y=2x^{2}+2x-5$ , $y=x^{2}+3x+7$ .

25.

The functions $f(x)=\cos(x)$ and $g(x)=\sin x$ intersect infinitely many times, forming an infinite number of repeated, enclosed regions. Find the areas of these regions.
26.

The functions $f(x)=\cos(2x)$ and $g(x)=\sin x$ intersect infinitely many times, forming an infinite number of repeated, enclosed regions. Find the areas of these regions.

In Exercises 27–32., find the area of the enclosed region in two ways:

(a)

by treating the boundaries as functions of $x$ , and
(b)

by treating the boundaries as functions of $y$ .

In Exercises 33–36., find the area triangle formed by the given three points.

33.

$(1,1)$ , $(2,3)$ , and $(3,3)$
34.

$(-1,1)$ , $(1,3)$ , and $(2,-1)$
35.

$(1,1)$ , $(-1,3)$ , and $(3,3)$
36.

$(0,0)$ , $(2,5)$ , and $(5,2)$