9.1 The Area Between Two Curves

Suppose two given curves $y = f (x)$ and $y = g (x)$ intersect at $x = a$ and $x = b$ and $f (x) \geq g (x) (a < x < b) .$ To find the area of the region bounded between these two curves, consider a vertical rectangle as shown in Figure 1. The height of this rectangle is $f (x) - g (x)$ and its width is $d x$ . The area of this infinitesimal rectangle is $d A = [f (x) - g (x)] d x,$ and the total area is $A = \int_{a}^{b} d A = \int_{a}^{b} [f (x) - g (x)] d x .$

Notice that we integrate from the smaller limit $a$ to the larger one $b$ , and $f (x)$ is greater than $g (x)$ in the entire interval of integration; otherwise the integral (and the area) becomes negative which is meaningless.
Because the curves $y = f (x)$ and $y = g (x)$ intersect at $x = a$ and $x = b$ , they have the same $y$ -values at these points. If in a problem $a$ and $b$ are not given, we can find them from solving the equation $f (x) = g (x)$ .

Example 1

Find the area of the region bounded by the curves

y = \frac{1}{2} x^{2}

and

y = 2

Figure 2

Solution 1

Using vertical rectangles, we realize that the height of the typical rectangle is

2 - \frac{1}{2} x^{2}

and its width is

d x

. So the element of area is

d A = (2 - \frac{1}{2} x^{2}) d x

Figure 3

Because these two curves intersect at $x = \pm 2$ : $\frac{1}{2} x^{2} = 2 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2,$ the total area bounded by $y = x^{2} / 2$ and $y = 2$ is then $\begin{aligned} A = \int_{- 2}^{2} d A & = \int_{- 2}^{2} (2 - \frac{1}{2} x^{2}) d x \\ = {[2 x - \frac{1}{6} x^{3}]}_{x = - 2}^{x = 2} \\ = 2 (4 - \frac{1}{6} \cdot 8) \\ = \frac{16}{3} . \end{aligned}$ Because the area is symmetric with respect to the $y$ -axis, we could simply integrate from $0$ to $2$ and multiply the result by $2$ . That is, $\begin{aligned} A & = 2 \int_{0}^{2} (2 - \frac{1}{2} x^{2}) d x \\ = 2 {[2 x - \frac{1}{6} x^{3}]}_{x = - 2}^{x = 2} \\ = 2 (4 - 8 / 6) = 16 / 3. \end{aligned}$

Sometimes, instead of vertical rectangles, we could or we should consider horizontal rectangles. To find the area of the region bounded by the curves $x = h (y)$ and $x = k (y)$ (with $k (y) \leq h (y)$ ) and the horizontal lines $y = c$ and $y = d$ ( $c \leq d$ ) (Figure 4), we consider horizontal rectangles of length $h (y) - k (y)$ and width $d y$ . The element of area is then $d A = [h (y) - k (y)] d y$ and the total area is $A = \int_{c}^{d} d A = \int_{c}^{d} [h (y) - k (y)] d y .$

For instance, in the above example, if we use horizontal rectangles, the length of the rectangle is the $x$ value (in terms of $y$ ) of the right curve ( $x = \sqrt{2 y}$ ) minus the $x$ value of the left curve ( $x = - \sqrt{2 y})$ . Because the width of the rectangle is $d y$ , the element of area is $\begin{aligned} d A & = [\sqrt{2 y} - (- \sqrt{2 y})] d y \\ = 2 \sqrt{2} \sqrt{y} d y . \end{aligned}$ As the horizontal rectangle sweeps across the region, $y$ varies between $0$ and $2$ . Therefore, the total area is $\begin{aligned} A & = \int_{0}^{2} 2 \sqrt{2} \sqrt{y} d y \\ = 2 \sqrt{2} \int_{0}^{2} y^{1 / 2} d y \\ = \frac{2 \sqrt{2}}{3 / 2} {y^{3 / 2} |}_{y = 0}^{y = 2} \\ = \frac{4 \sqrt{2}}{3} \sqrt{8} \\ = \frac{16}{3} . \end{aligned}$

Example 2

Find the area of the region bounded by the curves

y^{2} = x

and

y = x - 2

Figure 5

Solution 2

First we need to find where these curves intersect

x = y^{2}, x = y + 2 \Rightarrow y^{2} = y + 2

y^{2} - y - 2 = 0

\Rightarrow y = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{1 \pm \sqrt{1 + 4 (2)}}{2}

\Rightarrow y = 2, - 1

x = {y^{2} |}_{y = 2} = 4

x = {y^{2} |}_{y = - 1} = 1

Therefore, these curves intersect at

(2, 4)

and

(1, - 1)

To compute the area of the region, it is easier to consider horizontal thin rectangles (Figure 6). Because $x = y + 2$ is on the right, the length of the typical rectangle is $y + 2 - y^{2}$ and its width is $d y$ . Therefore, the element of area is $d A = (y + 2 - y^{2}) d y .$ Because $y$ varies between $- 1$ and $2$ , the area is $\begin{aligned} A & = \int_{- 1}^{2} d A = \int_{- 1}^{2} (y + 2 - y^{2}) d y \\ = {[\frac{1}{2} y^{2} + 2 y - \frac{y^{3}}{3}]}_{- 1}^{2} \\ = \frac{10}{3} - (- \frac{7}{6}) = \frac{9}{2} . \end{aligned}$

Figure 6

To compute the area, we can also consider thin vertical rectangles (Figure 7). However, we need to divide the region into two subregions. If $x > 1$ , then the height of the rectangle is $\sqrt{x} - (x - 2)$ and if $0 < x < 1$ , the height of the rectangle is $\sqrt{x} - (- \sqrt{x}) = 2 \sqrt{x}$ . Therefore $d A = {\begin{cases} \sqrt{x} - x + 2 & (if 1 \leq x \leq 4) \\ 2 \sqrt{x} & (if 0 \leq x \leq 1) \end{cases}$ and the total area is $\begin{aligned} A & = \int_{0}^{4} d A \\ = \int_{0}^{1} 2 \sqrt{x} d x + \int_{1}^{4} (\sqrt{x} - x + 2) d x \\ = {[\frac{4}{3} x^{3 / 2}]}_{0}^{1} + {[\frac{2}{3} x^{3 / 2} - \frac{1}{2} x^{2} + 2 x]}_{1}^{4} \\ = \frac{4}{3} + \frac{16}{3} - \frac{13}{6} = \frac{9}{2} . \end{aligned}$

Figure 7

Example 3

Find the area between the graphs of

y = \cos 2 x

and

y = \sin x

on the interval

[0, \frac{π}{2}]

Solution 3

It is a good idea to sketch these two curves. To graph

y = \cos 2 x

we start with the graph of

y = \cos x

and compress it horizontally by a factor of

2

(Figure 8).

Figure 8: Graphs of

\sin x

\cos 2 x

, and

\cos x

As we can see from Figure 9, the region between the curves consists of two parts. In one part, the curve $y = \cos 2 x$ is the upper curve and in the other part, the curve $y = \sin x$ is the upper one. To find exactly where these two curves intersect, we must solve the equation $\cos 2 x = \sin x$ . We do this by writing

$\begin{aligned} \cos 2 x & = \sin x \\ \cos^{2} x - \sin^{2} x & = \sin x (\cos 2 x = \cos^{2} x - \sin^{2} x) \\ 1 - 2 \sin^{2} x & = \sin x (\cos^{2} x + \sin^{2} x = 1) \\ 2 \sin^{2} x + \sin x - 1 & = 0 \end{aligned}$ The above equation is a quadratic equation in terms of $\sin x$ . That is, if $u = \sin x$ , then $2 u^{2} + u - 1 = 0$ $\Rightarrow u = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- 1 \pm \sqrt{1 + 8}}{4}$ so $\sin x = - 1 or \sin x = \frac{1}{2} .$ $\sin x = \frac{1}{2} \Rightarrow x = \frac{π}{6}$ The equation $\sin x = - 1$ has no solution in the interval $[0, \frac{π}{2}]$ .

Figure 9: Finding area between the graphs.

Therefore $d A = {\begin{cases} (\cos 2 x - \sin x) d x & 0 \leq x \leq \frac{π}{6} \\ (\sin x - \cos 2 x) d x & \frac{π}{6} \leq x \leq \frac{π}{2} \end{cases}$ and the desired area is: $\begin{aligned} \int_{0}^{\frac{π}{6}} d A + \int_{\frac{π}{6}}^{\frac{π}{2}} d A \\ = \int_{0}^{\frac{π}{6}} (\cos 2 x - \sin x) d x + \int_{\frac{π}{6}}^{\frac{π}{2}} (\sin x - \cos 2 x) d x \\ = {[\frac{1}{2} \sin 2 x + \cos 2 x]}_{0}^{\frac{π}{6}} + {[- \cos x - \frac{1}{2} \sin 2 x]}_{\frac{π}{6}}^{\frac{π}{2}} \\ = (\frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}) - (0 + 1) + (- (0) - \frac{1}{2} (0)) - (\frac{- \sqrt{3}}{2} - \frac{1}{2} \cdot \frac{\sqrt{3}}{2}) \\ = \frac{3 \sqrt{3}}{2} - 1. \end{aligned}$