2.1 Area Between Curves

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We compute the area between two curves.

Area Between Curves

In this section, we use definite integrals to find the area of a region in the $xy$ -plane bounded by two or more curves. Recall that if $f(x)$ is a positive, continuous function over the interval $[a, b]$ , then the area bounded below the curve $y= f(x)$ , above the $x$ -axis, and between the vertical lines $x = a$ and $x = b,$ is given by the definite integral, $\int _a^b f(x) \, dx,$ as shown in the figure below.

Combining this with a clever subtraction, we can find the area between two curves. If $f(x)$ and $g(x)$ are positive, continuous functions on the interval $[a,b]$ with $f(x) \geq g(x)$ , then the area between the curves is the difference between the area under $f(x)$ and the area under $g(x)$ .

We can write this formula as a single integral, as presented in the theorem below.

Suppose $f(x)$ and $g(x)$ are continuous on the interval $[a,b]$ . If $f(x) \geq g(x)$ for all $x$ in the interval $[a,b]$ , then the area of the region bounded by the graphs of the curves $y = f(x)$ and $y=g(x)$ and the vertical lines $x=a$ and $x=b$ is given by $\text {Area} = \int _a^b \left [f(x) -g(x)\right ] \; dx.$

The assumption that $f(x)$ and $g(x)$ are positive is unnecessary due to the following line of reasoning. The area of the region between $f(x)$ and $g(x)$ is the same as the area between $f(x) + k$ and $g(x) + k$ for any constant $k$ , i.e., shifting the region does not change its area. If $k$ is large enough then both $f(x) + k$ and $g(x) + k$ will be positive. The result follows by noting that $(f(x) + k) - (g(x) + k) = f(x) - g(x)$ .

Non-intersecting curves

In this section, the left and right vertical boundaries of the region are given, and the curves do not intersect between these vertical boundaries.

example 1 Find the area of the region between the parabola $y = x^2$ and the horizontal line $y = -2$ from $x = -1$ to $x = 2$ .
Since $x^2 > -2$ over the interval $[-1, 2]$ , the area of the region is $\begin{align*} \int_{-1}^2 \left[x^2 - (-2)\right] \; dx &= \int_{-1}^2 \left[x^2 + 2 \right] \; dx\\ &= \frac{x^3}{3} + 2x \bigg|_{-1}^2\\ &= \left(\frac83 + 4\right) - \left(-\frac13 - 2\right)\\ &= 9. \end{align*}$

(problem 1) Find the area of the region between the parabola $y = x^2 + 2$ and the horizontal line $y =1$ from $x = 0$ to $x = 3$ .
The top curve is: $y = \answer {x^2 + 2}$ .
The bottom curve is: $y = \answer {1}$ .
The definite integral that represents this area is:

$\dis \int _0^3 (-1-x^2)\; dx$
$\dis \int _0^3 (x^2 - 1) \; dx$
$\dis \int _0^3 (x^2 + 1) \; dx$

The area of the region is:

6 9 12

example 2 Find the area of the region bounded by curves $y = e^x$ and $y = \sin (x)$ from $x = 0$ to $x = \pi /2$ .

Since $e^x > \sin (x)$ over the interval $[0,\pi /2]$ , the area of the indicated region is

$\begin{align*} \int_0^{\pi/2} \left[e^x - \sin(x)\right] \; dx &= \left[e^x + \cos(x) \right] \bigg|_0^{\pi/2} \\ &= \left(e^{\pi/2} + 0 \right) - \left(1+1\right)\\ & = e^{\pi/2} - 2 \approx 2.81. \end{align*}$

(problem 2a) Find the area of the region bounded between the curves $y = 2e^x$ and $y = \cos (x)$ from $x = 0$ to $x = \pi /2$ .
The top curve is: $y = \answer {2e^x}$ .
The bottom curve is: $y = \answer {\cos (x)}$ .
The definite integral that represents this area is: $\displaystyle {\int _0^{\pi /2}} \answer {2e^x - \cos (x)} \; dx$
The area of the region is: $\answer {2e^{\pi /2} - 3}$ .

(problem 2b) Find the area of the region bounded between the curves $y = e^{2x}$ and $y = \sin (3x)$ from $x = 0$ to $x = \pi /3$ .
The top curve is: $y = \answer {e^{2x}}$
The bottom curve is: $y = \answer {\sin (3x)}$ .
The definite integral that represents this area is: $\displaystyle {\int _0^{\pi /3}} \answer {e^{2x} - \sin (3x)} \; dx$
The area of the region is: $\answer {e^{2\pi /3}/2 - 7/6}$ .

example 3 Find the area of the region bounded by curves $y = 5 - x^2$ and $y = x-4$ over the interval $[1,2]$ .

Since $5 - x^2 > x-4$ over the interval $[1,2]$ , the area is

$\begin{align*} \int_1^2 \left[(5-x^2) - (x-4)\right] \; dx &= \int_1^2 \left(9-x-x^2\right) \; dx \\ &= \left(9x - \frac12 x^2 - \frac13 x^3 \right) \bigg|_1^2 \\ &= \left(18 - 2 - \frac83 \right) - \left(9 - \frac12 - \frac13 \right)\\ &= 7 - \frac73 + \frac12 \\ & = \frac{31}{6}. \end{align*}$

(problem 3a) Find the area of the region bounded between the curves $y = x^2$ and $y = x - 2$ from $x = 0$ to $x = 2$ .
The top curve is: $y = \answer {x^2}$ .
The bottom curve is: $y = \answer {x - 2}$ .
The definite integral that represents this area is:

$\dis \int _0^2 (-x^2 +x -2)\; dx$
$\dis \int _0^2 (x^2 +x - 2) \; dx$
$\dis \int _0^2 (x^2 -x + 2) \; dx$

The area of the region is:

2 6 14/3

(problem 3b) Find the area of the region bounded between the curves $y = (2x+1)^3$ and $y = (2x+1)^2$ from $x = 0$ to $x = 1$ .
The top curve is: $y = \answer {(2x+1)^3}$ .
The bottom curve is: $y = \answer {(2x+1)^2}$ .
The definite integral that represents this area is:

$\int _0^1 \left [(2x+1)^2 -(2x+1)^3\right ]\; dx$
$\int _0^1 \left [(2x+1)^3 -(2x+1)^2\right ] \; dx$
$\int _0^1 (2x+1) \; dx$

The area of the region is:

use $u$ -substitution

17/3 9 10

Intersecting Curves

In this section, one or both of the endpoints of integration will be a point of intersection of the curves forming the top and bottom boundaries of the region. To find the points of intersection of the curves $y = f(x)$ and $y = g(x)$ , we solve the equation $f(x) = g(x)$ .

example 4 Find the area of the region bounded by the curves $y = \cos (x)$ and $y = \sin (x)$ from $x = 0$ to $x = \frac {\pi }{4}$ .

Since $\cos (x) \geq \sin (x)$ on the interval $\left [0, \frac {\pi }{4}\right ]$ (see the figure below), the area of the region is given by

$\begin{align*} \int_0^{\pi/4} [\cos(x) - \sin(x)] \; dx &= [\sin(x) + \cos(x)]\Big|_0^{\pi/4} \\ &= [\sin(\pi/4) + \cos(\pi/4)] - [\sin(0) + \cos(0)]\\ &= \left(\tfrac{\sqrt 2}{2} + \tfrac{\sqrt 2}{2} \right) - (0 + 1) \\ &= \sqrt 2 - 1. \end{align*}$

(problem 4) Find the area between the curves $y = \sin (x)$ and $y = \cos (x)$ over the interval $(\pi /4, 5\pi /4)$ .

The top curve is: $y = \answer {\sin (x)}$ .
The bottom curve is: $y = \answer {\cos (x)}$ .
The definite integral that represents this area is: $\displaystyle {\int _{\pi /4}^{5\pi /4}} \answer {\sin (x) - \cos (x)} \; dx$
The area of the region is: $\answer {2\sqrt 2}$ .

In some cases, the left and right boundaries of the region are determined by points of intersection between the curves in question.

example 5 Find the area of the region bounded by the curves $y = \sqrt x$ and $y = x^2$ .

We find the points of intersection of the curves by solving the equation $\sqrt x = x^2$ . We can square both sides, move the terms to one side, and factor, but the elementary nature of this equation allows up to find the two solutions by inspection. The curves intersect at $x = 0$ and at $x = 1$ . Noting that $\sqrt x \geq x^2$ on the interval $[0,1]$ , we can calculate the area of the region as

$\begin{align*} \int_0^1 \left(\sqrt x - x^2 \right) \; dx &= \int_0^1 \left(x^{1/2} - x^2 \right) \; dx \\ &= \left(\frac23 x^{3/2} - \frac13 x^3 \right) \bigg|_0^1 \\ &= \left( \tfrac23 - \tfrac13 \right) - (0-0) \\ &= \frac13. \end{align*}$

(problem 5a) Find the area bounded by the line $y = 4$ and the parabola $y = x^2$ .

The points of intersection (from smallest to largest) are: $x = \answer {-2}$ and $x = \answer {2}$ .
The top curve is: $y = \answer {4}$ .
The bottom curve is: $y = \answer {x^2}$ .
The area bounded by the curves is: $\answer {32/3}$ .

(problem 5b) Find the area bounded by the line $y = x$ and the parabola $y = x^2$ .

The points of intersection (from smallest to largest) are: $x = \answer {0}$ and $x = \answer {1}$ .
The top curve is: $y = \answer {x}$ .
The bottom curve is: $y = \answer {x^2}$ .
The area bounded by the curves is: $\answer {1/6}$ .

example 6 Find the area of the region between the parabola $y = x^2 + x - 3$ and the line $y = 2x - 1$ .
The left and right boundaries of the region between the line and the parabola are the points of intersection of the two curves. To find these points of intersection, we set them equal and solve for $x$ : $x^2 + x - 3 = 2x-1.$ Moving the terms to the left gives $x^2 -x -2 = 0.$ Then factor: $(x+1)(x-2) = 0,$ which gives $x = -1 \;\text {and}\; x= 2$ as the solutions.

To determine which curve is on top, we can use a test point in the interval $(-1, 2)$ . If $x = 0$ then the $y$ -coordinate on the line is $-1$ and the $y$ -coordinate on the parabola is $-3$ . Since $-1 > -3$ , the line is the top curve and the parabola is the bottom curve (see the figure below).

$\begin{align*} \int_{-1}^2 \left[(2x - 1) - (x^2 + x - 3)\right] \; dx &= \int_{-1}^2 \left(2 + x - x^2\right) \; dx \\ &= \left(2x + \frac12 x^2 - \frac13 x^3 \right) \bigg|_{-1}^2 \\ &= \left(4 + 2 - \frac83 \right) - \left(-2 + \frac12 + \frac13 \right)\\ &= 8 - \frac93 - \frac12 \\ & = \frac{9}{2}. \end{align*}$

(problem 6a) Find the area of the region bounded by the line $y = 5x-2$ and the parabola $y = x^2 +2$ .
The points of intersection (from smallest to largest) are at: $x = \answer {1}$ and $x = \answer {4}$ .
The top curve is: $y = \answer {5x-2}$ .
The bottom curve is: $y = \answer {x^2+2}$ .
The area bounded by the curves is: $\answer {9/2}$ .

(problem 6b) Find the area of the region bounded by the parabolas $y = x^2 - 4$ and $y = 8 - 2x^2$ .
The points of intersection (from smallest to largest) are: $x = \answer {-2}$ and $x = \answer {2}$ .
The top curve is: $y = \answer {8 - 2x^2}$ .
The bottom curve is: $y = \answer {x^2-4}$ .
The area bounded by the curves is: $\answer {32}$ .

In the next example, we explore the area bounded between two curves which are intertwined.

example 7 Find the area between the curves $y = \cos (x)$ and $y=\sin (x)$ over the interval $[0, \pi ]$ .

Since $\sin (\pi /4) = \cos (\pi /4) = \sqrt 2 / 2$ , the curves intersect at $\pi /4$ . This is the only point of intersection of the curves in the interval $[0, \pi ]$ . Observe that $\cos (x) \geq \sin (x)$ on the interval $[0, \pi /4],$ and that the reverse is true on the interval $[\pi /4, \pi ]$ . Thus, the area bounded between the two curves from $x = 0$ to $x = \pi$ is in the form of two separate regions, and hence we will need two separate integrals to compute the total area: $\text {Area} = \int _0^{\pi /4} \left [\cos (x) - \sin (x)\right ] \; dx + \int _{\pi /4}^\pi \left [\sin (x) -\cos (x)\right ] \; dx.$ We will compute these integrals separately and then add the results. For the first integral,

$\begin{align*} \int_0^{\pi/4} \left[\cos(x) - \sin(x)\right] \; dx &= \left[\sin(x) + \cos(x)\right]\bigg|_0^{\pi/4}\\ &= \left[\sin(\tfrac{\pi}{4}) + \cos(\tfrac{\pi}{4})\right]-\left[\sin(0) + \cos(0)\right]\\ &= \tfrac{\sqrt 2}{2} + \tfrac{\sqrt 2}{2} - 1\\ &= \sqrt{2} - 1. \end{align*}$ For the second integral, $\begin{align*} \int_{\pi/4}^\pi \left[\sin(x) -\cos(x)\right] \; dx &= \left[-\cos(x) - \sin(x) \right]\bigg|_{\pi/4}^{\pi} \\ &= \left[\cos(x) + \sin(x)\right]\bigg|_{\pi}^{\pi/4}\\ &= \left[\cos(\tfrac{\pi}{4}) + \sin(\tfrac{\pi}{4})\right]-\left[\cos(\pi) + \sin(\pi)\right]\\ &= \tfrac{\sqrt 2}{2} + \tfrac{\sqrt 2}{2} - (-1)\\ &= \sqrt{2} + 1. \end{align*}$ To obtain the total area, we now add the values of these definite integrals: $\text {Area} = (\sqrt 2 - 1) + (\sqrt 2 + 1) = 2\sqrt {2}.$

(problem 7) Find the area bounded between the curves $y = x^3$ and $y = x$ .
The $x$ -coordinates of the points of intersection are (in increasing order): $x= \answer {-1}, \quad x = \answer {0} \quad \mbox {and} \quad x = \answer {1}$

Two integrals are needed

The area is: $\answer {1/2}$

In the next example, we find the area of a region bounded by three curves. In this situation, we will need to use two separate integrals to determine the total area.

example 8 Find the area of the region bounded by the curves $y = 2-x^2, y = -2x-1,$ and $y = x$ , as shown in the figure below.

First, we need to find the points of intersection of the the curves. The intersection of the lines is at $-2x-1 = x$ $x = -\tfrac 13.$ Now the intersection of the parabola and the first line: $2-x^2 = -2x-1$ $0 = x^2 -2x -3$ $0 = (x+1)(x-3)$ $x = -1, \; x = 3.$

And the intersection of the parabola and the second line:

$2-x^2 = x$ $0 =x^2 +x -2$ $0 =(x+2)(x-1)$ $x = -2, \; x = 1.$

We can now see that the region is bounded by $x = -1$ on the left, and $x = 1$ on the right. Furthermore, the parabola, $y = 2 - x^2$ is the top curve, while the line $y = -2x-1$ is the bottom curve from $x = -1$ to $x = -\frac 13$ , and the line $y = x$ is the bottom curve from $x = -\frac 13$ to $x = 1$ . Hence the area is the sum of two integrals: $\text {Area} = \int _{-1}^{-1/3} \left [(2 - x^2) - (-2x-1)\right ] \; dx + \int _{-1/3}^1 \left [(2 - x^2) - (x)\right ] \; dx.$ The value of the first integral is

$\begin{align*} \int_{-1}^{-1/3} \left[(2 - x^2) - (-2x-1)\right] \; dx &= \int_{-1}^{-1/3} \left(3 - x^2 +2x\right) \; dx \\ &= \left(3x - \frac{x^3}{3} + x^2 \right)\bigg|_{-1}^{-1/3} \\ &= \left(-1 + \tfrac{1}{81} + \tfrac19 \right) - \left(-3 + \tfrac13 + 1 \right) \\ &= \tfrac{64}{81}. \end{align*}$

And the value of the second integral is

$\begin{align*} \int_{-1/3}^{1} \left[(2 - x^2) - (x)\right] \; dx &= \int_{-1/3}^{1} \left(2 - x^2 - x\right) \; dx \\ &= \left(2x - \frac{x^3}{3} -\frac{x^2}{2} \right)\bigg|_{-1/3}^{1} \\ &= \left(2 - \tfrac{1}{3} - \tfrac12 \right) - \left(-\tfrac23 + \tfrac{1}{81} -\tfrac{1}{18} \right) \\ &= \tfrac{152}{81}. \end{align*}$ Thus, the total area of the region is the sum of these two integrals: $\text {Area} = \frac {64}{81} + \frac {152}{81} = \frac {8}{3}.$

(problem 8) Find the area of the shaded region in the figure below.

The area of the shaded region is $\answer {8/3}$ .

Integrating with respect to $y$

In our next example, we consider curves of the form $x = f(y)$ . In this case, the formula for area is similar to the $y = f(x)$ situation. Instead of “top - bottom”, we have “right curve - left curve” as the integrand. Thus, if $x = f(y)$ and $x = g(y)$ are two curves with $f(y) \geq g(y)$ on the interval from $y = c$ to $y = d$ , then the area between them is given by the formula $\text {Area} = \int _c^d [f(y) - g(y)] \; dy.$

example 9 Find the area between the curves $x = y^2 + y$ and $x = 2y + 6$ .
First, we find the $y$ -coordinates of the points of intersection of the line $x = 2y+6$ and the parabola $x=y^2 + y$ (which opens to the right), by setting these formulas equal and solving for $y$ : $y^2 + y = 2y + 6$ $y^2 - y - 6 = 0$ $(y+2)(y-3) = 0$ $y = -2 \;\;\; \text {or} \;\;\; y = 3.$

The bigger, “right”, function is the line $x = 2y+6$ , as we can see by plugging in any value between $y = -2$ and $y=3$ . The area is given by

$\begin{align*} \text{Area} &= \int_{-2}^3 \left[(2y+6) - (y^2+y) \right] \; dy \\ &= \int_{-2}^3 \left(y - y^2 + 6\right) \; dy \\ &= \left(\frac{y^2}{2} - \frac{y^3}{3} + 6y \right) \bigg|_{-2}^{3} \\ &= \left(\tfrac92 - 9 + 18\right) - \left(2 + \tfrac83 - 12\right)\\ &= \tfrac{125}{6} \end{align*}$

(problem 9) Find the area of the region enclosed by the parabolas $x = y^2 - 1$ and $x = 7 - y^2$ .
The endpoints of integration are: $c = \answer {-2}\;\;\;$ and $\;\;\; d = \answer {2}$ .
The right curve is: $x = \answer {7 - y^2}$ and
the left curve is: $x = \answer {y^2 - 1}$ .
The definite integral that represents this area is: $\displaystyle {\int _{-2}^2} \answer {8 - 2y^2} \; dy$
The area of the region is: $\answer {64/3}$ .

Here is a detailed, lecture style video on the area between curves:

Area Between Curves

Non-intersecting curves

Intersecting Curves

Integrating with respect to y

Integrating with respect to $y$