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5.1: Area Between Two Curves

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

A student will be able to:

• Compute the area between two curves with respect to the $x-$ and $y-$axes.

In the last chapter, we introduced the definite integral to find the area between a curve and the $x-$ axis over an interval $[a, b].$ In this lesson, we will show how to calculate the area between two curves.

Consider the region bounded by the graphs $f$ and $g$ between $x = a$ and $x = b,$ as shown in the figures below. If the two graphs lie above the $x-$axis, we can interpret the area that is sandwiched between them as the area under the graph of $g$ subtracted from the area under the graph $f.$

Therefore, as the graphs show, it makes sense to say that

[Area under $f$ (Fig. 1a)] $-$ [Area under $g$ (Fig. 1b)] $=$ [Area between $f$ and $g$ (Fig. 1c)],

$\int_{a}^{b} f(x)dx - \int_{a}^{b} g(x) = \int_{a}^{b} [f(x)- g(x)] dx.$

This relation is valid as long as the two functions are continuous and the upper function $f(x) \ge g(x)$ on the interval $[a, b].$

The Area Between Two Curves (With respect to the $x-$axis)

If $f$ and $g$ are two continuous functions on the interval $[a, b]$ and $f(x) \ge g(x)$ for all values of $x$ in the interval, then the area of the region that is bounded by the two functions is given by

$A =\int_{a}^{b} [f(x) - g(x)] dx.$

Example 1:

Find the area of the region enclosed between $y = x^2$ and $y = x + 6.$

Solution:

We first make a sketch of the region (Figure 2) and find the end points of the region. To do so, we simply equate the two functions,

$x^2 = x + 6,$

and then solve for $x.$

$x^2 - x - 6 & = 0\\(x + 2)(x - 3) & = 0$

from which we get $x = -2$ and $x = 3.$

So the upper and lower boundaries intersect at points $(-2, 4)$ and $(3, 9).$

As you can see from the graph, $x + 6 \ge x^2$ and hence $f(x) = x + 6$ and $g(x) = x^2$ in the interval $[-2, 3].$ Applying the area formula,

$A & = \int_{a}^{b} [f(x) - g(x)] dx\\& = \int_{-2}^{3} [(x + 6) - (x^2)]dx.$

Integrating,

$A & = \left [\frac{x^2} {2} + 6x - \frac{x^3} {3} \right]_{-2}^{3}\\& = \frac{125} {6}.$

So the area between the two curves $f(x) = x + 6$ and $g(x) = x^2$ is $125/6.$

Sometimes it is possible to apply the area formula with respect to the $y-$coordinates instead of the $x-$coordinates. In this case, the equations of the boundaries will be written in such a way that $y$ is expressed explicitly as a function of $x$ (Figure 3).

The Area Between Two Curves (With respect to the $y-$axis)

If $w$ and $v$ are two continuous functions on the interval $[c, d]$ and $w(y) \ge v(y)$ for all values of $y$ in the interval, then the area of the region that is bounded by $x = v(y)$ on the left, $x = w(y)$ on the right, below by $y = c,$ and above by $y = d,$ is given by

$A = \int_{c}^{d} [w(y) - v(y)] dy.$

Example 2:

Find the area of the region enclosed by $x = y^2$ and $y = x - 6.$

Solution:

As you can see from Figure 4, the left boundary is $x = y^2$ and the right boundary is $y = x - 6.$ The region extends over the interval $-2 \le y \le 3.$ However, we must express the equations in terms of $y.$ We rewrite

$x & = y^2\\x & = y + 6$

Thus

$A & = \int_{-2}^{3} [y + 6 - y^2] dy\\& = \left [\frac{y^2} {2} + 6y - \frac{y^3} {3} \right]_{-2}^{3}\\& = \frac{125} {6}.$

For a video presentation of the area between two graphs (14.0)(16.0), see Math Video Tutorials by James Sousa, Area Between Two Graphs (6:12).

For an additional video presentation of the area between two curves (14.0)(16.0), see Just Math Tutoring, Finding Areas Between Curves (9:51).

Review Questions

In problems #1 - 7, sketch the region enclosed by the curves and find the area.

1. $y = x^2 , y = \sqrt{x},$ on the interval $[0.25, 1]$
2. $y = 0,$ $y = \cos 2x,$ on the interval $\left [\frac{\pi} {4} , \frac{\pi} {2} \right]$
3. $y = |- 1 + x| + 2,$ $y = \frac{-1} {5} x + 7$
4. $y = \cos x,$ $y = \sin x,$ $x = 0,$ $x = 2\pi$
5. $x = y^2,$ $y = x - 2,$ integrate with respect to $y$
6. $y^2 - 4x = 4,$ $4x - y = 16$ integrate with respect to $y$
7. $y = 8 \cos x,$ $y = sec^2 x,$ $-\pi / 3 \le x \le \pi / 3$
8. Find the area enclosed by $x = y^3$ and $x = y.$
9. If the area enclosed by the two functions $y = k \cos x$ and $y = kx^2$ is $2,$ what is the value of $k$?
10. Find the horizontal line $y = k$ that divides the region between $y = x^2$ and $y = 9$ into two equal areas.

1. $49/192$
2. $1/2$
3. $24$
4. $4 \sqrt{2}$
5. $9/2$
6. $30 \frac{3} {8}$
7. $6 \sqrt{3}$
8. $\frac{1} {2}$
9. $k \approx 1.83$
10. $9/ \sqrt[3]{4}$

Feb 23, 2012

Dec 03, 2014