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5.1: Area Between Two Curves

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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}.

Multimedia Links

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.

Review Answers

  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}

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Date Created:

Feb 23, 2012

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Aug 21, 2014
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