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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 \begin{align*}x-\end{align*}x and \begin{align*}y-\end{align*}yaxes.

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

Consider the region bounded by the graphs \begin{align*}f\end{align*}f and \begin{align*}g\end{align*}g between \begin{align*}x = a\end{align*}x=a and \begin{align*}x = b,\end{align*}x=b, as shown in the figures below. If the two graphs lie above the \begin{align*}x-\end{align*}xaxis, we can interpret the area that is sandwiched between them as the area under the graph of \begin{align*}g\end{align*}g subtracted from the area under the graph \begin{align*}f.\end{align*}f.

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

[Area under \begin{align*}f\end{align*}f (Fig. 1a)] \begin{align*}-\end{align*} [Area under \begin{align*}g\end{align*}g (Fig. 1b)] \begin{align*}=\end{align*}= [Area between \begin{align*}f\end{align*}f and \begin{align*}g\end{align*}g (Fig. 1c)],

\begin{align*} \int_{a}^{b} f(x)dx - \int_{a}^{b} g(x) = \int_{a}^{b} [f(x)- g(x)] dx.\end{align*}


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

The Area Between Two Curves (With respect to the \begin{align*}x-\end{align*}xaxis)

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

\begin{align*} A =\int_{a}^{b} [f(x) - g(x)] dx.\end{align*}


Example 1:

Find the area of the region enclosed between \begin{align*}y = x^2\end{align*}y=x2 and \begin{align*}y = x + 6.\end{align*}


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,

\begin{align*}x^2 = x + 6,\end{align*}

and then solve for \begin{align*}x.\end{align*}

\begin{align*}x^2 - x - 6 & = 0\\ (x + 2)(x - 3) & = 0\end{align*}

from which we get \begin{align*}x = -2\end{align*} and \begin{align*}x = 3.\end{align*}

So the upper and lower boundaries intersect at points \begin{align*}(-2, 4)\end{align*} and \begin{align*}(3, 9).\end{align*}

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

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


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

So the area between the two curves \begin{align*}f(x) = x + 6\end{align*} and \begin{align*}g(x) = x^2\end{align*} is \begin{align*}125/6.\end{align*}

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

The Area Between Two Curves (With respect to the \begin{align*}y-\end{align*}axis)

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

\begin{align*} A = \int_{c}^{d} [w(y) - v(y)] dy.\end{align*}

Example 2:

Find the area of the region enclosed by \begin{align*}x = y^2\end{align*} and \begin{align*}y = x - 6.\end{align*}


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

\begin{align*}x & = y^2\\ x & = y + 6\end{align*}


\begin{align*}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}.\end{align*}

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

Review Answers

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

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Date Created:

Feb 23, 2012

Last Modified:

Aug 19, 2015
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