Integrating Even and Odd Functions

Learning Outcomes

Apply the integrals of odd and even functions

We saw in Module 1: Functions and Graphs that an is a function in which $f(\text{−}x)=f(x)$ for all $x$ in the domain—that is, the graph of the curve is unchanged when $x$ is replaced with − $x$ . The graphs of even functions are symmetric about the $y$ -axis. An is one in which $f(\text{−}x)=\text{−}f(x)$ for all $x$ in the domain, and the graph of the function is symmetric about the origin.

Integrals of even functions, when the limits of integration are from − $a$ to $a$ , involve two equal areas, because they are symmetric about the $y$ -axis. Integrals of odd functions, when the limits of integration are similarly $\left[\text{−}a,a\right],$ evaluate to zero because the areas above and below the $x$ -axis are equal.

Integrals of Even and Odd Functions

For continuous even functions such that $f(\text{−}x)=f(x),$

\int_{− a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x .

${\displaystyle\int }_{\text{−}a}^{a}f(x)dx=2{\displaystyle\int }_{0}^{a}f(x)dx.$

For continuous odd functions such that $f(\text{−}x)=\text{−}f(x),$

\int_{− a}^{a} f (x) d x = 0.

${\displaystyle\int }_{\text{−}a}^{a}f(x)dx=0.$

It may be useful to recall how to quickly determine whether a function is even, odd or neither.

Recall: How to determine whether a function is even, odd or neither

Determine whether each of the following functions is even, odd, or neither.

$f(x)=-5x^4+7x^2-2$
$f(x)=2x^5-4x+5$
$f(x)=\large{\frac{3x}{x^2+1}}$

To determine whether a function is even or odd, we evaluate $f(−x)$ and compare it to $f(x)$ and $−f(x)$ .

$f(−x)=-5(−x)^4+7(−x)^2-2=-5x^4+7x^2-2=f(x)$ . Therefore, $f$ is even.
$f(−x)=2(−x)^5-4(−x)+5=-2x^5+4x+5$ . Now, $f(−x)\ne f(x)$ . Furthermore, noting that $−f(x)=-2x^5+4x-5$ , we see that $f(−x)\ne −f(x)$ . Therefore, $f$ is neither even nor odd.
$f(−x)=\frac{3(−x)}{((−x)^2+1)}=\frac{-3x}{(x^2+1)}=−\left[\frac{3x}{(x^2+1)}\right]=−f(x)$ . Therefore, $f$ is odd.

Example: Integrating an Even Function

Integrate the even function ${\displaystyle\int }_{-2}^{2}(3{x}^{8}-2)dx$ and verify that the integration formula for even functions holds.

Show Solution

The symmetry appears in the graphs in Figure 3. Graph (a) shows the region below the curve and above the $x$ -axis. We have to zoom in to this graph by a huge amount to see the region. Graph (b) shows the region above the curve and below the $x$ -axis. The signed area of this region is negative. Both views illustrate the symmetry about the $y$ -axis of an even function. We have

\begin{array}{ll} \int_{- 2}^{2} (3 x^{8} - 2) d x & = (\frac{x^{9}}{3} - 2 x) |_{- 2}^{2} \\ = [\frac{{(2)}^{9}}{3} - 2 (2)] - [\frac{{(- 2)}^{9}}{3} - 2 (- 2)] \\ = (\frac{512}{3} - 4) - (- \frac{512}{3} + 4) \\ = \frac{1000}{3} . \end{array}

$\begin{array}{ll}{\int }_{-2}^{2}(3{x}^{8}-2)dx\hfill & =(\frac{{x}^{9}}{3}-2x){|}_{-2}^{2}\hfill \\ \\ \\ & =\left[\frac{{(2)}^{9}}{3}-2(2)\right]-\left[\frac{{(-2)}^{9}}{3}-2(-2)\right]\hfill \\ & =(\frac{512}{3}-4)-(-\frac{512}{3}+4)\hfill \\ & =\frac{1000}{3}.\hfill \end{array}$

To verify the integration formula for even functions, we can calculate the integral from 0 to 2 and double it, then check to make sure we get the same answer.

\begin{array}{ll} \int_{0}^{2} (3 x^{8} - 2) d x & = (\frac{x^{9}}{3} - 2 x) |_{0}^{2} \\ = \frac{512}{3} - 4 \\ = \frac{500}{3} \end{array}

$\begin{array}{ll}{\int }_{0}^{2}(3{x}^{8}-2)dx\hfill & =(\frac{{x}^{9}}{3}-2x){|}_{0}^{2}\hfill \\ \\ & =\frac{512}{3}-4\hfill \\ & =\frac{500}{3}\hfill \end{array}$

Since $2·\frac{500}{3}=\frac{1000}{3},$ we have verified the formula for even functions in this particular example.

Two graphs of the same function f(x) = 3x^8 – 2, side by side. It is symmetric about the y axis, has x-intercepts at (-1,0) and (1,0), and has a y-intercept at (0,-2). The function decreases rapidly as x increases until about -.5, where it levels off at -2. Then, at about .5, it increases rapidly as a mirror image. The first graph is zoomed-out and shows the positive area between the curve and the x-axis over [-2,-1] and [1,2]. The second is zoomed-in and shows the negative area between the curve and the x-axis over [-1,1].

Figure 3. Graph (a) shows the positive area between the curve and the x-axis, whereas graph (b) shows the negative area between the curve and the x-axis. Both views show the symmetry about the y-axis.

Watch the following video to see the worked solution to Example: Integrating an Even Function.

Closed Captioning and Transcript Information for Video

Example: Integrating an Odd Function

Evaluate the definite integral of the odd function $-5 \sin x$ over the interval $\left[\text{−}\pi ,\pi \right].$

Show Solution

The graph is shown in Figure 4. We can see the symmetry about the origin by the positive area above the $x$ -axis over $\left[\text{−}\pi ,0\right],$ and the negative area below the $x$ -axis over $\left[0,\pi \right].$ We have

\begin{array}{ll} \int_{− π}^{π} - 5 \sin x d x & = - 5 (− \cos x) |_{− π}^{π} \\ = 5 \cos x |_{− π}^{π} \\ = [5 \cos π] - [5 \cos (− π)] \\ = - 5 - (- 5) \\ = 0. \end{array}

$\begin{array}{ll}{\int }_{\text{−}\pi }^{\pi }-5 \sin xdx\hfill & =-5(\text{−} \cos x){|}_{\text{−}\pi }^{\pi }\hfill \\ \\ \\ & =5 \cos x{|}_{\text{−}\pi }^{\pi }\hfill \\ & =\left[5 \cos \pi \right]-\left[5 \cos (\text{−}\pi )\right]\hfill \\ & =-5-(-5)\hfill \\ & =0.\hfill \end{array}$

A graph of the given function f(x) = -5 sin(x). The area under the function but above the x-axis is shaded over [-pi, 0], and the area above the function and under the x-axis is shaded over [0, pi].

Figure 4. The graph shows areas between a curve and the x-axis for an odd function.

Watch the following video to see the worked solution to Example: Integrating an Odd Function.

Closed Captioning and Transcript Information for Video

Try It

Integrate the function ${\displaystyle\int }_{-2}^{2}{x}^{4}dx.$

Hint

Show Solution

$\frac{64}{5}$

Module 1: Integration