Integration Formulas and the Net Change Theorem: Learn It 3

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Integration Formulas and the Net Change Theorem: Learn It 3

Integrating Even and Odd Functions

Recall that an is a function in which $f(\text{−}x)=f(x)$ for all $x$ in the domain. This means 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.

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

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

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.

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.

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.

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.4 Integration Formulas and the Net Change Theorem” here (opens in new window).

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

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 $\left[0,\pi \right].$

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.

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}

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

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.4 Integration Formulas and the Net Change Theorem” here (opens in new window).