# 5.4 Integration formulas and the net change theorem  (Page 4/8)

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## Integrating an odd function

Evaluate the definite integral of the odd function $-5\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x$ over the interval $\left[\text{−}\pi ,\pi \right].$

The graph is shown in [link] . 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

$\begin{array}{ll}{\int }_{\text{−}\pi }^{\pi }-5\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}xdx\hfill & =-5\left(\text{−}\text{cos}\phantom{\rule{0.1em}{0ex}}x\right){|}_{\text{−}\pi }^{\pi }\hfill \\ \\ \\ & =5\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x{|}_{\text{−}\pi }^{\pi }\hfill \\ & =\left[5\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\pi \right]-\left[5\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\text{−}\pi \right)\right]\hfill \\ & =-5-\left(-5\right)\hfill \\ & =0.\hfill \end{array}$ The graph shows areas between a curve and the x -axis for an odd function.

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

$\frac{64}{5}$

## Key concepts

• The net change theorem states that when a quantity changes, the final value equals the initial value plus the integral of the rate of change. Net change can be a positive number, a negative number, or zero.
• The area under an even function over a symmetric interval can be calculated by doubling the area over the positive x -axis. For an odd function, the integral over a symmetric interval equals zero, because half the area is negative.

## Key equations

• Net Change Theorem
$F\left(b\right)=F\left(a\right)+{\int }_{a}^{b}F\text{'}\left(x\right)dx$ or ${\int }_{a}^{b}F\text{'}\left(x\right)dx=F\left(b\right)-F\left(a\right)$

Use basic integration formulas to compute the following antiderivatives.

$\int \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)dx$

$\int \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)dx=\int {x}^{1\text{/}2}dx-\int {x}^{-1\text{/}2}dx=\frac{2}{3}{x}^{3\text{/}2}+{C}_{1}-2{x}^{1\text{/}2}+{C}_{2}=\frac{2}{3}{x}^{3\text{/}2}-2{x}^{1\text{/}2}+C$

$\int \left({e}^{2x}-\frac{1}{2}{e}^{x\text{/}2}\right)dx$

$\int \frac{dx}{2x}$

$\int \frac{dx}{2x}=\frac{1}{2}\text{ln}|x|+C$

$\int \frac{x-1}{{x}^{2}}dx$

${\int }_{0}^{\pi }\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x-\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)dx$

${\int }_{0}^{\pi }\text{sin}\phantom{\rule{0.1em}{0ex}}xdx-{\int }_{0}^{\pi }\text{cos}\phantom{\rule{0.1em}{0ex}}xdx=\text{−}\text{cos}\phantom{\rule{0.1em}{0ex}}x{|}_{0}^{\pi }-\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right){|}_{0}^{\pi }=\left(\text{−}\left(-1\right)+1\right)-\left(0-0\right)=2$

${\int }_{0}^{\pi \text{/}2}\left(x-\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)dx$

Write an integral that expresses the increase in the perimeter $P\left(s\right)$ of a square when its side length s increases from 2 units to 4 units and evaluate the integral.

$P\left(s\right)=4s,$ so $\frac{dP}{ds}=4$ and ${\int }_{2}^{4}4ds=8.$

Write an integral that quantifies the change in the area $A\left(s\right)={s}^{2}$ of a square when the side length doubles from S units to 2 S units and evaluate the integral.

A regular N -gon (an N -sided polygon with sides that have equal length s , such as a pentagon or hexagon) has perimeter Ns . Write an integral that expresses the increase in perimeter of a regular N -gon when the length of each side increases from 1 unit to 2 units and evaluate the integral.

${\int }_{1}^{2}Nds=N$

The area of a regular pentagon with side length $a>0$ is pa 2 with $p=\frac{1}{4}\sqrt{5+\sqrt{5+2\sqrt{5}}}.$ The Pentagon in Washington, DC, has inner sides of length 360 ft and outer sides of length 920 ft. Write an integral to express the area of the roof of the Pentagon according to these dimensions and evaluate this area.

A dodecahedron is a Platonic solid with a surface that consists of 12 pentagons, each of equal area. By how much does the surface area of a dodecahedron increase as the side length of each pentagon doubles from 1 unit to 2 units?

With p as in the previous exercise, each of the 12 pentagons increases in area from 2 p to 4 p units so the net increase in the area of the dodecahedron is 36 p units.

An icosahedron is a Platonic solid with a surface that consists of 20 equilateral triangles. By how much does the surface area of an icosahedron increase as the side length of each triangle doubles from a unit to 2 a units?

Write an integral that quantifies the change in the area of the surface of a cube when its side length doubles from s unit to 2 s units and evaluate the integral.

$18{s}^{2}=6{\int }_{s}^{2s}2xdx$

Write an integral that quantifies the increase in the volume of a cube when the side length doubles from s unit to 2 s units and evaluate the integral.

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