5.4 Integration formulas and the net change theorem  (Page 3/8)

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Chapter opener: iceboats (credit: modification of work by Carter Brown, Flickr)

As we saw at the beginning of the chapter, top iceboat racers ( [link] ) can attain speeds of up to five times the wind speed. Andrew is an intermediate iceboater, though, so he attains speeds equal to only twice the wind speed. Suppose Andrew takes his iceboat out one morning when a light 5-mph breeze has been blowing all morning. As Andrew gets his iceboat set up, though, the wind begins to pick up. During his first half hour of iceboating, the wind speed increases according to the function $v\left(t\right)=20t+5.$ For the second half hour of Andrew’s outing, the wind remains steady at 15 mph. In other words, the wind speed is given by

$v\left(t\right)=\left\{\begin{array}{lll}20t+5\hfill & \text{for}\hfill & 0\le t\le \frac{1}{2}\hfill \\ 15\hfill & \text{for}\hfill & \frac{1}{2}\le t\le 1.\hfill \end{array}$

Recalling that Andrew’s iceboat travels at twice the wind speed, and assuming he moves in a straight line away from his starting point, how far is Andrew from his starting point after 1 hour?

To figure out how far Andrew has traveled, we need to integrate his velocity, which is twice the wind speed. Then

Distance $={\int }_{0}^{1}2v\left(t\right)dt.$

Substituting the expressions we were given for $v\left(t\right),$ we get

$\begin{array}{cc}{\int }_{0}^{1}2v\left(t\right)dt\hfill & ={\int }_{0}^{1\text{/}2}2v\left(t\right)dt+{\int }_{1\text{/}2}^{1}2v\left(t\right)dt\hfill \\ & ={\int }_{0}^{1\text{/}2}2\left(20t+5\right)dt+{\int }_{1\text{/}3}^{1}2\left(15\right)dt\hfill \\ & ={\int }_{0}^{1\text{/}2}\left(40t+10\right)dt+{\int }_{1\text{/}2}^{1}30dt\hfill \\ & =\left[20{t}^{2}+10t\right]{|}_{0}^{1\text{/}2}+\left[30t\right]{|}_{1\text{/}2}^{1}\hfill \\ & =\left(\frac{20}{4}+5\right)-0+\left(30-15\right)\hfill \\ & =25.\hfill \end{array}$

Andrew is 25 mi from his starting point after 1 hour.

Suppose that, instead of remaining steady during the second half hour of Andrew’s outing, the wind starts to die down according to the function $v\left(t\right)=-10t+15.$ In other words, the wind speed is given by

$v\left(t\right)=\left\{\begin{array}{lll}20t+5\hfill & \text{for}\hfill & 0\le t\le \frac{1}{2}\hfill \\ -10t+15\hfill & \text{for}\hfill & \frac{1}{2}\le t\le 1.\hfill \end{array}$

Under these conditions, how far from his starting point is Andrew after 1 hour?

17.5 mi

Integrating even and odd functions

We saw in Functions and Graphs that an even function    is a function in which $f\left(\text{−}x\right)=f\left(x\right)$ 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 odd function    is one in which $f\left(\text{−}x\right)=\text{−}f\left(x\right)$ 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.

Rule: integrals of even and odd functions

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

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

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

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

Integrating an even function

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

The symmetry appears in the graphs in [link] . 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

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

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.

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

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

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