009B Sample Final 1, Problem 3

Consider the area bounded by the following two functions:

$y=\sin x$ and $y={\frac {2}{\pi }}x.$

a) Find the three intersection points of the two given functions. (Drawing may be helpful.)

b) Find the area bounded by the two functions.

1

Foundations:
Recall:
1. You can find the intersection points of two functions, say $f(x),g(x),$
by setting $f(x)=g(x)$ and solve for $x$.
2. The area between two functions, $f(x)$ and $g(x)$, is given by $\int _{a}^{b}f(x)-g(x)~dx$
for $a\leq x\leq b$ where $f(x)$ is the upper function and $g(x)$ is the lower function.

Solution:

2

(a)

Step 1:
First, we graph these two functions.
Insert graph here

Step 2:
Setting $\sin x={\frac {2}{\pi }}x$, we get three solutions $x=0,{\frac {\pi }{2}},{\frac {-\pi }{2}}$
So, the three intersection points are $(0,0),{\bigg (}{\frac {\pi }{2}},1{\bigg )},{\bigg (}{\frac {-\pi }{2}},-1{\bigg )}$.
You can see these intersection points on the graph shown in Step 1.

3

(b)

Step 1:
Using symmetry of the graph, the area bounded by the two functions is given by
$2\int _{0}^{\frac {\pi }{2}}{\bigg (}\sin(x)-{\frac {2}{\pi }}x{\bigg )}~dx$

Step 2:
Lastly, we integrate to get
${\begin{array}{rcl}\displaystyle {2\int _{0}^{\frac {\pi }{2}}{\bigg (}\sin(x)-{\frac {2}{\pi }}x{\bigg )}~dx}&{=}&\displaystyle {2{\bigg (}-\cos(x)-{\frac {x^{2}}{\pi }}{\bigg )}{\bigg |}_{0}^{\frac {\pi }{2}}}\\&&\\&=&\displaystyle {2{\bigg (}-\cos {\bigg (}{\frac {\pi }{2}}{\bigg )}-{\frac {1}{\pi }}{\bigg (}{\frac {\pi }{2}}{\bigg )}^{2}{\bigg )}}-2(-\cos(0))\\&&\\&=&\displaystyle {2{\bigg (}{\frac {-\pi }{4}}{\bigg )}+2}\\&&\\&=&\displaystyle {{\frac {-\pi }{2}}+2}\\\end{array}}$

Final Answer:
(a) $(0,0),{\bigg (}{\frac {\pi }{2}},1{\bigg )},{\bigg (}{\frac {-\pi }{2}},-1{\bigg )}$
(b) ${\frac {-\pi }{2}}+2$

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