009B Sample Final 1, Problem 3

Consider the area bounded by the following two functions:

$y=\cos x$  and  $y=2-\cos x,~0\leq x\leq 2\pi .$

(a) Sketch the graphs and find their points of intersection.

(b) Find the area bounded by the two functions.

Foundations:
Recall:
1. You can find the intersection points of two functions, say  $f(x),g(x),$
by setting  $f(x)=g(x)$  and solving 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:

(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.

(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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