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:
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  $\cos x=1-\cos x,$  we get  $2\cos x=2.$
Therefore, we have
$\cos x=1.$
In the interval  $0\leq x\leq 2\pi ,$  the solutions to this equation are
$x=0$  and  $x=2\pi .$
Plugging these values into our equations,
we get the intersection points  $(0,1)$  and  $(2\pi ,1).$
You can see these intersection points on the graph shown in Step 1.

(b)

Step 1:
The area bounded by the two functions is given by
$\int _{0}^{2\pi }(2-\cos x)-\cos x~dx.$

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

Final Answer:
(a)    $(0,1),(2\pi ,1)$
(b)    $4\pi$

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