# 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.
Step 2:
Setting  $\cos x=2-\cos x,$ 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

${\begin{array}{rcl}\displaystyle {\int _{0}^{2\pi }(2-\cos x)-\cos x~dx}&{=}&\displaystyle {\int _{0}^{2\pi }2-2\cos x~dx}\\&&\\&=&\displaystyle {(2x-2\sin x){\bigg |}_{0}^{2\pi }}\\&&\\&=&\displaystyle {(4\pi -2\sin(2\pi ))-(0-2\sin(0))}\\&&\\&=&\displaystyle {4\pi .}\\\end{array}}$ (a)    $(0,1),(2\pi ,1)$ (See Step 1 above for graph)
(b)    $4\pi$ 