# 009B Sample Final 1, Problem 3

Consider the area bounded by the following two functions:

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

(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  ${\displaystyle f(x),g(x),}$

by setting  ${\displaystyle f(x)=g(x)}$  and solving for  ${\displaystyle x.}$

2. The area between two functions,  ${\displaystyle f(x)}$  and  ${\displaystyle g(x),}$  is given by  ${\displaystyle \int _{a}^{b}f(x)-g(x)~dx}$

for  ${\displaystyle a\leq x\leq b,}$  where  ${\displaystyle f(x)}$  is the upper function and  ${\displaystyle g(x)}$  is the lower function.

Solution:

(a)

Step 1:
First, we graph these two functions.
Step 2:
Setting  ${\displaystyle \cos x=2-\cos x,}$  we get  ${\displaystyle 2\cos x=2.}$
Therefore, we have
${\displaystyle \cos x=1.}$
In the interval  ${\displaystyle 0\leq x\leq 2\pi ,}$  the solutions to this equation are
${\displaystyle x=0}$  and  ${\displaystyle x=2\pi .}$
Plugging these values into our equations,
we get the intersection points  ${\displaystyle (0,1)}$  and  ${\displaystyle (2\pi ,1).}$
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

${\displaystyle \int _{0}^{2\pi }(2-\cos x)-\cos x~dx.}$

Step 2:
Lastly, we integrate to get

${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{2\pi }(2-\cos x)-\cos x~dx}&{=}&\displaystyle {\int _{0}^{2\pi }2-2\cos x~dx}\\&&\\&=&\displaystyle {(2x-2\sin x){\bigg |}_{0}^{2\pi }}\\&&\\&=&\displaystyle {(4\pi -2\sin(2\pi ))-(0-2\sin(0))}\\&&\\&=&\displaystyle {4\pi .}\\\end{array}}}$

(a)    ${\displaystyle (0,1),(2\pi ,1)}$  (See Step 1 above for graph)
(b)    ${\displaystyle 4\pi }$