009B Sample Final 1, Problem 3

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Consider the area bounded by the following two functions:

  and  

(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
by setting and solving for .
2. The area between two functions, and , is given by
for , where is the upper function and is the lower function.


Solution:

(a)

Step 1:  
First, we graph these two functions.
Insert graph here
Step 2:  
Setting , we get three solutions:
So, the three intersection points are .
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
Step 2:  
Lastly, we integrate to get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (0,0),\bigg(\frac{\pi}{2},1\bigg),\bigg(\frac{-\pi}{2},-1\bigg)}
(b)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\pi}{2}+2}

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