009B Sample Final 1, Problem 3

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Consider the area bounded by the following two functions:

and

a) Find the three intersection points of the two given functions. (Drawing may be helpful.)

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, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)} and , is given by
for 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 a\leq x\leq b} , where 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 f(x)} is the upper function and 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 g(x)} is the lower function.

Solution:

(a)

Step 1:  
First, we graph these two functions.
9BF1 3 GP.png
|}
Step 2:  
Setting 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 \sin x=\frac{2}{\pi}x} , we get three solutions: 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 x=0,\frac{\pi}{2},\frac{-\pi}{2}.}
So, the three intersection points are 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)} .
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
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 2\int_0^{\frac{\pi}{2}}\bigg(\sin(x)-\frac{2}{\pi}x\bigg)~dx.}
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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