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:
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| Recall:
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1. You can find the intersection points of two functions, say
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- by setting
and solve for .
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2. The area between two functions, and , is given by
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- for
where is the upper function and is the lower function.
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Solution:
(a)
| Step 1:
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| First, we graph these two functions.
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| Insert graph here
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| Step 2:
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Setting , we get three solutions
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So, the three intersection points are .
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| You can see these intersection points on the graph shown in Step 1.
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(b)
| Step 1:
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| Using symmetry of the graph, the area bounded by the two functions is given by
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| Step 2:
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| Lastly, we integrate to get
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- 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}}
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| Final Answer:
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| (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)}
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| (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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