Science:Math Exam Resources/Courses/MATH101/April 2016/Question 07

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MATH101 April 2016

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Question 07
Find the average value of the function $f(x)=3\cos ^{3}x+2\cos ^{2}x$ on the interval $0\leq x\leq {\frac {\pi }{2}}$ . Simplify your answer completely.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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Hint
Consider substitution applied to integration of trig functions, and the trigonometric identities.

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Solution
By definition, the average value of the function $f(x)$ equals to ${\frac {1}{\pi /2-0}}\int _{0}^{\pi /2}(3\cos ^{3}x+2\cos ^{2}x)\,dx={\frac {2}{\pi }}{\bigg (}{\color {OliveGreen}\int _{0}^{\pi /2}3\cos ^{3}x\,dx}+{\color {OrangeRed}\int _{0}^{\pi /2}2\cos ^{2}x\,dx}{\bigg )}.$ For the first integral, we write $\cos ^{3}(x)=\cos ^{2}x.\cos x=(1-\sin ^{2}x)\cos x,$ , and use the substitution $u=\sin x$ , for which $du=\cos x\,dx$ ; we note that the endpoints $x=0$ and $x={\frac {\pi }{2}}$ become $u=0$ and $u=1$ , respectively. So, ${\begin{aligned}{\color {OliveGreen}\int _{0}^{\pi /2}3\cos ^{3}x\,dx}&=\int _{0}^{\pi /2}3(1-\sin ^{2}x)\cos x\,dx\\&=\int _{0}^{1}(3-3u^{2})\,du\\&=(3u-u^{3}){\big \|}_{0}^{1}=2.\end{aligned}}$ For the second integral, we use the trigonometric identity $\cos ^{2}x={\frac {1+\cos(2x)}{2}}$ to get ${\begin{aligned}{\color {OrangeRed}\int _{0}^{\pi /2}2\cos ^{2}x\,dx}&=\int _{0}^{\pi /2}{\big (}1+\cos(2x){\big )}\,dx\\&={\bigg (}x+{\frac {1}{2}}\sin(2x){\bigg )}{\bigg \|}_{0}^{\pi /2}={\frac {\pi }{2}}.\end{aligned}}$ Therefore, The average value of the function $f(x)$ = ${\frac {2}{\pi }}{\bigg (}\int _{0}^{\pi /2}3\cos ^{3}x\,dx+\int _{0}^{\pi /2}2\cos ^{2}x\,dx{\bigg )}={\frac {2}{\pi }}{\bigg (}2+{\frac {\pi }{2}}{\bigg )}={\color {blue}{\frac {4}{\pi }}+1}.$