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

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.

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

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $f(x)$ = ${\frac {2}{\pi }}{\bigg (}\int _{0}^{\pi /2}3\cos ^{3}x\,dx+\int _{0}^{\pi /2}2\cos ^{2}x\,dx{\bigg )}={\frac {2}{\pi }}{\bigg (}2+{\frac {\pi }{2}}{\bigg )}={\color {blue}{\frac {4}{\pi }}+1}.$