MATH101 April 2016
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Question 7

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?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

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

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Solution

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By definition, the average value of the function $f(x)$ equals to
${\frac {1}{\pi /20}}\int _{0}^{\pi /2}(3\cos ^{3}x+2\cos ^{2}x)\,dx={\frac {2}{\pi }}{\bigg (}{\color {OliveGreen}\int _{0}^{\pi /2}3\cos ^{3}x\,dx}+{\color {OrangeRed}\int _{0}^{\pi /2}2\cos ^{2}x\,dx}{\bigg )}.$
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,
${\begin{aligned}{\color {OliveGreen}\int _{0}^{\pi /2}3\cos ^{3}x\,dx}&=\int _{0}^{\pi /2}3(1\sin ^{2}x)\cos x\,dx\\&=\int _{0}^{1}(33u^{2})\,du\\&=(3uu^{3}){\big }_{0}^{1}=2.\end{aligned}}$
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 $f(x)$ = ${\frac {2}{\pi }}{\bigg (}\int _{0}^{\pi /2}3\cos ^{3}x\,dx+\int _{0}^{\pi /2}2\cos ^{2}x\,dx{\bigg )}={\frac {2}{\pi }}{\bigg (}2+{\frac {\pi }{2}}{\bigg )}={\color {blue}{\frac {4}{\pi }}+1}.$
