Science:Math Exam Resources/Courses/MATH101/April 2016/Question 07/Solution 1

By definition, the average value of the function $f(x)$ equals to

$\frac{1}{\pi /2-0}\underset{0}{\overset{\pi /2}{\int }}(3{\cos }^{3}x+2{\cos }^{2}x)dx=\frac{2}{\pi }(\underset{0}{\overset{\pi /2}{\int }}3{\cos }^{3}xdx+\underset{0}{\overset{\pi /2}{\int }}2{\cos }^{2}xdx).$

For the first integral, we write ${\cos }^3(x)={\cos }^{2}x.\cos x=(1-{\sin }^{}x)\cos x,$, and use the substitution $u=\sin x$, for which $du=\cos xdx$; we note that the endpoints $x=0$ and $x=\frac{\pi }{2}$ become $u=0$ and $u=1$, respectively. So,

$\begin{array}{rl}\underset{0}{\overset{\pi /2}{\int }}3{\cos }^{3}xdx& =\underset{0}{\overset{\pi /2}{\int }}3(1-{\sin }^{2}x)\cos xdx\\ & =\underset{0}{\overset{1}{\int }}(3-3u^2)du\\ & =(3u-u^3){|}_0^1=2.\end{array}$

For the second integral, we use the trigonometric identity ${\cos }^{2}x=\frac{1+\cos (2x)}{2}$ to get

$\begin{array}{rl}\underset{0}{\overset{\pi /2}{\int }}2{\cos }^{2}xdx& =\underset{0}{\overset{\pi /2}{\int }}(1+\cos (2x))dx\\ & =(x+\frac{1}{2}\sin (2x)){|}_0^{\pi /2}=\frac{\pi }{2}.\end{array}$

Therefore,

The average value of the function $f(x)$ = $\frac{2}{\pi }(\underset{0}{\overset{\pi /2}{\int }}3{\cos }^{3}xdx+\underset{0}{\overset{\pi /2}{\int }}2{\cos }^{2}xdx)=\frac{2}{\pi }(2+\frac{\pi }{2})=\frac{4}{\pi }+1.$