Chapter 5, Problem 38RE

To determine

To Draw: the graph for the function $f\left(x\right)={\cos }^{2}x\sin x$.

To Evaluate: The value of the integral function ${\displaystyle \underset{0}{\overset{2\pi }{\int }}f\left(x\right)}dx$

Answer to Problem 38RE

The value of the integral function is 0.

Explanation of Solution

Given information:

The function is $f\left(x\right)={\cos }^{2}x\sin x$.

The region lies between $x=0$ and $x=2\pi$.

Calculation:

Show the function as follows:

$f\left(x\right)={\cos }^{2}x\sin x$ (1)

Substitute 0 for x in Equation (1).

$\begin{array}{c}f\left(0\right)={\cos }^2\left(0\right)\sin \left(0\right)\\ =1\times 0\\ =0\end{array}$

Substitute $\frac{\pi }{6}$ for x in Equation (1).

$\begin{array}{c}f\left(\frac{\pi }{6}\right)={\cos }^2\left(\frac{\pi }{6}\right)\sin \left(\frac{\pi }{6}\right)\\ =0.75\times 0.5\\ =0.38\end{array}$

Calculate the remaining values of the function as shown below.

Show the x and $f\left(x\right)$ values as in Table (1).

x	${\cos }^{2}x$	$\sin x$	$f\left(x\right)$
$0$	1.00	0.00	0.00
$\frac{\pi }{6}$	0.75	0.50	0.38
$\frac{\pi }{3}$	0.25	0.87	0.22
$\frac{\pi }{2}$	0.00	1.00	0.00
$\frac{2\pi }{3}$	0.25	0.87	0.22
$\frac{5\pi }{6}$	0.75	0.50	0.38
$\pi$	1.00	0.00	0.00
$\frac{7\pi }{6}$	0.75	-0.50	-0.38
$\frac{4\pi }{3}$	0.25	-0.87	-0.22
$\frac{3\pi }{2}$	0.00	-1.00	0.00
$\frac{5\pi }{3}$	0.25	-0.87	-0.22
$\frac{11\pi }{6}$	0.75	-0.50	-0.38
$2\pi$	1.00	0.00	0.00

Table 1

Sketch the function as shown in Figure 1.

Refer to Figure 1.

The value of the integral function ${\displaystyle {\int }_0^{2\pi }f\left(x\right)dx}$ is 0.

Consider $u=\cos x$ (2)

Differentiate both sides of the Equation (2).

$du=-\sin xdx$ (3)

Calculate the lower limit value of u using Equation (2).

Substitute 0 for x in Equation (2).

$\begin{array}{c}u=\cos \left(0\right)\\ =1\end{array}$

Calculate the upper limit value of u using Equation (2).

Substitute $2\pi$ for x in Equation (2).

$\begin{array}{c}u=\cos \left(2\pi \right)\\ =1\end{array}$

Substitute u for $\cos x$ and $-du$ for $\sin xdx$ in Equation (1) as shown below.

$\begin{array}{c}f\left(u\right)=u^2\left(-du\right)\\ =-u^{2}du\end{array}$ (4)

Find the integral value of the function as shown below.

$\begin{array}{c}{\displaystyle \underset{0}{\overset{2\pi }{\int }}f\left(x\right)dx}={\displaystyle \underset{1}{\overset{1}{\int }}\left(-u^{2}du\right)}\\ =-{\left(\frac{u^3}{3}\right)}_1^1\\ =-\left(\frac{1^3}{3}-\frac{1^3}{3}\right)\\ =0\end{array}$

Therefore, the integral value of the function is 0.