Question

Consider the function $f\left(x\right)=4{\sin }^{3}x{\cos }^{2}x$.

(a) Find the signed area between the graph of $f\left(x\right)$and the x-axis on $[-\pi ,\pi ]$, shown next at the left.

(b) Find the absolute area between the graph of $f\left(x\right)$and the x-axis on $[-\pi ,\pi ]$.

(c) Find the average value of $f\left(x\right)$on $\left[0,\frac{\mathrm{\pi }}{2}\right]$, shown next at the right.

Short answer

Part (a) The signed area between the graph of $f\left(x\right)$and the x-axis on $\left[\mathrm{\pi },-\mathrm{\pi }\right]$, is 0.

Part (b) The absolute area between the graph of $f\left(x\right)$and the x-axis on $\left[\mathrm{\pi },-\mathrm{\pi }\right]$is $\frac{32}{15}$.

Part (c) The average value of $f\left(x\right)$on $\left[0,\frac{\mathrm{\pi }}{2}\right]$is $\frac{16}{15\mathrm{\pi }}$.

Step-by-step solution

Part (a) Step 1. Given Information.

The function is$f\left(x\right)=4{\sin }^{3}x{\cos }^{2}x$.

Part (a) Step 2 Calculation

To find the signed area between the graph of $f\left(x\right)$, integrate the function with limit $\left[-\mathrm{\pi },\mathrm{\pi }\right]$.

role="math" localid="1649312825934" ${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}4{\sin }^{3}x{\cos }^{2}xdx$

Here, $f\left(x\right)$is odd function and it is continuous on $\left[-\mathrm{\pi },\mathrm{\pi }\right]$.

so,

role="math" localid="1649314891156" ${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}4{\sin }^{3}x{\cos }^{2}xdx=0$.

Part (b). Step 1. Explanation.

To find the absolute area between the graph of $f\left(x\right)$, integrate the function.

${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left|4{\sin }^{3}x{\cos }^{2}x\right|dx$

Divide the integral in parts. so find limits.

$4{\sin }^{3}x{\cos }^{2}x=0\Rightarrow x=0$So limits becomes $\left[-\mathrm{\pi },0\right]$and $\left[0,\mathrm{\pi }\right]$.

role="math" localid="1649315342621" ${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}4{\sin }^{3}x{\cos }^{2}xdx={\int }_{-\mathrm{\pi }}^0-4{\sin }^{3}x{\cos }^{2}xdx+{\int }_0^{\mathrm{\pi }}4{\sin }^{3}x{\cos }^{2}xdx$

Substitute $u=\cos x$so, $du=-\sin xdx\Rightarrow -du=\sin xdx$

role="math" localid="1649320212180" $$\begin{gathered} {\int }_{-\mathrm{\pi }}^0-4{\sin }^{3}x{\cos }^{2}xdx+{\int }_0^{\mathrm{\pi }}4{\sin }^{3}x{\cos }^{2}xdx={\int }_{-\mathrm{\pi }}^0-4\left(1-{\cos }^{2}x\right)\sin x{\cos }^{2}xdx+{\int }_0^{\mathrm{\pi }}4\left(1-{\cos }^{2}x\right)\sin x{\cos }^{2}xdx \\ =-4{\int }_{-1}^1-\left(1-u^2\right)u^{2}du+4{\int }_1^{-1}-\left(1-u^2\right)u^{2}du \\ =4{\int }_{-1}^1{u}^2-u^{4}du-4{\int }_1^{-1}{u}^2-u^{4}du \\ =4{\int }_{-1}^1{u}^2-u^{4}du+4{\int }_{-1}^1{u}^2-u^{4}du \end{gathered}$$

Part (b)  Step 2. Calculation.

Further simplify as follows,

$$\begin{gathered} 4{\int }_{-1}^1{u}^2-u^{4}du+4{\int }_{-1}^1{u}^2-u^{4}du=4\left({\left[\frac{u^3}{3}\right]}_{-1}^1-{\left[\frac{u^5}{5}\right]}_{-1}^1\right)+4\left({\left[\frac{u^3}{3}\right]}_{-1}^1-{\left[\frac{u^5}{5}\right]}_{-1}^1\right) \\ =4\times \frac{4}{15}+4\times \frac{4}{15} \\ =\frac{16}{15}+\frac{16}{15} \\ =\frac{32}{15} \end{gathered}$$

Part (c) Step 1. Explanation.

To find the average value of f(x) on $\left[0,\frac{\mathrm{\pi }}{2}\right]$use formula as below,

The average value of $f\left(x\right)$on $\left[a,b\right]$, $A\left(x\right)=\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx$

So, integral becomes,

role="math" localid="1649323440694" $\frac{1}{{\displaystyle \frac{\mathrm{\pi }}{2}}-0}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}4{\sin }^{3}x{\cos }^{2}xdx=\frac{2}{{\displaystyle \mathrm{\pi }}}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}4{\sin }^{3}x{\cos }^{2}xdx$

Substitute $u=\cos x$, so, $du=-\sin xdx\Rightarrow -du=\sin xdx$

role="math" localid="1649325556289" $$\begin{gathered} \frac{2}{{\displaystyle \mathrm{\pi }}}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}4{\sin }^{3}x{\cos }^{2}xdx=\frac{8}{\mathrm{\pi }}{\int }_1^0-\left(1-u^2\right)u^{2}du \\ =-\frac{8}{\mathrm{\pi }}{\int }_1^0{u}^2-u^{4}du \\ =\frac{8}{\mathrm{\pi }}{\int }_0^1{u}^2-u^{4}du \\ =\frac{8}{\mathrm{\pi }}\left({\left[\frac{u^3}{3}\right]}_0^1-{\left[\frac{u^5}{5}\right]}_0^1\right) \\ =\frac{8}{\mathrm{\pi }}\times \frac{2}{15} \\ =\frac{16}{15\mathrm{\pi }} \end{gathered}$$