91影视

The function is$f\left(x\right)={\sin }^{5}x$.

To find the area between the graphs of $f\left(x\right)={\sin }^{5}x$and $g\left(x\right)=\sin x$on [0, 蟺]

Area = role="math" localid="1649396617544" ${鈭�}_0^{\mathrm{蟺}}(f\left(x\right)-g\left(x\right))dx$

Rewrite the integral,

role="math" localid="1649397574380" $$\begin{gathered} {鈭�}_0^{\mathrm{蟺}}({\sin }^{5}x-\sin x)dx={鈭�}_0^{\mathrm{蟺}}{\sin }^{5}xdx-{鈭�}_0^{\mathrm{蟺}}\sin xdx \\ ={鈭�}_0^{\mathrm{蟺}}{\left({\sin }^{2}x\right)}^2\sin xdx-{\left[-\cos x\right]}_0^{\mathrm{蟺}} \end{gathered}$$

Use trigonometric identity.

${鈭�}_0^{\mathrm{蟺}}{\left({\sin }^{2}x\right)}^2\sin xdx-\left(2\right)={鈭�}_0^{\mathrm{蟺}}{\left(1-{\cos }^{2}x\right)}^2\sin xdx-\left(2\right)$

Substitute role="math" localid="1649398532405" $u=\cos x$so, role="math" localid="1649398720697" $du=-\sin xdx鈬�-du=\sin xdx$

$$\begin{gathered} {鈭�}_1^{-1}-{\left(1-u^2\right)}^{2}du-\left(2\right)={鈭�}_{-1}^1{\left(1-u^2\right)}^{2}du-\left(2\right) \\ ={鈭�}_{-1}^{1}1-2u^2+u^{4}du-\left(2\right) \\ ={鈭�}_{-1}^{1}1du-{鈭�}_{-1}^{1}2u^{2}du+{鈭�}_{-1}^1{u}^{4}du-\left(2\right) \\ ={\left[u\right]}_{-1}^1-2{\left[\frac{2}{3}\right]}_{-1}^1+{\left[\frac{2}{5}\right]}_{-1}^1-\left(2\right) \\ =\frac{16}{15} \end{gathered}$$

To find the area between the graphs of $f\left(x\right)={\sin }^{5}x$and $h\left(x\right)={\sin }^{5}2x$on [0, 蟺], and intersection point is $\frac{\mathrm{蟺}}{3}$,

Area =role="math" localid="1649408548966" ${鈭�}_0^{\raisebox{1ex}{$\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}f\left(x\right)-h\left(s\right)dx+{鈭�}_{\raisebox{1ex}{$\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{\mathrm{蟺}}f\left(x\right)-h\left(x\right)dx$

role="math" localid="1649408543011" ${鈭�}_0^{\raisebox{1ex}{$\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\sin }^{5}x-{\sin }^{5}2xdx+{鈭�}_{\raisebox{1ex}{$\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{\mathrm{蟺}}{\sin }^{5}x-{\sin }^{5}2xdx$

Use Pythagorean identity ${\sin }^{2}x=1-{\cos }^{2}x$

role="math" localid="1649409981785" ${鈭�}_0^{\raisebox{1ex}{$\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\sin }^{5}x-{\sin }^{5}2xdx+{鈭�}_{\raisebox{1ex}{$\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{\mathrm{蟺}}{\sin }^{5}x-{\sin }^{5}2xdx={鈭�}_0^{\raisebox{1ex}{$\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(1-{\cos }^{2}x\right)}^2\sin xdx-{鈭�}_0^{\raisebox{1ex}{$\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\sin }^{5}2xdx+{鈭�}_{\raisebox{1ex}{$\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{\mathrm{蟺}}{\left(1-{\cos }^{2}x\right)}^{2}dx-{鈭�}_{\raisebox{1ex}{$\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{\mathrm{蟺}}{\sin }^{5}2xdx$

Substitute $u=\cos x$so $du=-\sin xdx鈬�-du=\sin xdx$and

$w=2x$, so, $dw=2dx鈬�\frac{dw}{2}=dx$

role="math" localid="1649410050597" ${鈭�}_1^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(1-u^2\right)}^{2}du-{鈭�}_0^{\raisebox{1ex}{$2\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\frac{{\sin }^{5}w}{2}dw+{鈭�}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{-1}-{\left(1-u^2\right)}^{2}du-{鈭�}_{\raisebox{1ex}{$2\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{2\mathrm{蟺}}\frac{{\sin }^{5}w}{2}dw$

role="math" localid="1649410269127" $={鈭�}_1^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(1-u^2\right)}^{2}du-\frac{1}{2}{鈭�}_0^{\raisebox{1ex}{$2\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\sin }^{5}wdw+{鈭�}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{-1}-{\left(1-u^2\right)}^{2}du-\frac{1}{2}{鈭�}_{\raisebox{1ex}{$2\mathrm{蟺}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{2\mathrm{蟺}}{\sin }^{5}wdw$

Substitute so $dv=-\sin wdw鈬�-dv=\sin wdw$

role="math" localid="1649410979991" $$\begin{gathered} ={鈭�}_1^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(1-u^2\right)}^{2}du-\frac{1}{2}{鈭�}_1^{-1/2}-{\left(1-v^2\right)}^{2}dv+{鈭�}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{-1}-{\left(1-u^2\right)}^{2}du-\frac{1}{2}{鈭�}_{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^1-{\left(1-v^2\right)}^{2}dv \\ =1.07 \end{gathered}$$