Question Number 133556 by leena12345 last updated on 22/Feb/21
Commented by leena12345 last updated on 23/Feb/21
$${help} \\ $$
Answered by bemath last updated on 23/Feb/21
![V=π∫_4 ^5 (−x^2 +9x−20)^2 dx V=π∫_4 ^5 (x^4 +81x^2 +400−18x^3 +40x^2 −360x)dx =π [(x^5 /5)+27x^3 +400x−(9/2)x^4 +((40)/3)x^3 −180x^2 ]_4 ^5 = (π/(30))](https://www.tinkutara.com/question/Q133566.png)
$$\mathrm{V}=\pi\underset{\mathrm{4}} {\overset{\mathrm{5}} {\int}}\left(−\mathrm{x}^{\mathrm{2}} +\mathrm{9x}−\mathrm{20}\right)^{\mathrm{2}} \mathrm{dx} \\ $$$$\mathrm{V}=\pi\underset{\mathrm{4}} {\overset{\mathrm{5}} {\int}}\left(\mathrm{x}^{\mathrm{4}} +\mathrm{81x}^{\mathrm{2}} +\mathrm{400}−\mathrm{18x}^{\mathrm{3}} +\mathrm{40x}^{\mathrm{2}} −\mathrm{360x}\right)\mathrm{dx} \\ $$$$=\pi\:\left[\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{5}}+\mathrm{27x}^{\mathrm{3}} +\mathrm{400x}−\frac{\mathrm{9}}{\mathrm{2}}\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{40}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} −\mathrm{180x}^{\mathrm{2}} \:\right]_{\mathrm{4}} ^{\mathrm{5}} \\ $$$$=\:\frac{\pi}{\mathrm{30}} \\ $$