Question Number 135062 by liberty last updated on 09/Mar/21
$$\mathrm{Let}\:\mathrm{f}\left(\mathrm{0}\right)\:=\:\mathrm{a}\:;\:\mathrm{f}\left(\mathrm{3}\right)=\mathrm{0}\:\mathrm{and}\:\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{e}^{\mathrm{x}^{\mathrm{4}} } \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\int_{\mathrm{0}} ^{\:\mathrm{3}} \mathrm{x}^{\mathrm{2}} \:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:? \\ $$
Answered by john_santu last updated on 10/Mar/21
![integration by parts { ((u=f(x)⇒du=f ′(x)dx)),((v=(x^3 /3))) :} J = [ ((x^3 f(x))/3) ]_0 ^3 −(1/3)∫_0 ^( 3) x^3 f ′(x) dx J = 0−(1/3)∫_0 ^( 3) x^3 e^(x^4 ) dx J=−(1/(12))∫_0 ^( 3) e^x^4 d(x^4 )=−(1/(12))[ e^x^4 ]_0 ^3 J=−(1/(12)) [e^(81) −1 ]](https://www.tinkutara.com/question/Q135063.png)
$${integration}\:{by}\:{parts}\: \\ $$$$\:\begin{cases}{{u}={f}\left({x}\right)\Rightarrow{du}={f}\:'\left({x}\right){dx}}\\{{v}=\frac{{x}^{\mathrm{3}} }{\mathrm{3}}}\end{cases} \\ $$$$\mathbb{J}\:=\:\left[\:\frac{{x}^{\mathrm{3}} \:{f}\left({x}\right)}{\mathrm{3}}\:\right]_{\mathrm{0}} ^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\mathrm{3}} {x}^{\mathrm{3}} \:{f}\:'\left({x}\right)\:{dx}\: \\ $$$$\mathbb{J}\:=\:\mathrm{0}−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\mathrm{3}} {x}^{\mathrm{3}} \:{e}^{{x}^{\mathrm{4}} \:} {dx}\: \\ $$$$\mathbb{J}=−\frac{\mathrm{1}}{\mathrm{12}}\int_{\mathrm{0}} ^{\:\mathrm{3}} {e}^{{x}^{\mathrm{4}} } \:{d}\left({x}^{\mathrm{4}} \right)=−\frac{\mathrm{1}}{\mathrm{12}}\left[\:{e}^{{x}^{\mathrm{4}} } \:\right]_{\mathrm{0}} ^{\mathrm{3}} \\ $$$$\mathbb{J}=−\frac{\mathrm{1}}{\mathrm{12}}\:\left[{e}^{\mathrm{81}} −\mathrm{1}\:\right]\: \\ $$