Integration Questions

Question Number 187531 by sciencestudentW last updated on 18/Feb/23

$$\underset{\mathrm{0}} {\overset{\infty} {\int}}{x}^{\mathrm{2}} {e}^{−{x}} {dx}=? \\$$

Answered by a.lgnaoui last updated on 18/Feb/23

$${posons}\:\:\:{t}={e}^{{x}} \:\:{x}={lnt}\:\:\:{dx}=\frac{{dt}}{{t}} \\$$$${x}^{\mathrm{2}} {e}^{−{x}} =\left({lnt}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{{t}}=\frac{{lnt}}{{t}}×{lnt} \\$$$$\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−{x}} {dx}=\int_{\mathrm{1}} ^{\infty} \left[\left(\frac{\mathrm{1}}{\mathrm{2}}{lnt}\right)^{\mathrm{2}} \right]'{lnt} \\$$$${U}=\frac{\mathrm{1}}{\mathrm{2}}\left[{lnt}\right]^{\mathrm{2}} \:\:\:\:{V}={lnt}\:\:\:\:\:{V}^{'} =\frac{\mathrm{1}}{{t}} \\$$$$\left.{I}=\frac{\mathrm{1}}{\mathrm{2}}\left({lnt}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left.{lnt}\right)^{\mathrm{2}} }{{t}}{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left({lnt}\right)^{\mathrm{3}} \right]_{\mathrm{1}} ^{\infty} −{J} \\$$$$\mathrm{2}{I}=\frac{\mathrm{1}}{\mathrm{2}}\left[\left({lnt}\right)^{\mathrm{2}} \right]_{\mathrm{1}} ^{\infty} \Rightarrow\:{I}=+\infty \\$$

Answered by ARUNG_Brandon_MBU last updated on 18/Feb/23

$$\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−{x}} {dx}=\Gamma\left(\mathrm{3}\right)=\mathrm{2}!=\mathrm{2} \\$$

Answered by ARUNG_Brandon_MBU last updated on 18/Feb/23

$${I}=\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−{x}} {dx} \\$$$$\begin{cases}{\mathrm{u}\left({x}\right)={x}^{\mathrm{2}} }\\{\mathrm{v}'\left({x}\right)={e}^{−{x}} }\end{cases}\Rightarrow\begin{cases}{\mathrm{u}'\left({x}\right)=\mathrm{2}{x}}\\{\mathrm{v}\left({x}\right)=−{e}^{−{x}} }\end{cases} \\$$$${I}=\left[−{x}^{\mathrm{2}} {e}^{−{x}} \right]_{\mathrm{0}} ^{\infty} +\mathrm{2}\int_{\mathrm{0}} ^{\infty} {xe}^{−{x}} {dx} \\$$$$\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {xe}^{−{x}} {dx}=\mathrm{2}\left[−{xe}^{−{x}} \right]_{\mathrm{0}} ^{\infty} +\mathrm{2}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {dx} \\$$$$\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {dx}=\mathrm{2}\left[−{e}^{−{x}} \right]_{\mathrm{0}} ^{\infty} =\mathrm{2} \\$$