∫_0 ^∞ x^2 e^(−x) dx=?


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

$\underset{0}{\int^{\infty}} x^{2} e^{- x} d x = ?$
	Answered by a.lgnaoui last updated on 18/Feb/23

	$p o s o n s t = e^{x} x = l n t d x = \frac{d t}{t}$ $x^{2} e^{- x} = {(l n t)}^{2} \times \frac{1}{t} = \frac{l n t}{t} \times l n t$ $\int_{0}^{\infty} x^{2} e^{- x} d x = \int_{1}^{\infty} {[{(\frac{1}{2} l n t)}^{2}]}^{'} l n t$ $U = \frac{1}{2} {[l n t]}^{2} V = l n t V^{^{'}} = \frac{1}{t}$ ${I = \frac{1}{2} {(l n t)}^{2} - \frac{1}{2} \int \frac{{l n t)}^{2}}{t} d t = \frac{1}{2} {(l n t)}^{3}]}_{1}^{\infty} - J$ $2 I = \frac{1}{2} {[{(l n t)}^{2}]}_{1}^{\infty} \Rightarrow I = + \infty$
	Answered by ARUNG_Brandon_MBU last updated on 18/Feb/23

	$\int_{0}^{\infty} x^{2} e^{- x} d x = Γ (3) = 2! = 2$
	Answered by ARUNG_Brandon_MBU last updated on 18/Feb/23
	$I=∫_0 ^∞ x^2 e^(−x) dx { ((u(x)=x^2 )),((v′(x)=e^(−x) )) :}⇒ { ((u′(x)=2x)),((v(x)=−e^(−x) )) :} I=[−x^2 e^(−x) ]_0 ^∞ +2∫_0 ^∞ xe^(−x) dx =2∫_0 ^∞ xe^(−x) dx=2[−xe^(−x) ]_0 ^∞ +2∫_0 ^∞ e^(−x) dx =2∫_0 ^∞ e^(−x) dx=2[−e^(−x) ]_0 ^∞ =2$
	$I = \int_{0}^{\infty} x^{2} e^{- x} d x$ ${\begin{cases} u (x) = x^{2} \\ v^{'} (x) = e^{- x} \end{cases} \Rightarrow {\begin{cases} u^{'} (x) = 2 x \\ v (x) = - e^{- x} \end{cases}$ $I = {[- x^{2} e^{- x}]}_{0}^{\infty} + 2 \int_{0}^{\infty} {x e}^{- x} d x$ $= 2 \int_{0}^{\infty} {x e}^{- x} d x = 2 {[- {x e}^{- x}]}_{0}^{\infty} + 2 \int_{0}^{\infty} e^{- x} d x$ $= 2 \int_{0}^{\infty} e^{- x} d x = 2 {[- e^{- x}]}_{0}^{\infty} = 2$
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