How do you evaluate the integral $\int x^{2} e^{x} d x$ $int x^2e^xdx$ from $- \infty$ $-oo$ to 0?

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Answer 1 · 2016-09-28T12:29:00

$= 2$ $=2$

you can IBP it

Indefinite Integral
$\int x^{2} e^{x} d x = \int x^{2} \frac{d}{d x} (e^{x}) d x$ $int x^2 e^x dx = int x^2 d/dx(e^x) dx$

$= \int x^{2} \frac{d}{d x} (e^{x}) d x$ $= int x^2 d/dx(e^x) dx$

$= x^{2} e^{x} - \int \frac{d}{d x} (x^{2}) e^{x} d x$ $= x^2 e^x - int d/dx( x^2) e^x dx$

$= x^{2} e^{x} - \int 2 x e^{x} d x$ $= x^2 e^x - int 2 x e^x dx$

$= x^{2} e^{x} - \int 2 x \frac{d}{d x} (e^{x}) d x$ $= x^2 e^x - int 2 x d/dx( e^x) dx$

$= x^{2} e^{x} - (2 x e^{x} - \int \frac{d}{d x} (2 x) e^{x} d x)$ $= x^2 e^x - ( 2 x e^x - int d/dx(2 x) e^x dx )$

$= x^{2} e^{x} - (2 x e^{x} - 2 \int e^{x} d x)$ $= x^2 e^x - ( 2 x e^x - 2int e^x dx )$

$= x^{2} e^{x} - 2 x e^{x} + 2 e^{x} + C$ $= x^2 e^x - 2 x e^x + 2 e^x + C$

$= e^{x} (x^{2} - 2 x + 2) + C$ $= e^x (x^2 - 2 x + 2 ) + C$

Definite Integral

$= {[e^{x} (x^{2} - 2 x + 2)]}_{- \infty}^{x}$ $= [ e^x (x^2 - 2 x + 2 ) ]_(-oo)^0$

$= 2$ $= 2$

How do you evaluate the integral int x^2e^xdx∫x2exdxint x^2e^xdx from -oo−∞-oo to 0?