How do you evaluate the integral $((x^2)+1) e^-x dx$ from 0 to 1?

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Answer 1 · 2018-04-11T17:52:19

$3/e(e-2)$ .

Let, $I=int(x^2+1)e^-xdx$ .

Using the following Rule of Integration by Parts (IBP) :

IBP : $intuv'dx=uv-intvu'dx$ .

We take, $u=x^2+1, and, v'=e^-x$ .

$:. u'=2x, and, v=inte^-xdx=e^-x/-1=-e^-x$ .

$:. I=-(x^2+1)e^-x-int(-e^-x*2x)dx$ ,

$rArr I=-(x^2+1)e^-x+2I_1," where, "I_1=intxe^-xdx$ .

We once again use IBP for $I_1$ ; this time, we choose,

$u=x, and, v'=e^-x :. u'=1, and, v=-e^-x$ .

$:. I_1=-xe^-x-int(-e^-x*1)dx$ ,

$: I_1=-xe^-x-e^-x$ .

Utilising $I_1$ in $I$ , we get,

$I=-(x^2+1)e^-x+2{-xe^-x-e^-x}$ ,

$rArr I=-(x^2+2x+3)e^-x+C$ .

$:. int_0^1(x^2+1)e^-xdx=-[(x^2+2x+3)e^-x]_0^1$ ,

$=-[(1+2+3)e^-1-(0+0+3)e^-0]$ .

$rArr int_0^1(x^2+1)e^-xdx=-(6/e-3)=3/e(e-2)$ .

Enjoy Maths.!

How do you evaluate the integral ((x^2)+1) e^-x dx((x^2)+1) e^-x dx from 0 to 1?