How do you evaluate the integral ((x^2)+1) e^-x dx((x2)+1)exdx from 0 to 1?

1 Answer
Apr 11, 2018

3/e(e-2)3e(e2).

Explanation:

Let, I=int(x^2+1)e^-xdxI=(x2+1)exdx.

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.!