How do you determine if the improper integral converges or diverges #intx^2 e^-x dx# from 0 to infinity?

We will make use of integration by parts to find a solution to the indefinite integral, and then apply the limits to evaluate the improper definite integral.

#intx^2e^(-x)dx#

Integration by parts 1:

Let #u = x^2# and #dv = e^-xdx#
Then #du = 2xdx# and #v = -e^-x#

Applying the integration by parts formula #intudv = uv-intvdu#:

#intx^2e^-xdx = -x^2e^-x+2intxe^-xdx#

Integration by parts 2:

Let #u = x# and #dv = e^-xdx#
Then #du = dx# and #v = -e^-x#

Applying the formula to the remaining integral:

#intxe^-xdx = -xe^-x+inte^-xdx#

#= -xe^-x-e^-x+C#

Substituting this back in, we have

#intx^2e^-xdx = -x^2e^-x+2(-xe^-x-e^-x)+C#

#=-e^(-x)(x^2+2x+2)+C#

Now we can check the definite integral:

#int_0^oox^2e^-xdx = lim_(M->oo)int_0^Mx^2e^-xdx#

#=lim_(M->oo)[-e^-x(x^2+2x+2)]_0^M#

#=lim_(M->oo)-(M^2+2M+2)/e^M+2#

Intuitively we can say at this point that the first term will go to #0#, as the exponential #e^x# in the denominator will grow much faster than the polynomial in the numerator. This will give us an answer of #2#. Still, we can prove this using L'Hopital's rule to show that the initial term does converge to #0#.

#lim_(x->oo)(x^2+2x+2)/e^x = (d/dx(x^2+2x+2))/(d/dxe^x)#

#=lim_(x->oo)(2x+2)/e^x#

#=lim_(x->oo)(d/dx(2x+2))/(d/dxe^x)#

#=lim_(x->oo)(2/e^x)#

#2/oo#

#=0#

With that, using that if two functions converge at a limit, then the limit of their sum is equal to the sum of their limits, we have:

#lim_(M->oo)2-(M^2+2M+2)/e^M = lim_(M->oo)2-lim_(M->oo)(M^2+2M+2)/e^M#

#=2+0#

#+2#

Therefore the integral converges to #2#.