How do you evaluate the definite integral #int_0^(oo)x^3e^(-x)dx#?

first, integrate the indefinite integral #intx^3e^-xdx#

with some integration by parts, you should get #-(x^3+3x^2+6x+6)/e^x+C# (althought the C doesnt matter here)

the answer is #-(x^3+3x^2+6x+6)/e^x# as #xrarroo# subtracted by #-(x^3+3x^2+6x+6)/e^x# at #x=0#

since #e^x# increases much faster than the polynomial in the numerator, #-(x^3+3x^2+6x+6)/e^x# as #xrarroo# is #0#.
#-(x^3+3x^2+6x+6)/e^x# at #x=0# is #-6# from direct substitution.

so the answer is #0-(-6)=6#

Filling in some of the extra steps...

You can use integration by parts.

#uv-intvdu#

Let:

#u=x^3 " " du=3x^2dx#

#dv=e^(-x)" " v=-e^(-x)#

Substituting into the above expression:

#=>-x^3e^(-x)+3intx^2e^(-x)dx#

Apply again:

#u=x^2 " " du=2xdx#

#dv=e^(-x)" " v=-e^(-x)#

#=>-x^3e^(-x)+3[-x^2e^(-x)+2intxe^(-x)dx]#

Finally:

#u=x " " du=dx#

#dv=e^(-x)" " v=-e^(-x)#

#=>-x^3e^(-x)+3[-x^2e^(-x)+2{-xe^(-x)+inte^(-x)dx}]#

#-x^3-3x^2-6x-6#

Simplify:

#-e^(-x)(x^3+3x^2+6x+6)#

Evaluate from #0->oo#.

Note that #e^(-x)=1/e^x#. When we plug in "#oo#", think of this as one divided by a very big number (to say the least). This is approximately zero.

#=>0-[-1/e^0((0)^3+3(0^2)+6(0)+6)]#

Note that #e^0=1#.

#=>1(6)=6#