How do you evaluate the integral #int x^2e^x#?

1 Answer
Jun 1, 2017

Answer:

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

Explanation:

#int x^2 e^x dx#
#= x^2*e^x - int 2x*e^x#
#= x^2*e^x -( 2*x * e^x- int 2*e^x)#
#= x^2*e^x-(2*x * e^x - 2*e^x) +C#
#= e^x(x^2-2x+2)+C#

Explanation:
Integrate by parts :
#int x^2 e^x dx#
Let #u = x^2#, #v = e^x#. You will get #(d v) = e^x*dx#, #(du)=2x*dx#, .

To reduce the power of #x# in the integral, you may rewrite it in this form #int u * d v #. This new expression is equivalent to the original one. (by replacing #x^2# with u and #e^x *dx# with #(d v)#.)

#int u*d v = u*v - int du * v#
#int x^2 e^x dx = x^2*e^x - int 2x*e^x#

Now, let's do the partial integration again with #m = 2*x# this time. In this case #d m = 2 *dx#

#int m*d v = m*v - int dm * v#
#int 2x*e^x = 2*x * e^x- int 2*e^x = 2*x * e^x - 2*e^x +C#
(include the arbritary constant #C# in the expression)