Why Does My Integration by Parts Solution Differ from the Textbook's?

[bfd]

Ok so I was attempting to solve an integration by parts problem and somewhere along the line I got stuck. Here's the problem:

\int^{\infty}_{2} {x^2 e^{-x} - 2xe^{-x}

After using integration by parts twice I came up with this:

2xe^{-x} - x^{2}e^{-x} + 2e^{-x} \vert^{\infty}_{2}

But this differs greatly with what my book has to say. They came up with an answer around .5. Any ideas on this one?

 

[Allday]

Hi bfd,

Its hard to tell what you did without seeing the steps, but I got what your book got. (I assumed the dx went after the second term in the integral). Break the problem up into two integrals. One with x^2*exp(-x) as the integrand and the other with -2*x*exp(-x) as the integrand. Preform integration by parts on the first integral. This should give you two terms one of which cancels the second integral out. Then you have to evaluate the first term at the limits 2 and infinity. Again one term will be zero and that should leave you with an answer.

Regrads

 

[bfd]

Hi bfd,

Its hard to tell what you did without seeing the steps, but I got what your book got. (I assumed the dx went after the second term in the integral). Break the problem up into two integrals. One with x^2*exp(-x) as the integrand and the other with -2*x*exp(-x) as the integrand. Preform integration by parts on the first integral. This should give you two terms one of which cancels the second integral out. Then you have to evaluate the first term at the limits 2 and infinity. Again one term will be zero and that should leave you with an answer.

Regrads

Hi thanks for the reply. I wasn't sure you could break up the integral so I just took it all as one problem. I'm pretty rusty when it comes to calculus. I will try that