# Question to Calculus Tutorial

• June 7th 2009, 11:20 AM
coobe
Question to Calculus Tutorial
- hello again :)

im working my way thourgh the awesome calculus tutorial but now im stuck at one point:

$\int x^2 e^x
$
• June 7th 2009, 11:29 AM
TheEmptySet
Quote:

Originally Posted by coobe
hello again :)

im working my way thourgh the awesome calculus tutorial but now im stuck at one point:

$\int x^2 e^x
$

You will need integration by parts twice

The first time gives

$\int x^2 e^xdx=x^2e^x-2\int xe^xdx$

and the 2nd is

$\int x^2 e^xdx=x^2e^x-2(xe^x-\int e^xdx)$

So finally you get

$\int x^2 e^xdx=x^2e^x-2xe^x+2e^x +C$
• June 7th 2009, 11:45 AM
coobe
Question to Calculus Tutorial
$\int x^2 e^x$
$\\u = x^2\\$
$\\v'=e^x\\$
$\\u'=2x\\$
$\\v =e^x\\$
$\\x^2e^x-\int 2xe^xdx = x^2 e^x -2\int x e^x$

everything clear so far. but now it seems the author is missing an $e^x$
the given solution: $x^2-2xe^x+2e^x+C$

is it just a typo or am i missing something ?
• June 7th 2009, 11:58 AM
Chop Suey
You're right. It should've been:
$x^2e^x - 2xe^x + 2e^x + C$

You can always to differentiate your answer and see if you get your original integrand. That way you can make sure if you got it right or not.
• June 7th 2009, 03:05 PM
HallsofIvy
Quote:

Originally Posted by coobe
hello again :)

im working my way thourgh the awesome calculus tutorial but now im stuck at one point:

$\int x^2 e^x$
$\\u = x^2\\$
$\\v'=e^x\\$
$\\u'=2x\\$
$\\v =e^x\\$
$\\x^2e^x-\int 2xe^xdx = x^2 e^x -2\int x e^x$

everything clear so far. but now it seems the author is missing an $e^x$
the given solution: $x^2-2xe^x+2e^x+C$

is it just a typo or am i missing something ?

You haven't finished. Integrate $\int x e^x dx$ by using "integration by parts" again.