Question to Calculus Tutorial

**coobe** · June 7th 2009, 10:20 AM

- hello again

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

$\int x^2 e^x<br />$

**TheEmptySet** · June 7th 2009, 10:29 AM

Originally Posted by coobe

hello again

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

$\int x^2 e^x<br />$

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

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

**coobe** · June 7th 2009, 10:45 AM

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

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 ?

**Chop Suey** · June 7th 2009, 10:58 AM

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.

**HallsofIvy** · June 7th 2009, 02:05 PM

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.

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