Question to Calculus Tutorial

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June 7th 2009, 10:20 AM
coobe

Question to Calculus Tutorial

- hello again :)

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

$\int x^2 e^x<br />$
June 7th 2009, 10:29 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:

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

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

$\int x^2 e^xdx=x^2e^x-2xe^x+2e^x +C$
June 7th 2009, 10:45 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, 10:58 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, 02:05 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.