# Thread: Question to Calculus Tutorial

1. ## 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
$

2. 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

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

3. ## 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 ?

4. 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.

5. 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.