Integration by parts, where am I going wrong?

[Kaldanis]

Homework Statement

$\int_{1}^{2} x^2 e^{x} dx$

Homework Equations

Integrating by parts. Writing out chain rule, integrating both sides and rearranging gives ∫f(x)g'(x) dx = f(x)g(x) - ∫f'(x)g(x) dx

The Attempt at a Solution

$\int_{1}^{2} x^2 e^{x} dx = \left[x^2 e^x\right]_{1}^{2} - \int_{1}^{2} 2x e^{x} dx$

Applying again gives:

$\int_{1}^{2} x^2 e^{x} dx = \left[x^2 e^x\right]_{1}^{2} - \left(\left[2x e^x\right]_{1}^{2} - \int_{1}^{2} 2 e^{x} dx\right)$

Integrating last term gives:

$\int_{1}^{2} x^2 e^{x} dx = \left[x^2 e^x\right]_{1}^{2} - \left(\left[2x e^x\right]_{1}^{2} - \left[2e^x\right]_{1}^{2}\right)$

When I simplify this I get -2e² + 3e, however I know the answer should be 2e²-e.

Can someone please point out where my mistake is?

 

[vela]

You got the calculus part right. You just didn't simplify it correctly, but we can't tell you where since you didn't show us that part of your work.

 

[Kaldanis]

You got the calculus part right. You just didn't simplify it correctly, but we can't tell you where since you didn't show us that part of your work.

Ah yeah, I was getting frustrated with trying to write it out in latex! But thanks for letting me know I've at least got all that part right, I'll try again a few more times and then write it out here if I can't get it

 

[jedishrfu]

you didnt show you last step so I think you made a sign error there.

(4ee - e) - ( 4ee - 2e - (2ee - 2e) ) ==> 4ee - e - 4ee + 2e + 2ee - 2e ==> 2ee - e

 

[Kaldanis]

Thank you :D I got it. I screwed up a + and - my first 50 times of trying to solve this.

(2²e²-1²e) - ( (2*2e² - 2e) - (2e² - 2e) )
= (4e² - e) - ( (4e² - 2e) - (2e² - 2e) )
= (4e² - e) - ( 4e² - 2e -2e² + 2e )
= 4e² - e - 4e² + 2e + 2e² - 2e
= 2e² - e