1. ## Integral Problem

Hi, I am having some trouble with this integral problem,

I believe it is integration by parts. I don't know the "[m@th][/m@th]" tags so if someone could point me to a link that would be great too.
Thanks,
Matt

2. Originally Posted by matt3D
Hi, I am having some trouble with this integral problem,

I believe it is integration by parts.
Thanks,
Matt
yes, integration by parts.

let $u = t$ (the part you differentiate) and $dv = e^{-t}~dt$

can you take it from here?

3. I get stuck after -te^-t - int from 0 to 2 -e^-t dt

4. Originally Posted by matt3D
I get stuck after -te^-t - int from 0 to 2 -e^-t dt
stuck with what? finding $\int_0^2-e^{-t}~dt$?

5. Originally Posted by Jhevon
stuck with what? finding $\int_0^2-e^{-t}~dt$?
Yea!

6. Originally Posted by matt3D
Yea!
i am really surprised by this

you manage to do $\int e^{-t}~dt$ but you can't do $\int-e^{-t}~dt$? i don't get you

Note that $\int-e^{-t}~dt = - \int e^{-t}~dt$ which you know how to find, because you did it already

is it the limits that are bothering you? you know how to use the fundamental theorem of calculus?

7. Okay, I get $-te^{-t}-\int_0^2-e^{-t}~dt$
And I am stuck with the new integral to the right. Yes, I do know the fundamental theorem of calculus. Could you just show me the steps to the whole problem? I really appreciate your time and help.

8. Originally Posted by matt3D
Okay, I get $-te^{-t}-\int_0^2-e^{-t}~dt$
And I am stuck with the new integral to the right. Could you just show me the steps to the whole problem? I really appreciate your time and help.
you're lucky i have to leave now, so i can't spend the time having you work things out

$\int_0^2 te^{-t}~dt = -te^{-t} - \int_0^2 -e^{-t}~dt$

$= -te^{-t} + \int_0^2 e^{-t}~dt$

$= -te^{-t} - e^{-t} \bigg|_0^2$

$= -(t + 1)e^{-t} \bigg|_0^2$

$= -(2 + 1)e^{-2} - [-(0 + 1)e^0]$ .............by the fundamental theorem of calculus

$= -3e^{-2} + 1$

9. Thank you so much!

10. Originally Posted by matt3D
I don't know the "[m@th][/m@th]" tags so if someone could point me to a link that would be great too.
Thanks,
Matt
see here