# Integral Help!

• Apr 6th 2009, 09:11 PM
Mudd_101
Integral Help!
Hey, can you show me how to do the integral for this question

(t^2)(e^(-t^3)

Thanks
• Apr 6th 2009, 09:20 PM
Chris L T521
Quote:

Originally Posted by Mudd_101
Hey, can you show me how to do the integral for this question

(t^2)(e^(-t^3)

Thanks

Using the substitution $\displaystyle u=-t^3$, I leave it for you to show that $\displaystyle \int t^2e^{-t^3}\,dt\xrightarrow{u=-t^3}{}-\tfrac{1}{3}\int e^u\,du$

Can you continue with the calculations?
• Apr 6th 2009, 09:32 PM
Mudd_101
I almost got the answer but I'm negative off uhh I put my steps down their if you could Pleaassse look at them but uhh I'm new to integrals so I think they will be difficult to follow cuz I don't really know what I'm doing.

okay so I have t^2 e^u

u = -t^3
-3t^2 dt = du
t^2 dt= (-1/3)du

so then I have: (-1/3)du *e^u
then I did the integral of e^u (leaving behind the -1/3 from the du)
(-1/3) ((e^(-t^3))/(-1))

The answer in my text is (-1/3)((e^(-t^3))
• Apr 6th 2009, 09:45 PM
Chris L T521
Quote:

Originally Posted by Mudd_101
I almost got the answer but I'm negative off uhh I put my steps down their if you could Pleaassse look at them but uhh I'm new to integrals so I think they will be difficult to follow cuz I don't really know what I'm doing.

okay so I have t^2 e^u

u = -t^3
-3t^2 dt = du
t^2 dt= (-1/3)du

so then I have: (-1/3)du *e^u
then I did the integral of e^u (leaving behind the -1/3 from the du)
(-1/3) ((e^(-t^3))/(-1))

The answer in my text is (-1/3)((e^(-t^3))

I'm not sure how you got that -1...

When you evaluate $\displaystyle -\tfrac{1}{3}\int e^u\,du$, you end up with $\displaystyle -\tfrac{1}{3}e^u+C$. Now back substitute to get $\displaystyle \boxed{-\tfrac{1}{3}e^{-t^3}+C}$
• Apr 6th 2009, 09:55 PM
Mudd_101
ooo thanks