Integration by Parts

1. May 26, 2005

cogs24

hi guys
just doing some revision and im stuck on this question

*integral sign* x^2 . exponential ^ -3x . dx

I know i have to use integration by parts, but i just cant seem to get it out
any ideas?
thanx

2. May 26, 2005

dextercioby

HINT:Take $$dv=e^{-3x} \ dx$$ and $$u=x^2$$.

U'll figure out what to do next.

Daniel.

3. May 26, 2005

cogs24

but, what is the integral of exponential ^ -3x?

4. May 26, 2005

cogs24

is it -1/3 *exponential* ^ -3x?

5. May 26, 2005

dextercioby

Yes,as you can check by differentiation.

Daniel.

6. May 26, 2005

Jameson

The tabular method would work great on this question. Are you familiar with this?

7. May 26, 2005

cogs24

ok, once i sub u, du, v and dv into the integral by parts formula, i have to assign, u and dv again to:
-1/3 exponential ^ -3x and 2x

so is dv assigned to the exponential again, like in the first case?

8. May 26, 2005

cogs24

no, i am not aware of the tabular method, sorry

9. May 26, 2005

cogs24

ok, ive got my answer, unfortunately i dont kow if its right, we dont get solutions for this exercise

-1/3 *exponential* ^ -3x . x^2 - 1/9*exponential*^-3x . 2x - 2/27*exponential*^-3x

10. May 26, 2005

whozum

Maple agrees, good job.

11. May 26, 2005

dextercioby

Daniel.

12. May 27, 2005

cogs24

arbitary constant would be + c

13. May 27, 2005

Jameson

In case you were interested in this method, http://marauder.millersville.edu/~bikenaga/calculus/parts/partspf.html [Broken] is a link to the explanation. If one of the two terms will eventually differentiate to zero, this method is much less time consuming.

Jameson

Last edited by a moderator: May 2, 2017
14. May 28, 2005

cogs24

i see, thanx for that link, i will probably apply this method from now on