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calculus/integral question

Started by jlewicki Sep 25, 2006 at 9:30 PM 2 replies 800+ views

Original Post

Sep 25, 2006 at 9:30 PM

i'm stumped on a problem atm: integral(x(3x+2)^2) i tried this: let u = 3x+2 du = (3)dx 1/3du = dx but there is an x left over, should i do this: since u = 3x+2, x = u-2/3, and substitute that? integral(x(3x+2)^2) = (1/3)integral( (u-2/3) (u)^2)

Zipster

511

Zipster

Sep 25, 2006 at 9:40 PM

You don't need to do any fancy integration by parts, just expand everything out and you should get something simpler.

TheAdmiral

Sep 26, 2006 at 1:12 PM

I shan't give any answers either, as this looks like homework. But indeed, your integrand is a simple polynomial, so expanding into terms and integrating as normal makes for a problem about as simple as they get. In fact, if you perform that substitution, the equation doesn't get any simpler and you're left with a messy substitution to get back into x coordinates.

But for completeness's sake, yes - you would need to transform the remaining x. A substitution is only useful if you can express the entire expression in the new variable. You have the right idea, but be careful with the inversion:

u = 3x + 2 => x = (u - 2) / 3

so

Int[x(3x + 2)² . dx] = (1/3)Int[((u - 2) / 3)u² . du]

I'm not sure if you made a geniune error or if you were just ignoring the operator precedence.

Regards
Admiral

grhodes_at_work

Mod

197

grhodes_at_work

Sep 29, 2006 at 8:37 AM

Closed. This is homework, or indistinguishable from it. Forum FAQ

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