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lolokay · Feb 15, 2009

$\int_{0}^{a} f(2a-x)dx\\ let\ u = 2a-x\\ du = -dx\\ when\ x=0,\ u=2a,\ when\ x=a,\ u=a\\ \int_{2a}^{a}f(u)[-du]\\ \int_{a}^{2a}f(u)du\ =\int_{a}^{2a}f(x)dx\\ \int_{0}^{a}f(x)dx+\int_{a}^{2a}f(x)dx\\ =\int_{0}^{2a}f(x)dx$
you should start with the LHS side though

shaon0 · Feb 15, 2009

RHS=S [ 0 to a] f(x) dx + S [a to 2a] f(x) dx
= F(a)-F(0)+F(2a)-F(a)
= F(2a)+F(0)

LHS=F(a)-F(a)+F(0)+F(2a)
= F(2a)+F(0)
= RHS

shuning · Feb 15, 2009

just wondering, what topic does this question fall under? it's not quite integration is it?

lolokay · Feb 15, 2009

yeah, it's integration

Aerath · Feb 15, 2009

Yeah integration.

shuning · Feb 15, 2009

lolokay said:
yeah, it's integration

cant find it in my txt book T.T

oh well... i want to die!!! my HW is too hard, i cant even go thru 1 whole question!!! fuck!!!

Aerath · Feb 15, 2009

Just ask the questions here. I know I, amongst many others would be more than happy to help out.

shuning · Feb 15, 2009

Aerath said:
Just ask the questions here. I know I, amongst many others would be more than happy to help out.

nice, can u please add me on msn? my msn is shuning_zhao@hotmail.com

thanks very much ^^

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