I tried doing this, but ended up at a dead end.

Use integration by parts to show that the integral from 0 to 1 of f''(x)g(x)dx = the integral from 0 to 1 of f(x)g''(x)dx.

suppose f(0)=g(0)=0 and f(1)=g(1)=0. give examples for g(x) and f(x).

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- Feb 12th 2009, 12:25 PMvalkyrieprove using integration by parts
I tried doing this, but ended up at a dead end.

Use integration by parts to show that the integral from 0 to 1 of f''(x)g(x)dx = the integral from 0 to 1 of f(x)g''(x)dx.

suppose f(0)=g(0)=0 and f(1)=g(1)=0. give examples for g(x) and f(x). - Feb 12th 2009, 12:52 PMScott H
Are $\displaystyle f(0)=g(0)=f(1)=g(1)=0$ suppositions of the first problem?

With integration by parts, we obtain a new integral by integrating one factor and differentiating the other. It's a way to bump one factor up and the other down, so to speak. For example,

$\displaystyle \int f'(x)g(x)\,dx=f(x)g(x) - \int f(x)g'(x)\,dx.$

The problem given can be solved by integrating $\displaystyle f$ and differentiating $\displaystyle g$ twice.