Ok so I don't understand what I'm supposed to be doing in this problem:

if f(0) = g(0) = 0 and f" and g" are continuous, show that

$\displaystyle \int_0^a f(x)g''(x) \, dx = f(a)g'(a) - f'(a)g(a) + \int_0^a f''(x)g(x) \, dx$

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- Sep 27th 2011, 10:30 AMourdemiseIntegration by parts problem.
Ok so I don't understand what I'm supposed to be doing in this problem:

if f(0) = g(0) = 0 and f" and g" are continuous, show that

$\displaystyle \int_0^a f(x)g''(x) \, dx = f(a)g'(a) - f'(a)g(a) + \int_0^a f''(x)g(x) \, dx$ - Sep 27th 2011, 10:38 AMskeeterRe: Integration by parts problem, please help?
$\displaystyle \int_0^a f(x) \cdot g''(x) \, dx$

$\displaystyle u = f(x)$ , $\displaystyle dv = g''(x) \, dx$

$\displaystyle du = f'(x) \, dx$ , $\displaystyle v = g'(x)$

$\displaystyle \left[f(x) \cdot g'(x)\right]_0^a - \int_0^a g'(x) \cdot f'(x) \, dx$

do parts again w/ the last integral - Sep 27th 2011, 10:46 AMSironRe: Integration by parts problem, please help?
You have to prove that:

$\displaystyle \int_{0}^{a} f(x)g''(x)dx=f(a)g'(a)-f'(a)g(a)+\int_{0}^{a}f''(x)g(x)dx$

If we use integration by parts on this integral: $\displaystyle \int_{0}^{a}f(x)g''(x)dx$ then, let $\displaystyle f(x)=u$ with $\displaystyle dv=g''(x) \Rightarrow v=g'(x)$

Therefore we get:

$\displaystyle \int_{0}^{a}f(x)d[g'(x)]=[f(x)\cdot g'(x)]_{0}^{a}-\int_{0}^{a} f'(x)g'(x)dx$

Now use integration by parts again to compute $\displaystyle \int_{0}^{a}f'(x)g'(x)dx$

EDIT: sorry, I didn't see skeeter's post :)