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
$\displaystyle \int_0^a f(x)g''(x) \, dx = f(a)g'(a) - f'(a)g(a) + \int_0^a f''(x)g(x) \, dx$
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
$\displaystyle \int_0^a f(x)g''(x) \, dx = f(a)g'(a) - f'(a)g(a) + \int_0^a f''(x)g(x) \, dx$
$\displaystyle \int_0^a f(x) \cdot g''(x) \, dx$
$\displaystyle u = f(x)$ , $\displaystyle dv = g''(x) \, dx$
$\displaystyle du = f'(x) \, dx$ , $\displaystyle v = g'(x)$
$\displaystyle \left[f(x) \cdot g'(x)\right]_0^a - \int_0^a g'(x) \cdot f'(x) \, dx$
do parts again w/ the last integral
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