# Thread: Integration by parts problem.

1. ## Integration 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

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

Originally Posted by ourdemise
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

$\int(from 0 to a) f(x)g"(x)dx = f(a)g'(a) - f'(a)g(a) + \int(from 0 to a)f"(x)g(x)dx$
$\int_0^a f(x) \cdot g''(x) \, dx$

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

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

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

do parts again w/ the last integral

$\int_{0}^{a} f(x)g''(x)dx=f(a)g'(a)-f'(a)g(a)+\int_{0}^{a}f''(x)g(x)dx$
If we use integration by parts on this integral: $\int_{0}^{a}f(x)g''(x)dx$ then, let $f(x)=u$ with $dv=g''(x) \Rightarrow v=g'(x)$
$\int_{0}^{a}f(x)d[g'(x)]=[f(x)\cdot g'(x)]_{0}^{a}-\int_{0}^{a} f'(x)g'(x)dx$
Now use integration by parts again to compute $\int_{0}^{a}f'(x)g'(x)dx$