$f$ twice differentiable, $f(a)=f(b)=g(a)=g(b)=0$ $\implies$ $\int_a^b f''(x)g(x)dx=\int_a^bf(x)g''(x)dx$

Question

$f:[a,b]\rightarrow \mathbb{R}$ twice continuously differentiable,
$f(a)=f(b)=g(a)=g(b)=0$ $\implies$ $\int_a^b f''(x)g(x)dx=\int_a^bf(x)g''(x)dx$

I think this has something to do with integration by parts, but I cannot see it yet.

Hints?

Answer 1

Let $u = g(x)$. Then $du = g'(x) \, dx$, $dv = f''(x) \, dx$, and $v = f'(x)$, and you have $$ \int_a^b f''(x) g(x) \, dx = g(x)f'(x)\bigg|_{x=a}^{x=b} - \int_a^b f'(x)g'(x) \, dx. $$ The first term on the RHS is zero because $g(a) = g(b) = 0$. We don't know what $f'(a)$ and $f'(b)$ are but who cares - they're being multiplied by zero.

So we get $$ \int_a^b f''(x) g(x) \, dx = - \int_a^b f'(x)g'(x) \, dx. $$

Integrate by parts again but be careful how you choose $u$. If you choose $u$ incorrectly you'll go in circles.

Answer 2

Note $$ \int_a^b f''(x)g(x)dx=\int_a^bg(x)d(f'(x))=f'(x)g(x)|_{a}^b-\int_a^bf'(x)g'(x)dx=-\int_a^bf'(x)g'(x)dx.$$ Similarly $$ \int_a^bf(x)g''(x)dx=-\int_a^bf'(x)g'(x)dx.$$