Question

Fun integral problem

Answer 1

\[\Large \int\limits_a^b f''(x)g(x)dx +\int\limits_b^a g''(x)f(x)dx\]

Evaluate

Answer 2

Good luck! :)

Answer 3

*watching this *

Answer 4

*drum roll*

Answer 5

like to make a conjecture
\[(f(x))'|_a^b \cdot g(x)-(g(x))'|_a^b f(x)\]

Answer 6

oops missing pluggin a and b into those one things

Answer 7

Sturm-Liouville operators are self-adjoint, meaning $$\langle Lf,g\rangle=\langle f,Lg\rangle$$ over a suitable Hilbert space

Answer 8

\[[(f(x))' g(x)]_a^b-[(g(x))' f(x)]_a^b\]

Answer 9

you can verify this by just integrating by parts, although in this case it is not as simple as there are no suitable boundary conditions: $$\begin{align*}\int_a^b f''(x) g(x)\, dx&=f'(x)g(x)\bigg|_a^b -\int_a^b f'(x) g'(x)\, dx\\&=\bigg[f'(x)g(x)-f(x)g'(x)\bigg]_a^b+\int_a^b f(x) g''(x)\, dx\end{align*}$$

Answer 10

$$\large \int\limits_a^b f''(x)g(x)dx +\int\limits_b^a g''(x)f(x)dx
\\= \large \int\limits_a^b f''(x)g(x)dx -\int\limits_a^b g''(x)f(x)dx
\\= \large \int\limits_a^b [f''(x)g(x) -g''(x) f(x)]dx
$$

Answer 11

oh integration by parts
awesome @oldrin.bataku

Answer 12

anyways, since $$\int_b^a f(x)g''(x) \, dx=-\int_a^b f(x)g''(x)\, dx$$ the remaining integral cancels out and we have $$\bigg[f'(x) g(x)-f(x)g'(x)\bigg]_a^b$$

Answer 13

Without using IBP (Which is admittedly better since it's systematic and doesn't rely on tricks) we can us this cheap trick is if you can recognize:

\[\frac{d}{dx}[f'(x)g(x)-f(x)g'(x)] = f''(x)g(x)-f(x)g''(x) \]

Similarly you can use this sort of trick to integrate:

\[\int \vec v \times \vec v '' d t = \vec v \times \vec v' \]

Fun stuff :D

Answer 14

nice problem :)