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Fun integral problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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\[\Large \int\limits_a^b f''(x)g(x)dx +\int\limits_b^a g''(x)f(x)dx\] Evaluate
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Good luck! :)
ikram002p
  • ikram002p
*watching this *

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  • Empty
*drum roll*
freckles
  • freckles
like to make a conjecture \[(f(x))'|_a^b \cdot g(x)-(g(x))'|_a^b f(x)\]
freckles
  • freckles
oops missing pluggin a and b into those one things
anonymous
  • anonymous
Sturm-Liouville operators are self-adjoint, meaning $$\langle Lf,g\rangle=\langle f,Lg\rangle$$ over a suitable Hilbert space
freckles
  • freckles
\[[(f(x))' g(x)]_a^b-[(g(x))' f(x)]_a^b\]
anonymous
  • anonymous
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: $$\begin{align*}\int_a^b f''(x) g(x)\, dx&=f'(x)g(x)\bigg|_a^b -\int_a^b f'(x) g'(x)\, dx\\&=\bigg[f'(x)g(x)-f(x)g'(x)\bigg]_a^b+\int_a^b f(x) g''(x)\, dx\end{align*}$$
perl
  • perl
$$\large \int\limits_a^b f''(x)g(x)dx +\int\limits_b^a g''(x)f(x)dx \\= \large \int\limits_a^b f''(x)g(x)dx -\int\limits_a^b g''(x)f(x)dx \\= \large \int\limits_a^b [f''(x)g(x) -g''(x) f(x)]dx $$
freckles
  • freckles
oh integration by parts awesome @oldrin.bataku
anonymous
  • anonymous
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 $$\bigg[f'(x) g(x)-f(x)g'(x)\bigg]_a^b$$
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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
perl
  • perl
nice problem :)

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