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  • one year ago

Fun integral problem

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  1. Empty
    • one year ago
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    \[\Large \int\limits_a^b f''(x)g(x)dx +\int\limits_b^a g''(x)f(x)dx\] Evaluate

  2. Empty
    • one year ago
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    Good luck! :)

  3. ikram002p
    • one year ago
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    *watching this *

  4. Empty
    • one year ago
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    *drum roll*

  5. freckles
    • one year ago
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    like to make a conjecture \[(f(x))'|_a^b \cdot g(x)-(g(x))'|_a^b f(x)\]

  6. freckles
    • one year ago
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    oops missing pluggin a and b into those one things

  7. anonymous
    • one year ago
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    Sturm-Liouville operators are self-adjoint, meaning $$\langle Lf,g\rangle=\langle f,Lg\rangle$$ over a suitable Hilbert space

  8. freckles
    • one year ago
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    \[[(f(x))' g(x)]_a^b-[(g(x))' f(x)]_a^b\]

  9. anonymous
    • one year ago
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    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: $$\begin{align*}\int_a^b f''(x) g(x)\, dx&=f'(x)g(x)\bigg|_a^b -\int_a^b f'(x) g'(x)\, dx\\&=\bigg[f'(x)g(x)-f(x)g'(x)\bigg]_a^b+\int_a^b f(x) g''(x)\, dx\end{align*}$$

  10. perl
    • one year ago
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    $$\large \int\limits_a^b f''(x)g(x)dx +\int\limits_b^a g''(x)f(x)dx \\= \large \int\limits_a^b f''(x)g(x)dx -\int\limits_a^b g''(x)f(x)dx \\= \large \int\limits_a^b [f''(x)g(x) -g''(x) f(x)]dx $$

  11. freckles
    • one year ago
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    oh integration by parts awesome @oldrin.bataku

  12. anonymous
    • one year ago
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    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 $$\bigg[f'(x) g(x)-f(x)g'(x)\bigg]_a^b$$

  13. Empty
    • one year ago
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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

  14. perl
    • one year ago
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    nice problem :)

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