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David.Butler

  • 2 years ago

\int _{ 0 }^{ pi }{ \frac { sin\theta }{ \sqrt { pi\quad -\quad \theta } } } using\quad \frac { 1 }{ \sqrt { pi\quad -\quad \theta } } \quad as\quad comparison

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  1. ChelseaSweets
    • 2 years ago
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    what???

  2. David.Butler
    • 2 years ago
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    \[\int\limits _{ 0 }^{ \pi }{ \frac { \sin\theta }{ \sqrt { \pi\quad -\quad \theta } } } using\quad \frac { 1 }{ \sqrt { \pi\quad -\quad \theta } } \quad as\quad comparison\]

  3. ChelseaSweets
    • 2 years ago
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    sorry idk how to do it, good luck though!

  4. David.Butler
    • 2 years ago
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    Hahaha, thanks.

  5. sirm3d
    • 2 years ago
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    you're asking if the integral is convergent or divergent?

  6. David.Butler
    • 2 years ago
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    Yeah

  7. sirm3d
    • 2 years ago
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    if \[\int_0^{\pi}\frac{d\theta}{\sqrt{\pi-\theta}}\] is convergent, you can conclude that the integral in question is also convergent.

  8. David.Butler
    • 2 years ago
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    How do I know that integral you stated above is convergent? That's the part that is troubling me.

  9. sirm3d
    • 2 years ago
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    why don't you evaluate the definite integral? if it is a value, then the integral is convergent.

  10. David.Butler
    • 2 years ago
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    Oh okay, I thought it was improper. I just realized that it was definite. Thanks!

  11. sirm3d
    • 2 years ago
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    it is an improper integral, but after integration, the function is continuous in that interval, so you can treat it like an ordinary definite integral.

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