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Diyadiya

  • 4 years ago

Integration

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  1. Diyadiya
    • 4 years ago
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    \[\int\limits\limits_{-a}^{a} \sqrt{\frac{a-x}{a+x}}\]

  2. FoolForMath
    • 4 years ago
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    Lets say, \( f(x)= \sqrt{\frac{a-x}{a+x}} \) , now f(X) is even or odd?

  3. Diyadiya
    • 4 years ago
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    even or odd?

  4. Diyadiya
    • 4 years ago
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    i dont know :/@Fool

  5. DHASHNI
    • 4 years ago
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    multiply and divide by the numerator

  6. FoolForMath
    • 4 years ago
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    There is a property of definite integral which I think could be used here.

  7. DHASHNI
    • 4 years ago
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    then the equation becomes solvable

  8. Diyadiya
    • 4 years ago
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    hm letme see

  9. anonymous
    • 4 years ago
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    hmm i think i know the answer, but not sure why it should be easy

  10. Mr.crazzy
    • 4 years ago
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    let x=acosθ

  11. Diyadiya
    • 4 years ago
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    can you post it ?

  12. DHASHNI
    • 4 years ago
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    |dw:1326825738809:dw|

  13. DHASHNI
    • 4 years ago
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    |dw:1326825797080:dw|

  14. FoolForMath
    • 4 years ago
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    I was talking about this: http://en.wikibooks.org/wiki/Calculus/Definite_integral#Even_and_odd_functions

  15. DHASHNI
    • 4 years ago
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    now you substitute x=acos O

  16. anonymous
    • 4 years ago
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    answer is \[a\pi\] but not sure why yet

  17. rld613
    • 4 years ago
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    lol @sat

  18. anonymous
    • 4 years ago
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    well i don't think this beast is even or odd

  19. anonymous
    • 4 years ago
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    would be odd without the radical

  20. anonymous
    • 4 years ago
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    ahh, i think dhashni has it!

  21. FoolForMath
    • 4 years ago
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    No sat after you break it as dash showed, the first is even and the second is odd.

  22. anonymous
    • 4 years ago
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    ahh, and second integral is 0 for sure

  23. FoolForMath
    • 4 years ago
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    so the first term reduces to zero.

  24. anonymous
    • 4 years ago
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    hmm

  25. anonymous
    • 4 years ago
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    i got zero for the second one, maybe i am wrong

  26. FoolForMath
    • 4 years ago
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    how second? the first is even and the second is odd.

  27. anonymous
    • 4 years ago
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    no i am right, second on is zero

  28. anonymous
    • 4 years ago
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    if even, then \[\int_{-a}^a f(x)dx=2\int_0^af(x)dx\]

  29. FoolForMath
    • 4 years ago
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    Oh yes ... my apologies, yes the the second is zero.

  30. FoolForMath
    • 4 years ago
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    For the first one we can use the step.

  31. anonymous
    • 4 years ago
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    if odd then zero, and in any case we can actually compute the second integral right? \[\int\frac{x}{\sqrt{a^2-x^2}}dx=-\sqrt{a^2-x^2}\]

  32. anonymous
    • 4 years ago
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    which is 0 and a and -a, so that one is gone.

  33. FoolForMath
    • 4 years ago
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    So as the second one is odd and hence zero, we have to evaluate the first one.

  34. anonymous
    • 4 years ago
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    and it is somewhat of a pain because it is an improper integral, but you will get \[a\tan^{-1}(\frac{x}{\sqrt{a^2-x^2}})\] and then taking the limit as x goes to 0 you get the limit as x goes to zero of arctan(x) =pi/2 from one direction, -pi/2 from other direction subtract and get pi and that explains the answer of \[a\pi\] whrew

  35. FoolForMath
    • 4 years ago
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    sat, that's great :D

  36. anonymous
    • 4 years ago
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    sorry, i meant the limits as x goes to a, not as x goes to zero

  37. anonymous
    • 4 years ago
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    x goes to a, \[\frac{x}{\sqrt{a^2-x^2}}\] goes to infinity and \[\arctan(\infty)\] goes to \[\frac{\pi}{2}\] is what i meant

  38. Diyadiya
    • 4 years ago
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    Lol OkaYy :D Thaanks a Lot :D

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