Diyadiya
  • Diyadiya
Integration
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Diyadiya
  • Diyadiya
\[\int\limits\limits_{-a}^{a} \sqrt{\frac{a-x}{a+x}}\]
anonymous
  • anonymous
Lets say, \( f(x)= \sqrt{\frac{a-x}{a+x}} \) , now f(X) is even or odd?
Diyadiya
  • Diyadiya
even or odd?

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Diyadiya
  • Diyadiya
i dont know :/@Fool
anonymous
  • anonymous
multiply and divide by the numerator
anonymous
  • anonymous
There is a property of definite integral which I think could be used here.
anonymous
  • anonymous
then the equation becomes solvable
Diyadiya
  • Diyadiya
hm letme see
anonymous
  • anonymous
hmm i think i know the answer, but not sure why it should be easy
anonymous
  • anonymous
let x=acosĪø
Diyadiya
  • Diyadiya
can you post it ?
anonymous
  • anonymous
|dw:1326825738809:dw|
anonymous
  • anonymous
|dw:1326825797080:dw|
anonymous
  • anonymous
I was talking about this: http://en.wikibooks.org/wiki/Calculus/Definite_integral#Even_and_odd_functions
anonymous
  • anonymous
now you substitute x=acos O
anonymous
  • anonymous
answer is \[a\pi\] but not sure why yet
anonymous
  • anonymous
lol @sat
anonymous
  • anonymous
well i don't think this beast is even or odd
anonymous
  • anonymous
would be odd without the radical
anonymous
  • anonymous
ahh, i think dhashni has it!
anonymous
  • anonymous
No sat after you break it as dash showed, the first is even and the second is odd.
anonymous
  • anonymous
ahh, and second integral is 0 for sure
anonymous
  • anonymous
so the first term reduces to zero.
anonymous
  • anonymous
hmm
anonymous
  • anonymous
i got zero for the second one, maybe i am wrong
anonymous
  • anonymous
how second? the first is even and the second is odd.
anonymous
  • anonymous
no i am right, second on is zero
anonymous
  • anonymous
if even, then \[\int_{-a}^a f(x)dx=2\int_0^af(x)dx\]
anonymous
  • anonymous
Oh yes ... my apologies, yes the the second is zero.
anonymous
  • anonymous
For the first one we can use the step.
anonymous
  • anonymous
if odd then zero, and in any case we can actually compute the second integral right? \[\int\frac{x}{\sqrt{a^2-x^2}}dx=-\sqrt{a^2-x^2}\]
anonymous
  • anonymous
which is 0 and a and -a, so that one is gone.
anonymous
  • anonymous
So as the second one is odd and hence zero, we have to evaluate the first one.
anonymous
  • anonymous
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
anonymous
  • anonymous
sat, that's great :D
anonymous
  • anonymous
sorry, i meant the limits as x goes to a, not as x goes to zero
anonymous
  • anonymous
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
Diyadiya
  • Diyadiya
Lol OkaYy :D Thaanks a Lot :D

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