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Diyadiya
 4 years ago
Integration
Diyadiya
 4 years ago
Integration

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Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.2\[\int\limits\limits_{a}^{a} \sqrt{\frac{ax}{a+x}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Lets say, \( f(x)= \sqrt{\frac{ax}{a+x}} \) , now f(X) is even or odd?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0multiply and divide by the numerator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There is a property of definite integral which I think could be used here.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then the equation becomes solvable

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm i think i know the answer, but not sure why it should be easy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1326825738809:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1326825797080:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I was talking about this: http://en.wikibooks.org/wiki/Calculus/Definite_integral#Even_and_odd_functions

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now you substitute x=acos O

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0answer is \[a\pi\] but not sure why yet

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well i don't think this beast is even or odd

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0would be odd without the radical

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahh, i think dhashni has it!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No sat after you break it as dash showed, the first is even and the second is odd.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahh, and second integral is 0 for sure

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the first term reduces to zero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i got zero for the second one, maybe i am wrong

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how second? the first is even and the second is odd.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no i am right, second on is zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if even, then \[\int_{a}^a f(x)dx=2\int_0^af(x)dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh yes ... my apologies, yes the the second is zero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For the first one we can use the step.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if odd then zero, and in any case we can actually compute the second integral right? \[\int\frac{x}{\sqrt{a^2x^2}}dx=\sqrt{a^2x^2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0which is 0 and a and a, so that one is gone.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So as the second one is odd and hence zero, we have to evaluate the first one.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and it is somewhat of a pain because it is an improper integral, but you will get \[a\tan^{1}(\frac{x}{\sqrt{a^2x^2}})\] 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
 4 years ago
Best ResponseYou've already chosen the best response.0sorry, i meant the limits as x goes to a, not as x goes to zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x goes to a, \[\frac{x}{\sqrt{a^2x^2}}\] goes to infinity and \[\arctan(\infty)\] goes to \[\frac{\pi}{2}\] is what i meant

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.2Lol OkaYy :D Thaanks a Lot :D
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