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

FoolForMathBest ResponseYou've already chosen the best response.1
Lets say, \( f(x)= \sqrt{\frac{ax}{a+x}} \) , now f(X) is even or odd?
 2 years ago

DHASHNIBest ResponseYou've already chosen the best response.2
multiply and divide by the numerator
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
There is a property of definite integral which I think could be used here.
 2 years ago

DHASHNIBest ResponseYou've already chosen the best response.2
then the equation becomes solvable
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
hmm i think i know the answer, but not sure why it should be easy
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
I was talking about this: http://en.wikibooks.org/wiki/Calculus/Definite_integral#Even_and_odd_functions
 2 years ago

DHASHNIBest ResponseYou've already chosen the best response.2
now you substitute x=acos O
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
answer is \[a\pi\] but not sure why yet
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
well i don't think this beast is even or odd
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
would be odd without the radical
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
ahh, i think dhashni has it!
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
No sat after you break it as dash showed, the first is even and the second is odd.
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
ahh, and second integral is 0 for sure
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
so the first term reduces to zero.
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
i got zero for the second one, maybe i am wrong
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
how second? the first is even and the second is odd.
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
no i am right, second on is zero
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
if even, then \[\int_{a}^a f(x)dx=2\int_0^af(x)dx\]
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
Oh yes ... my apologies, yes the the second is zero.
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
For the first one we can use the step.
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
if odd then zero, and in any case we can actually compute the second integral right? \[\int\frac{x}{\sqrt{a^2x^2}}dx=\sqrt{a^2x^2}\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
which is 0 and a and a, so that one is gone.
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
So as the second one is odd and hence zero, we have to evaluate the first one.
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
and 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
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.1
sat, that's great :D
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
sorry, i meant the limits as x goes to a, not as x goes to zero
 2 years ago

satellite73Best ResponseYou've already chosen the best response.3
x goes to a, \[\frac{x}{\sqrt{a^2x^2}}\] goes to infinity and \[\arctan(\infty)\] goes to \[\frac{\pi}{2}\] is what i meant
 2 years ago

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