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

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

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

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

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3hmm i think i know the answer, but not sure why it should be easy

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

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3answer is \[a\pi\] but not sure why yet

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3well i don't think this beast is even or odd

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3would be odd without the radical

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3ahh, i think dhashni has it!

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3ahh, and second integral is 0 for sure

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3i got zero for the second one, maybe i am wrong

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3no i am right, second on is zero

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3if even, then \[\int_{a}^a f(x)dx=2\int_0^af(x)dx\]

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

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3if odd then zero, and in any case we can actually compute the second integral right? \[\int\frac{x}{\sqrt{a^2x^2}}dx=\sqrt{a^2x^2}\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3which is 0 and a and a, so that one is gone.

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3and 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

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3sorry, i meant the limits as x goes to a, not as x goes to zero

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.3x 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
 2 years ago
Best ResponseYou've already chosen the best response.2Lol OkaYy :D Thaanks a Lot :D
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