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anonymous
 4 years ago
Make a substitution to express the integrand as a rational function and then evaluate the integral.
Integrate[sqrt x/(x36), {x, 4, 9}]
anonymous
 4 years ago
Make a substitution to express the integrand as a rational function and then evaluate the integral. Integrate[sqrt x/(x36), {x, 4, 9}]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{4}^{9} \sqrt x/(x36) dx\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328291169857:dw

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.2Substitute \(u=\sqrt{x} \implies x=u^2 \implies dx=2udu \) to get \[\int_4^9 \frac{u}{u^236}2udu=2\int_4^9 \frac{u^2}{u^236}du=\cdots\] Can you take it from here?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328291206918:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I did mine like math's and each time I keep coming up with 26 log(3)+log(64)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And thats not an option,

cristiann
 4 years ago
Best ResponseYou've already chosen the best response.0sqrtx=t=>x=t^2, dx=2tdt x=4=>t=2 x=9=>t=3 Integrate[2t^2/(t^236), {t, 2, 3}] 6ln26ln3+2

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.2Oh sorry, I made a mistake. Change the limits.

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.2At x=9 u=3, and at x=4 u=2. So your limits should be from 2 to 3.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0See, either way I do it, if I completely solve my solution it comes out negative, and the options I get to choose from are positive...

cristiann
 4 years ago
Best ResponseYou've already chosen the best response.0Also, please note that between 4 and 9 the expression x36 is not null (otherwise your integral would be at a different level)

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.2\[2\int \frac{u^236+36}{u^236}du=2\int(1+\frac{3}{u6}\frac{3}{u+6})du=2u+6\ln{\frac{6u}{6+u}}.\]

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.2Now plug the limits 2 to 3, and you should find your answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There was supposed to be an integral sign on that last park, correct?
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