anonymous
  • anonymous
Make a substitution to express the integrand as a rational function and then evaluate the integral. Integrate[sqrt x/(x-36), {x, 4, 9}]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{4}^{9} \sqrt x/(x-36) dx\]
myininaya
  • myininaya
|dw:1328291169857:dw|
Mr.Math
  • Mr.Math
Substitute \(u=\sqrt{x} \implies x=u^2 \implies dx=2udu \) to get \[\int_4^9 \frac{u}{u^2-36}2udu=2\int_4^9 \frac{u^2}{u^2-36}du=\cdots\] Can you take it from here?

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myininaya
  • myininaya
|dw:1328291206918:dw|
myininaya
  • myininaya
go math! :)
anonymous
  • anonymous
I did mine like math's and each time I keep coming up with 2-6 log(3)+log(64)
anonymous
  • anonymous
And thats not an option,
cristiann
  • cristiann
sqrtx=t=>x=t^2, dx=2tdt x=4=>t=2 x=9=>t=3 Integrate[2t^2/(t^2-36), {t, 2, 3}] 6ln2-6ln3+2
Mr.Math
  • Mr.Math
Oh sorry, I made a mistake. Change the limits.
Mr.Math
  • Mr.Math
At x=9 u=3, and at x=4 u=2. So your limits should be from 2 to 3.
anonymous
  • anonymous
See, 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...
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cristiann
  • cristiann
Also, please note that between 4 and 9 the expression x-36 is not null (otherwise your integral would be at a different level)
Mr.Math
  • Mr.Math
\[2\int \frac{u^2-36+36}{u^2-36}du=2\int(1+\frac{3}{u-6}-\frac{3}{u+6})du=2u+6\ln{\frac{6-u}{6+u}}.\]
Mr.Math
  • Mr.Math
Now plug the limits 2 to 3, and you should find your answer.
anonymous
  • anonymous
There was supposed to be an integral sign on that last park, correct?
Mr.Math
  • Mr.Math
No, not really.

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