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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/(x-36), {x, 4, 9}]

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  1. anonymous
    • 4 years ago
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    \[\int\limits_{4}^{9} \sqrt x/(x-36) dx\]

  2. myininaya
    • 4 years ago
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    |dw:1328291169857:dw|

  3. Mr.Math
    • 4 years ago
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    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?

  4. myininaya
    • 4 years ago
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    |dw:1328291206918:dw|

  5. myininaya
    • 4 years ago
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    go math! :)

  6. anonymous
    • 4 years ago
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    I did mine like math's and each time I keep coming up with 2-6 log(3)+log(64)

  7. anonymous
    • 4 years ago
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    And thats not an option,

  8. cristiann
    • 4 years ago
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    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

  9. Mr.Math
    • 4 years ago
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    Oh sorry, I made a mistake. Change the limits.

  10. Mr.Math
    • 4 years ago
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    At x=9 u=3, and at x=4 u=2. So your limits should be from 2 to 3.

  11. anonymous
    • 4 years ago
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    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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  12. cristiann
    • 4 years ago
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    Also, please note that between 4 and 9 the expression x-36 is not null (otherwise your integral would be at a different level)

  13. Mr.Math
    • 4 years ago
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    \[2\int \frac{u^2-36+36}{u^2-36}du=2\int(1+\frac{3}{u-6}-\frac{3}{u+6})du=2u+6\ln{\frac{6-u}{6+u}}.\]

  14. Mr.Math
    • 4 years ago
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    Now plug the limits 2 to 3, and you should find your answer.

  15. anonymous
    • 4 years ago
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    There was supposed to be an integral sign on that last park, correct?

  16. Mr.Math
    • 4 years ago
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    No, not really.

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