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missylulu Group Title

please please please help!!! double integral of 2/(x+y)^2 dydx, 2< x < 3, 1<y <5

  • one year ago
  • one year ago

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  1. KingGeorge Group Title
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    First step, you just need to find the following integral where you just pretend \(x\) is some constant. \[\int\limits_1^5 \frac{2}{(x+y)^2} dy\]Can you do this yourself?

    • one year ago
  2. missylulu Group Title
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    is it 2ln(x+5) -2ln(x+1) ???

    • one year ago
  3. KingGeorge Group Title
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    I think you kind of have the idea. That would be the integral if the denominator did not have a power of 2. So you would get \[\int\limits_1^5 \frac{2}{(x+y)^2} dy=\left[-\frac{2}{x+y}\right]^5_1=-\frac{2}{x+5}-\left(-\frac{2}{x+1}\right)\]

    • one year ago
  4. KingGeorge Group Title
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    Does that make sense? Do you see how I did the integral?

    • one year ago
  5. missylulu Group Title
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    yes!!

    • one year ago
  6. KingGeorge Group Title
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    If you simplify that, you should get\[-\frac{2}{x+5}-\left(-\frac{2}{x+1}\right)=\frac{8}{5+6 x+x^2}=\frac{8}{(x+5)(x+1)}\]With this, I imagine you could use partial fractions, and then you'll end up with two integrals of x which should be fairly straightforward. I'll let you work out the rest on your own. If you get stuck, let me know and I can help some more.

    • one year ago
  7. missylulu Group Title
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    okok is itpossible to do it by usub?

    • one year ago
  8. missylulu Group Title
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    will you ever use integration by parts for double integrals?

    • one year ago
  9. KingGeorge Group Title
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    If you did \(u=x+3\), then you would get \((u+2)(u-2)=u^2-4\) in the denominator. I suppose you could then do something with the trig substitutions. I really think that it might be better if you do the partial fractions.

    • one year ago
  10. KingGeorge Group Title
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    And you can do integration by parts in double integrals, but in my experience, you don't usually see it simply because you don't need it.

    • one year ago
  11. zepdrix Group Title
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    Hmm I'm confused, how did you end up needing to use partial fractions King? Why not integrate in X from where you left off after evaluating the upper and lower limits of y?\[\large \frac{2}{x+1}-\frac{2}{x+5}\]

    • one year ago
  12. sirm3d Group Title
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    i agree with @zepdrix . the resulting expression is ripe for integration.

    • one year ago
  13. KingGeorge Group Title
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    Oops. @zepdrix is correct. We don't even need to do partial fractions. \[\int\limits_2^3\frac{2}{x+1}dx-\int\limits_2^3\frac{2}{x+5}dx\]Integrating this should work.

    • one year ago
  14. KingGeorge Group Title
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    And this step is where the natural logs start to come in.

    • one year ago
  15. missylulu Group Title
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    thanks guys. is the answer... 2(-ln8 + ln4 +ln7 +ln3) @KingGeorge :(

    • one year ago
  16. KingGeorge Group Title
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    Hmm... I'm getting something a little different. If you had a \(-\ln(3)\) at the end, that's what I'm getting.

    • one year ago
  17. missylulu Group Title
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    2(ln4 - ln3 - ln8 + ln7)!?

    • one year ago
  18. sirm3d Group Title
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    \[\Huge \color{blue} \checkmark \]

    • one year ago
  19. sirm3d Group Title
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    if you simplify further, \(2(\ln 7 - \ln 6)\)

    • one year ago
  20. missylulu Group Title
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    Ok!! THANK YOU!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

    • one year ago
  21. KingGeorge Group Title
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    Looks great.

    • one year ago
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