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anonymous

  • 5 years ago

Anyone want to help with the integral of x/(x+2)^1/2 from -1 to 2?

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  1. anonymous
    • 5 years ago
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    I just don't know how to set it up, what should u be?

  2. anonymous
    • 5 years ago
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    say u = 1/(sqrt(x-2) then you have integral of u/u^2-2. use partial fractions to solve.

  3. anonymous
    • 5 years ago
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    sorry,u = 1/sqrt(x+2)

  4. anonymous
    • 5 years ago
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    lemme know how it works out.

  5. anonymous
    • 5 years ago
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    ok thx I'll try it out

  6. anonymous
    • 5 years ago
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    did you get du=-1/2(x+2)^3/2? or did I do something wrong

  7. anonymous
    • 5 years ago
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    thats right, just checked it

  8. anonymous
    • 5 years ago
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    ok great! . i'd actually not done it. I just looked at the problem and did some calculation in my head. sorry.

  9. anonymous
    • 5 years ago
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    np, wait though not done yet

  10. anonymous
    • 5 years ago
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    so next du=-1(x+2)/2(x+2)^1/2 right?

  11. anonymous
    • 5 years ago
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    umm, you can solve it entirely in u and plug in the values of u, since this is a definite integral.

  12. anonymous
    • 5 years ago
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    no need to deal with x at all

  13. anonymous
    • 5 years ago
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    wait how come? I thought we had to make du look like the remaining terms in the integral, sorry I've just learned integrals today

  14. anonymous
    • 5 years ago
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    Okay. let me get back to you with this. I will take out my pen and paper and write it down. Do ask someone else too.

  15. anonymous
    • 5 years ago
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    ok

  16. anonymous
    • 5 years ago
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    Ok, found it

  17. anonymous
    • 5 years ago
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    hey!

  18. anonymous
    • 5 years ago
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    Ok, so first Let \(u = (x+2)^{1/2} \implies du = (1/2)(x+2)^{-1/2}dx\) \[ \implies dx = 2(x+2)^{1/2} du\]

  19. anonymous
    • 5 years ago
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    So then \[\int_{-1}^2\frac{x}{(x+2)^{1/2}}dx = \int_1^2[u^2-2]du\]

  20. anonymous
    • 5 years ago
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    Whoops forgot the 2

  21. anonymous
    • 5 years ago
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    wow, ok let me try that out thanks, do u know of any rules to find these? or to you just have to guess the u sometimes?

  22. anonymous
    • 5 years ago
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    and does the integral change from -1 to 2, to 1 to 2, or is that a typo?

  23. anonymous
    • 5 years ago
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    The integral changes because \(u = (x+2)^{1/2}\) so when x = -1, u = 1 and when x = 2, u = 2

  24. anonymous
    • 5 years ago
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    right right, forgot about that, when you said you forgot the 2, is it ]2u^2-2]

  25. anonymous
    • 5 years ago
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    \(2[u^2 - 2]\)

  26. anonymous
    • 5 years ago
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    ah right, that would make more sense, well thx again

  27. anonymous
    • 5 years ago
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    Do you see how I arrived at \(2[u^2-2]\)?

  28. anonymous
    • 5 years ago
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    Cause I kinda skipped some steps there.

  29. anonymous
    • 5 years ago
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    well no, but I understand that it's the right equation to the integral

  30. anonymous
    • 5 years ago
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    I can show the steps.

  31. anonymous
    • 5 years ago
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    there are steps to it?

  32. anonymous
    • 5 years ago
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    Yes certainly.

  33. anonymous
    • 5 years ago
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    awesome

  34. anonymous
    • 5 years ago
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    Since \( u = (x+2)^{1/2} \implies x = u^2 -2\) \(\implies dx = 2u\ du\) \[\implies \frac{x}{(x+2)^{1/2}}dx = \frac{u^2-2}{u} *2u\ du\] \[=2[u^2-2]\ du\]

  35. anonymous
    • 5 years ago
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    awesome thanks a lot, I don't think we even learned that yet

  36. anonymous
    • 5 years ago
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    do you always implicitly differentiate x? (x'=dx and not 1?)

  37. anonymous
    • 5 years ago
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    I found that out nvm

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