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tomiko

double integral question.

  • 10 months ago
  • 10 months ago

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  1. tomiko
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    can someone please help me solve this? http://gyazo.com/d88e00524337c49ec6734dcb2b9ad5b1

    • 10 months ago
  2. amistre64
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    its nice they give you a box to work in

    • 10 months ago
  3. amistre64
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    let y+1 = some constant, say k \[\int_{0}^{2}\frac{1}{(x+k)^3}dx\] \[\int_{0}^{2}(x+k)^{-3}dx\]

    • 10 months ago
  4. amistre64
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    whats your solution to this part?

    • 10 months ago
  5. Euler271
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    |dw:1370890838800:dw|

    • 10 months ago
  6. tomiko
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    @amistre64 solution \[-\frac{ (x + k) ^ 2 }{ 2} + C\]

    • 10 months ago
  7. tomiko
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    i forgot the (-) on the 2 up there

    • 10 months ago
  8. amistre64
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    that would be the indefinite integral yes; lets apply the limits of 0 to 2, recall that \[\int_{a}^{b}f(x)~dx=F(b)-F(a)\]

    • 10 months ago
  9. amistre64
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    \[\left(-\frac{ (2 + k) ^{-2} }{ 2}\right)-\left(-\frac{ (0 + k) ^{-2} }{ 2} \right)\] \[-\frac{ (2 + k) ^{-2} }{ 2}+\frac{ ( k) ^{-2} }{ 2} \] \[\frac12(k^{-2}-(2 + k) ^{-2})~:~k=y+1 \] \[\frac12((y+1)^{-2}-(y+3) ^{-2})\] integrate that from 0 to 1

    • 10 months ago
  10. amistre64
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    \[\frac12\left(\int_{0}^{1}(y+1)^{-2}dy-\int_{0}^{1}(y+3)^{-2}dy\right)\]

    • 10 months ago
  11. tomiko
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    from your 2nd step, am i not to replace k = 2? and k = 0? @amistre64

    • 10 months ago
  12. amistre64
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    no, k is just some constant with respect to x your integrating "x" from 0 to 2 .... not "k" from 0 to 2

    • 10 months ago
  13. tomiko
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    why not with respect to "y"?

    • 10 months ago
  14. amistre64
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    because your order of integration is dxdy dx first, then dy

    • 10 months ago
  15. amistre64
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    \[\int_{0}^{1}\left(\int_{0}^{2}\frac{1}{(x+y+1)^3}~dx\right)dy\] the inside stuff is dx .. anything that is not xed is constant, therefore y+1 = k \[...\left(\int_{0}^{2}\frac{1}{(x+k)^3}~dx\right)...\]

    • 10 months ago
  16. amistre64
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    \[...\left(\int_{0}^{2}\frac{1}{(x+k)^3}~dx\right)...\] \[...\left(-\frac{1}{2(2+k)^2}+\frac{1}{2(0+k)^2}\right)...\] replace k with y+1 \[...\left(-\frac{1}{2(2+y+1)^2}+\frac{1}{2(y+1)^2}\right)...\] \[\frac12\int_{0}^{1}\left(\frac{1}{(y+1)^2}-\frac{1}{(y+3)^2}\right)dy\]

    • 10 months ago
  17. tomiko
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    hmmm....i got 1/24 as final answer....

    • 10 months ago
  18. amistre64
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    you most likely added things incorrectly

    • 10 months ago
  19. amistre64
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    \[\frac12\int_{0}^{1}\left(\frac{1}{(y+1)^2}-\frac{1}{(y+3)^2}\right)dy\] \[\frac12\left(\frac{1}{-(y+1)}-\frac{1}{-(y+3)}\right)\] \[\frac12\left(\frac{1}{-(1+1)}-\frac{1}{-(1+3)}\right)-\frac12\left(\frac{1}{-(0+1)}-\frac{1}{-(0+3)}\right)\] \[\frac12\left(\frac14-\frac12\right)-\frac12\left(\frac13-1\right)\] \[\frac12\left(\frac14-\frac12-\frac13+1\right)\] \[\frac12\left(\frac{3-6-4+12}{12}\right)\] \[\frac12\left(\frac{5}{12}\right)\]

    • 10 months ago
  20. tomiko
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    ouch!...i am redoing this thing!

    • 10 months ago
  21. amistre64
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    :) the algebra/arithmetic is usually the culprit

    • 10 months ago
  22. tomiko
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    exactly. i did it a second time and got 11/24. i'm doing it again!!

    • 10 months ago
  23. tomiko
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    @amistre64 got it finally. my friend also got 5/24 :). thank you very much!!

    • 10 months ago
  24. amistre64
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    :) good job

    • 10 months ago
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