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tomiko

  • one year ago

need help with this double integral question :) http://gyazo.com/a3e6e7133a02d92949f4a14f0cb3961e

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  1. tomiko
    • one year ago
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    @amistre64

  2. FutureMathProfessor
    • one year ago
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    What are you stuck with?

  3. FutureMathProfessor
    • one year ago
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    \[\int\limits_{\pi}^{2\pi}\int\limits_{0}^{1}xsin(xy)dxdy\]

  4. amistre64
    • one year ago
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    since your limits are constants; you can prolly more easily swap it for dydx

  5. amistre64
    • one year ago
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    which proff did ... ironically :)

  6. FutureMathProfessor
    • one year ago
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    IRONICALLY?!?! LOOOOOOOOOOL

  7. amistre64
    • one year ago
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    \[\int\limits_{\pi}^{2\pi}\int\limits_{0}^{1}xsin(xy)~dydx\] \[\int\limits_{\pi}^{2\pi}\left(\int\limits_{0}^{1}xsin(xy)~dy\right)dx\]

  8. amistre64
    • one year ago
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    doh ... loathesome integrals!!

  9. FutureMathProfessor
    • one year ago
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    @amistre64 doesn't the zero to one have to stick with the dx differential though?

  10. amistre64
    • one year ago
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    \[\int_{x=0}^{x=1}\left(\int_{y=\pi}^{y=2\pi}xsin(xy)~dy\right)dx\]

  11. amistre64
    • one year ago
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    it tends to help me out if i label the integrals ....

  12. FutureMathProfessor
    • one year ago
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    Isn't the general integral of that xsin(xy)/y|

  13. amistre64
    • one year ago
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    \[\int_{x=0}^{x=1}\left(\int_{y=\pi}^{y=2\pi}xsin(xy)~dy\right)dx\] \[\int_{x=0}^{x=1}-cos(2\pi~x)+cos(\pi~x)~dx\]

  14. FutureMathProfessor
    • one year ago
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    I thought you didn't have to do parts integration since your X term out front evidently doesn't have a Y in it?

  15. amistre64
    • one year ago
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    you still have to integrate your limits respectively x = [0,1] has to address your dx y = [pi, 2pi] has to address your dy

  16. amistre64
    • one year ago
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    the dx, dy parts tells you what you focus on while leaving the other term a constant

  17. FutureMathProfessor
    • one year ago
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    LOLOL I forgot about that

  18. FutureMathProfessor
    • one year ago
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    I didn't realize you were evaluating limits in your 2nd step of that response

  19. amistre64
    • one year ago
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    let x = k int k sin(ky) dy -> -cos(ky) -cos(x 2pi) - -cos(x pi)

  20. amistre64
    • one year ago
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    int cos(x pi) - cos(x 2pi) dx sin(x pi)/pi - sin(x 2pi)/2pi (sin(pi)/pi - sin(2pi)/2pi) - (sin(0 pi)/pi - sin(0 2pi)/2pi) (0 - 0) - (0 - 0) so with any luck :)

  21. tomiko
    • one year ago
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    thanks @amistre64

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