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akl3644

  • 3 years ago

Calculate an integral

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  1. akl3644
    • 3 years ago
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  2. vf321
    • 3 years ago
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    Have you tried u-sub?

  3. vf321
    • 3 years ago
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    try u = 9-x^2

  4. jamesHayek
    • 3 years ago
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    No, sub u=x^2

  5. vf321
    • 3 years ago
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    I'm pretty sure mine still works... Both are fine...

  6. akl3644
    • 3 years ago
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    my substitution become like this...

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  7. vf321
    • 3 years ago
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    Well you can trig sub if you want but that makes it needlessly complicated IMO.

  8. akl3644
    • 3 years ago
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    emm what u suggest ?

  9. vf321
    • 3 years ago
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    I told u. u = 9-x^2.

  10. vf321
    • 3 years ago
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    Do you know how to u-sub?

  11. jamesHayek
    • 3 years ago
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    \[u=x^2; \frac{ 1 }{ 2 }du = xdx\]

  12. vf321
    • 3 years ago
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    Again, either mine or james' work, since he keeps insisting....

  13. jamesHayek
    • 3 years ago
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    you will get: \[\frac{ 1 }{ 2 }\int\limits_{0}^{1.6}(9-u)^{3/2}u du\]

  14. jamesHayek
    • 3 years ago
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    for the integrand \[(9-u)^{3/2}u\] sub s=9-u and ds=-du

  15. jamesHayek
    • 3 years ago
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    at this point you can expand the integrand and solve.

  16. vf321
    • 3 years ago
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    @jamesHayek I msged you but I'll tell you again: We don't want to give them most of the problem.

  17. jamesHayek
    • 3 years ago
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    vf321, this is a site to help. Not bicker. Show your work to help her.

  18. vf321
    • 3 years ago
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    @jamesHayek help, not do homework for. But fine, this is not getting us anywhere. Let's stop arguing.

  19. akl3644
    • 3 years ago
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    i am following ur step now. thx. trying to get the right answer

  20. akl3644
    • 3 years ago
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    i will try both method

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