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anonymous

  • one year ago

Another integral problem!

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    So I know that the integral is -cos^3 (x) / 3 + c for the integral of cos^2 (x) * sin(x)

  3. anonymous
    • one year ago
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    So the first part would be zero because 2/3 - 2/3 is 0

  4. anonymous
    • one year ago
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    I mean 1/3 - 1/3

  5. anonymous
    • one year ago
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    But what about the second part? Wouldn't that just be adding 1/3 and 1/3 together to get 2/3?

  6. ganeshie8
    • one year ago
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    For the second part, you need to honor the absolute bars, for what values of `x`, is the expression `cos^2x*sinx` negative ?

  7. anonymous
    • one year ago
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    When x is pi/2 to 3pi/2

  8. ganeshie8
    • one year ago
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    |dw:1436014008840:dw|

  9. anonymous
    • one year ago
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    Oh wait I read it wrong

  10. anonymous
    • one year ago
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    OH okay.

  11. ganeshie8
    • one year ago
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    so simply evaluate the integral between (0, pi) and multiply by 2

  12. anonymous
    • one year ago
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    Ohhh which is 4/3!!

  13. ganeshie8
    • one year ago
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    yes!

  14. anonymous
    • one year ago
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    And then for the last part we just multiply 4/3 by 30~

  15. ganeshie8
    • one year ago
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    Excelent!

  16. anonymous
    • one year ago
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    ^however how does this work? just multiplying it by 30...

  17. anonymous
    • one year ago
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    Because I don't think you would be able to do that in a another integral about has bounds...

  18. ganeshie8
    • one year ago
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    you must be knowing that definite integral can be interpreted as "signed area under the curve" ?

  19. anonymous
    • one year ago
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    Ohhhhh yeah.

  20. ganeshie8
    • one year ago
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    Look at the graph, it clearly repeats every 2pi, yes ?

  21. anonymous
    • one year ago
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    yeah.... it's summer.... thank you!

  22. ganeshie8
    • one year ago
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    np :)

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