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Math Help - Definite integral

  1. #1
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    Definite integral

    Evaluate each Definite integral.
    ∫ lower limit 1, upper limit 8. 3-y^(1/3) / y^(2/3) ; dy
    I did
    ∫(3*y^(-2/3) - y^(-1/3)) dy from 1 to 8

    9*y^(1/3) - 3/2*y^(2/3) eval. from 1 to 8

    = 9(2) - 6 - 9 + 3/2

    = 3 + 3/2 = 9/2
    Is that right? Thanks
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  2. #2
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    I cant really make out properly what the integral is. If you have to integrate 3-(y^1/3)/(y^2/3), the answer will be different. Can you type the problem out properly again. With the proper parentheses?

    The answer is correct with your evaluation though.
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  3. #3
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    Quote Originally Posted by rowdy3 View Post
    Evaluate each Definite integral.
    ∫ lower limit 1, upper limit 8. 3-y^(1/3) / y^(2/3) ; dy
    I did
    ∫(3*y^(-2/3) - y^(-1/3)) dy from 1 to 8

    9*y^(1/3) - 3/2*y^(2/3) eval. from 1 to 8

    = 9(2) - 6 - 9 + 3/2

    = 3 + 3/2 = 9/2
    Is that right? Thanks
    You can check your answer here: integrate (3 - y^(1/3))/y^(2/ 3) from y = 1 to y = 8 - Wolfram|Alpha
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  4. #4
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    Quote Originally Posted by Maanasi View Post
    I cant really make out properly what the integral is. If you have to integrate 3-(y^1/3)/(y^2/3), the answer will be different. Can you type the problem out properly again. With the proper parentheses?

    The answer is correct with your evaluation though.
    ∫ lower limit 1, upper limit 8. (3-y^(1/3)) / y^(2/3) ; dy
    I'll try to scan it.
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  5. #5
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    Quote Originally Posted by rowdy3 View Post
    ∫ lower limit 1, upper limit 8. (3-y^(1/3)) / y^(2/3) ; dy
    I'll try to scan it.
    In the link in my reply:

    1. Your integral is clear. (In fact, with a small amount of effort it is clear from your OP as well).

    2. Your question is answered.
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