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Thread: Substitution

  1. #1
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    Substitution

    Can someone please explain to me how the final answer has -16?
    Here is my work

    Integral 3r square root(8-r)

    u= 8-r

    dr= -du

    3 integral (u-8) square root u du

    3 integral u^3/2 - 8^1/2 du

    3[2/5u^5/2 -6u^3/2] du + C

    6/5(8-r)^5/2-?(8-r)^3/2 + C

    The question mark represents the 16. Not sure how to get it and I prob messed up earlier in the problem.
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  2. #2
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    Quote Originally Posted by McDiesel View Post
    Can someone please explain to me how the final answer has -16?
    Here is my work

    Integral 3r square root(8-r)

    u= 8-r

    dr= -du

    3 integral (u-8) square root u du

    3 integral u^3/2 - 8u^1/2 du

    3[2/5u^5/2 -8(2/3)u^3/2] du + C

    6/5(8-r)^5/2-16(8-r)^3/2 + C
    Does that clear it up?
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  3. #3
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    Yes, thank you
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