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Math Help - Intergrals

  1. #1
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    Intergrals

    The intergral of (4-2x)^3 dx?
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  2. #2
    MHF Contributor harish21's Avatar
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    substitute u=4-2x
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    Quote Originally Posted by homerb View Post
    The intergral of (4-2x)^3 dx?
    Use substitution:

    Let:   u=4-2x, \text{ then } du=-2\,dx
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  4. #4
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    thats what I did and I got -1/8*(4-2x)^4 + c, but in the book says -2(x-4)^4, am I doing something wrong?
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  5. #5
    MHF Contributor harish21's Avatar
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    Your answer is correct.. its needs simplification..

    Quote Originally Posted by homerb View Post
    thats what I did and I got -1/8*(4-2x)^4 + c, but in the book says -2(x-4)^4, am I doing something wrong?
    thats the same thing.. take -2 as a common factor in the numerator and simplify..

    -\dfrac{1}{8} (4-2x)^4 + C = -\dfrac{1}{8} [-2(x-2)]^4 + C = ...
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