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Math Help - Help with a U substitution problem.

  1. #1
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    Exclamation Help with a U substitution problem.

    I can't figure out how to do this but Ive been assured its easily doable.

    Calculate: (int) 2x(4x+7)8 dx

    So I'm assuming we try to get 4x on its own and find 2x in terms of u?

    A solution to this would help me heaps here I just cant see it!
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  2. #2
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    Re: Help with a U substitution problem.

    \int 2x(4x+7)^8 \, dx

    u = 4x+7

    x = \frac{u-7}{4}

    dx = \frac{du}{4}


    \int 2x(4x+7)^8 \, dx

    \int \frac{u-7}{8} \cdot u^8 \, du

    \frac{1}{8} \int u^9 -7u^8 \, du

    can you finish?
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  3. #3
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    Help with a U substitution problem.

    Thanks! Im not sure if I can Im really having trouble with integration. Would appreciate the rest.
    Last edited by camjerlams; January 5th 2013 at 03:39 PM.
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  4. #4
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    Re: Help with a U substitution problem.

    If you're at the level of doing integrals by substitution, then this should be basic stuff.

    what is the antiderivative of u^9 ?

    what is the antiderivative of u^8 ?
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  5. #5
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    Re: Help with a U substitution problem.

    I havent studied math for about 10 years so i struggle with the basic stuff.

    so antiderivative of u^9=1/10u^10
    and for u^8=1/9u^9

    so 1/8 (1/10 u^10 - 7/9 u^9) ??
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  6. #6
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    Re: Help with a U substitution problem.

    Quote Originally Posted by camjerlams View Post
    I havent studied math for about 10 years so i struggle with the basic stuff.

    so antiderivative of u^9=1/10u^10
    and for u^8=1/9u^9

    so 1/8 (1/10 u^10 - 7/9 u^9) + C
    was that so difficult?

    now finish it by back-substituting (4x+7) for u ...
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  7. #7
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    Re: Help with a U substitution problem.

    Thanks skeeter, big help, I know I need practice
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  8. #8
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    Re: Help with a U substitution problem.

    note you don't have to resort to u-substitution, you COULD just multiply out (4x+7)8, and then multiply every term by 2x, and then integrate term-by term. this is kind of "the hard way", u-substitutions are meant to make your life EASIER (short-cuts are good, right?).

    linear substitutions u = ax+b are always "do-able", we just have a "fudge factor" of a constant to worry about:

    du = a dx, so

    dx = (1/a) du <--take the 1/a outside the integral sign.
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  9. #9
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    Re: Help with a U substitution problem.

    so, 1/8((1/10u^10) - (7/9u^9))

    =1/8((4x+7)^10/10 - 7/9(4x+7)^9)
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  10. #10
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    Re: Help with a U substitution problem.

    don't forget to add the constant of integration....
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