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Thread: Integration Problem

  1. #1
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    Integration Problem

    Integrate:
    $\displaystyle
    6(e^{4-3x})^2
    $

    Is it correct if i do this: $\displaystyle 6(e^{8-6x})$ and then solve?

    and then let u = 8-6x and du=-6

    then the answer comes to $\displaystyle -e^{8-6x} + C$
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  2. #2
    Moo
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    Hello,

    It's all correct
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  3. #3
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    thank you, here is another one (I have a test on Friday so I'm trying to do as many problems as I can)

    Integrate: $\displaystyle \frac{3}{4}-\frac{1}{\sqrt{6x}}$

    I let u = 6x and du = 6dx

    the answer I end up with is:
    $\displaystyle

    -\frac{1}{4}\sqrt{6x}+C
    $
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  4. #4
    Moo
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    Hello,

    And what about integrating the constant ? The integral of 1 is x, the integral of 3/4 will be...

    But according to your answer, would the integral rather be $\displaystyle \int \frac 34\times \frac{1}{\sqrt{6x}} ~dx$ ? In which case, it's correct - again
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  5. #5
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    oops no you were right I did it wrong but thanks for pointing that out
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  6. #6
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    the answer should be:

    $\displaystyle
    \frac{3}{4}x-\frac{1}{2}\sqrt{6x}+C
    $

    correct?
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  7. #7
    Moo
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    No, it looks like you forgot du=6 dx
    Otherwise, it's correct.
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