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Math Help - Slight mistake with Integral, I can't see why though.

  1. #1
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    Slight mistake with Integral, I can't see why though.

    I am having a small problem with this question,
    \int\frac{1}{2}(e^x-e^{-x})dx
    My workings are:
    Breaking it down into two terms gives:
    \int(e^{x})dx=\frac{1}{1}e^{1x}+c=e^{x}+c and \int(-e^{-x})dx=\frac{1}{-1}e^{-1x}+c=-e^{-x}+c
    Combining it all together then gives:
    \int\frac{1}{2}(e^x-e^{-x})dx=\frac{1}{2}(e^{x}-e^{-x})+c
    My mistake is inside the brackets, I am getting -e^{-x} when the solution gives +e^{-x}
    Can anybody see where I am going wrong?
    Many thanks.
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  2. #2
    Super Member Quacky's Avatar
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    Quote Originally Posted by Hydralisk View Post
    I am having a small problem with this question,
    \int(-e^{-x})dx=\frac{1}{-1}e^{-1x}+c=-e^{-x}+c
    Here is the mistake.

    When you integrate -e^{-x}, you get \frac{-1}{-1}e^{-x}+c Can you see why?
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  3. #3
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    I'm really sorry but I can't see why, I don't understand why the numerator is negative.
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  4. #4
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    I think I've got it, it's because the e is negative, ie -e=-1e
    So \frac{1}{-1}(-1e)=\frac{-1}{-1 }e
    Is this correct?
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  5. #5
    Super Member Quacky's Avatar
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    Quote Originally Posted by Hydralisk View Post
    I think I've got it, it's because the e is negative, ie -e=-1e
    So \frac{1}{-1}(-1e)=\frac{-1}{-1 }e
    Is this correct?
    Yes, and obviously this gives +e^{-x}

    You had integrated e^{-x} in the initial post, whereas you were trying to integrate -e^{-x}
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  6. #6
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    That's clear now, thank you!
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