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Math Help - Incorrect Indefinite Integral Answers

  1. #1
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    Incorrect Indefinite Integral Answers

    I'm having difficulty with this question, I've been able to successfully get it as far as this (which is correct according to the solutions page) but I cannot get the actual stage of finding the indefinite integral part correct.
    \int(1-2e^{-x/2}+e^{x/2})dx
    Finding the indefinite integrals for each term at a time gives me:
    \int(1)dx=x+c
    \int(-2e^{-x/2})dx=\int(-2)(e^{(-1)x/2})dx=(-2)\frac{1}{-1}e^{(-1)x/2}+c=2e^{-x/2}+c
    \int(e^{x/2})dx=\int(e^{(1)x/2})dx=\frac{1}{1}e^{(1)x/2}+c=e^{x/2}+c
    Putting this all together gives me:
    \int(1-2e^{-x/2}+e^{x/2})dx=x+2e^{-x/2}+e^{x/2}+c
    But the correct solution is:
    \int(1-2e^{-x/2}+e^{x/2})dx=x+4e^{-x/2}+2e^{x/2}+c
    Can anybody see where I am making my mistakes?
    Many thanks.
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  2. #2
    Super Member Matt Westwood's Avatar
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    Third line, e^{x/2} is e^{x \times 1/2} and not what you got, so you need to times its integral by 2. Similarly for the line above except it's minus it.
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  3. #3
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    Matt that's clear now, thank you very much.
    Shayne.
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