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Math Help - evaluate the integral using FTC

  1. #1
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    evaluate the integral using FTC

    ∫10 (e^(-9x)) / (e^(-9x) + 6)) dx
    8

    Evaluate the integral using the fundamental theorem of calculus.

    I don't know how to do rational numbers, i know that it is complex
    because of the quotient rule for derivatives

    anyway, here's mywork so far

    ∫from 10 to 8 (e^(-9x))/(e^(-9x)) + ∫from 10 to 8 (e^(-9x))/6

    ∫from 10-8 (1) + ∫from 10 to 8 (e^(-9x))/6

    2 + 1/6 ∫from 10 to 8 (e^(-9x))

    2 + 1/6 [((e^(-90+1))/(-90+1)) - ((e^(-72+1))/(-72+1))]

    2 + 1/6 [(e^(-89)/(-89)) - (e^(-71)/(-71))]

    this is wrong,
    please help
    thanks,
    brittany
    Last edited by williamb; April 13th 2009 at 11:40 AM.
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  2. #2
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    Just let u=e^{-9x}\Rightarrow du=-9e^{-9x}\,dx.

    As for your work...

    Quote Originally Posted by williamb View Post
    ∫from 10 to 8 (e^(-9x))/(e^(-9x)) + ∫from 10 to 8 (e^(-9x))/6
    You cannot split fractions in this way. \frac a{b+c} is not, in general, equal to \frac ab+\frac ac.

    Also, you are integrating from 8 to 10, not 10 to 8 (unless you wrote it wrong at the top of your post).

    ...
    2 + 1/6 [((e^(-90+1))/(-90+1)) - ((e^(-72+1))/(-72+1))]
    Okay, even if we assume that your previous work was correct, this line does not make sense. You can only apply the power rule when the variable is not in the exponent. For an exponential function like e^{-9x}, you need to use the following rules.

    \frac d{dx}\left[e^x\right]=e^x

    \int e^x\,dx=e^x+C

    If the base is something other than e, you can use

    \frac d{dx}\left[a^x\right]=(\ln a)a^x

    \int a^x\,dx=\frac{a^x}{\ln a}+C
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