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Math Help - Convergent Improper integrals

  1. #1
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    Convergent Improper integrals




    So I set up everything to find the limit as b approaches infinite. After integration, I have;

    \frac{-e^{-4x}}{16}(4x+1)

    The problem I have is when applying the FTC with b. I know it ends up being zero, but I don't understand how. There are 2 x values there to replace with infinity, how can I evaluate the outcome?

    Also, I think I'm doing something wrong with replacing the x values with 7 because the result is a complicated number; -1.2532*10^{-12}
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  2. #2
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    Quote Originally Posted by Archduke01 View Post





    So I set up everything to find the limit as b approaches infinite. After integration, I have;

    \frac{-e^{-4x}}{16}(4x+1)


    Good. So now you must evaluate \lim_{b\to\infty}\left[\frac{-e^{-4x}}{16}(4x+1)\right]^b_7=\lim_{b\to\infty}\frac{-e^{-4b}}{16}(4b+1)+\frac{-e^{-28}}{16}(29)=\frac{29}{16}\,e^{-28}

    Tonio


    The problem I have is when applying the FTC with b. I know it ends up being zero, but I don't understand how. There are 2 x values there to replace with infinity, how can I evaluate the outcome?


    Also, I think I'm doing something wrong with replacing the x values with 7 because the result is a complicated number; -1.2532*10^{-12}
    .
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