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Math Help - Integration help required~~

  1. #16
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    Re: Integration help required~~

    So when I integrated them, I got
    ln (u) -6ln(1+6u)-6ln(1/1+6u)

    is this right? if it's right do i just subsitute the u back to the original value?
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  2. #17
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    Re: Integration help required~~

    Let's back up a step, are you sure your partial fraction decomposition is correct?

    Once you are sure you have the correct decomposition, when you integrate, make the limits of integration reflect the substitution and you won't need to back-substitute for the original variable.
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  3. #18
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    Re: Integration help required~~

    I checked my A,B and C and C should actually be -6 instead of 6. is that right? And if it's right do i substitute my u with my upper and lower limit?
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  4. #19
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    Re: Integration help required~~

    Yes, C = -6. Then if x_1 and x_2 are the original lower and upper limits respectively, change the limits to u(x_1) and u(x_2).
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  5. #20
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    Re: Integration help required~~

    My upper limit is 12 and lower limit is so i substituted them as (ln12-6ln(1+6(12))-6ln(1/1+6(12))-(ln7-6ln(1+6ln(7)-6ln(1/1+6(7)) and the answer I got is 0.539 (in 3 sig fig). What did i do wrong? I used the graphics calculator to check the answer of the original function and it is 97.1 ... What did i do wrong? did I jump a step?
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  6. #21
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    Re: Integration help required~~

    Quote Originally Posted by ghfjdksla94 View Post
    So when I integrated them, I got
    ln (u) -6ln(1+6u)-6ln(1/1+6u)

    is this right? if it's right do i just subsitute the u back to the original value?
    First of all, you need absolute value signs around the stuff inside the logarithms.

    The final term is also incorrect. \displaystyle \begin{align*} \frac{1}{(1 + 6u)^2} = (1 + 6u)^{-2} \end{align*}, now integrate...
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  7. #22
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    Re: Integration help required~~

    Then, is the correct term -(1/6)(1+6u)^-1 ? and by absolute values for logarithms, you mean l l <-- these lines around them?
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  8. #23
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    Re: Integration help required~~

    Quote Originally Posted by ghfjdksla94 View Post
    Then, is the correct term -(1/6)(1+6u)^-1 ? and by absolute values for logarithms, you mean l l <-- these lines around them?
    Yes, that is correct
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  9. #24
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    Re: Integration help required~~

    so my final equation just before substituting u into my upper and lower limit is:
    (ln lul -6ln l1+6ul +1/6 (1+6u)^-1 ) ? Then multiply the answer by 10.5984 which is the constant we brought out to the in front of the integral?
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  10. #25
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    Re: Integration help required~~

    Quote Originally Posted by ghfjdksla94 View Post
    so my final equation just before substituting u into my upper and lower limit is:
    (ln lul -6ln l1+6ul +1/6 (1+6u)^-1 ) ? Then multiply the answer by 10.5984 which is the constant we brought out to the in front of the integral?
    It should actually be \displaystyle \begin{align*} \ln{|u|} - \ln{|1 + 6u|} + \frac{1}{1 + 6u} \end{align*}.
    Last edited by Prove It; September 16th 2012 at 04:17 AM.
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  11. #26
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    Re: Integration help required~~

    Quote Originally Posted by Prove It View Post
    It should actually be \displaystyle \begin{align*} \ln{|u|} - \ln{|1 + 6u|} + \frac{1}{1 + 6u} \end{align*}.
    So, I substitute the u into my upper limit 12 and subtracted it by the same equation this time u substituted into my lower limit 7?

    I got an answer of 1.8 X 10^-4. I putted all the logarithm numbers as an absolute... what is wrong now??
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