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Thread: Integration problem

  1. #1
    ixi
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    Integration problem

    It's the integration of k(1+x)^-4 from 0 to infinity which is equal to 1. Solve for k.

    My solution is ---
    Let u=1+x du=dx
    Then, it's k(-3(1+x)^-3) from 0 to infinity, which means k=1/3

    But the solution manual says it's
    (-k/3)(1/(1+3)^3) from 0 to infinity giving k=3


    I'm not sure why it's -k/3 and not -3k

    I thought when integration u^-4 it becomes -3u^-3
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  2. #2
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    Quote Originally Posted by ixi View Post
    It's the integration of k(1+x)^-4 from 0 to infinity which is equal to 1. Solve for k.

    My solution is ---
    Let u=1+x du=dx
    Then, it's k(-3(1+x)^-3) from 0 to infinity, which means k=1/3

    But the solution manual says it's
    (-k/3)(1/(1+3)^3) from 0 to infinity giving k=3


    I'm not sure why it's -k/3 and not -3k

    I thought when integration u^-4 it becomes -3u^-3
    take the derivative of $\displaystyle -3u^{-3}$ ... what do you get?


    $\displaystyle
    \lim_{b \to \infty} \int_0^b \frac{k}{(x+1)^4} \, dx = 1
    $

    $\displaystyle
    \lim_{b \to \infty} \left[-\frac{1}{3}\frac{k}{(x+1)^3} \right]_0^b = 1
    $

    $\displaystyle
    -\frac{k}{3} \lim_{b \to \infty} \left[\frac{1}{(b+1)^3} - \frac{1}{(0+1)^3} \right] = 1
    $

    $\displaystyle
    \frac{k}{3} \cdot \frac{1}{1} = 1
    $

    $\displaystyle
    k = 3
    $
    Last edited by skeeter; Jun 1st 2010 at 12:22 PM. Reason: fixed error
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  3. #3
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by ixi View Post
    It's the integration of k(1+x)^-4 from 0 to infinity which is equal to 1. Solve for k.

    My solution is ---
    Let u=1+x du=dx
    Then, it's k(-3(1+x)^-3) from 0 to infinity, which means k=1/3

    But the solution manual says it's
    (-k/3)(1/(1+3)^3) from 0 to infinity giving k=3


    I'm not sure why it's -k/3 and not -3k

    I thought when integration u^-4 it becomes -3u^-3

    $\displaystyle k \int_0^{ \infty } (1+x)^{-4} dx = 1 $

    $\displaystyle - \frac{k}{3} (1+x)^{-3} |_0^{ \infty } = 1 $

    $\displaystyle - \frac{k}{3} \lim_{ r \to \infty } (1+x)^{-3} |_0^{ R } = 1$

    $\displaystyle - \frac{k}{3} ( 0 - 1 ) = 1 $

    $\displaystyle k = 3 $
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by skeeter View Post
    take the derivative of $\displaystyle -3u^{-3}$ ... what do you get?





    $\displaystyle

    \lim_{b \to \infty} \int_0^b \frac{k}{(x+1)^4} \, dx = 1

    $



    $\displaystyle

    \lim_{b \to \infty} \left[-\frac{1}{3}\frac{k}{(x+1)^3} \right]_0^b = 1

    $





    $\displaystyle -\frac{k}{3} \lim_{b \to \infty} \left[\frac{1}{(b+1)^3} - \frac{1}{(1+1)^3} \right] = 1 $





    $\displaystyle

    \frac{k}{3} \cdot \frac{1}{8} = 1

    $



    $\displaystyle

    k = 24

    $
    The integral is from 0 to b, not 1 to b!
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  5. #5
    MHF Contributor
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    Quote Originally Posted by AllanCuz View Post
    The integral is from 0 to b, not 1 to b!
    thank you
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