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Math Help - Integration: General Power Formula

  1. #1
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    Integration: General Power Formula

    Could you explain where I went wrong with this one?

    28-1-16

    (int) 0.8((3 + 2 ln(u))^3) du/u

    Attempt:
    u = 3 + 2 ln u
    du = 1 / u

    Therefore answer should be:
    (.8 [3 + 2 ln u]^3) / 4

    the ti-89 comes up with another answer when this integration is performed, which is long and bizarre.
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  2. #2
    Senior Member x3bnm's Avatar
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    Re: Integration: General Power Formula

    \begin{align*}\int \frac{0.8(3 + 2 \ln(u))^3}{u} du =& \int 0.8\left(\frac{27}{u} + \frac{54 \ln(u)}{u} + \frac{36 \ln^{2}(u)}{u} + \frac{8\ln^{3}(u)}{u} \right) \,\,du.....\text{[by expanding cube]}\\=& 0.8\left(\int \frac{27}{u}\,\,du +\int \frac{54 \ln(u)}{u}\,\,du + \int \frac{36 \ln^{2}(u)}{u}\,\,du + \int \frac{8\ln^{3}(u)}{u}\,\,du \right).....(1)  \end{align*}

    Let m = \ln(u)\,\, \therefore dm = \frac{1}{u}\,\,du

    So (1) becomes:

    \begin{align*}0.8\left(\int \frac{27}{u}\,\,du +\int \frac{54 \ln(u)}{u}\,\,du + \int \frac{36 \ln^{2}(u)}{u}\,\,du + \int \frac{8\ln^{3}(u)}{u}\,\,du \right) =& 0.8\left( 27\ln|u| + \int 54m\,\, dm + 12\ln^{3}|u|+ 2\ln^{4}|u| \right) + C\\ =& 0.8\left( 27\ln|u| + 27m^2 +  12\ln^{3}|u|+ 2\ln^{4}|u| \right) + C\\=& 0.8\left( 27\ln|u| + 27\ln^{2}|u| +  12\ln^{3}|u|+ 2\ln^{4}|u| \right) + C....\text{[because }m = \ln(u)\text{]}\\=& 21.6\ln|u| + 21.6\ln^{2}|u| + 9.6 \ln^{3}|u| + 1.6 \ln^{4}|u|  + C........\text{[Answer]}\end{align*}

    You can check the answer here to make sure.
    int (0.8(3 + 2 ln(u))^3)/u - Wolfram|Alpha
    Last edited by x3bnm; May 18th 2013 at 04:40 PM.
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  3. #3
    Senior Member x3bnm's Avatar
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    Re: Integration: General Power Formula

    Quote Originally Posted by togo View Post
    Could you explain where I went wrong with this one?

    28-1-16

    (int) 0.8((3 + 2 ln(u))^3) du/u

    Attempt:
    u = 3 + 2 ln u
    du = 1 / u

    Therefore answer should be:
    (.8 [3 + 2 ln u]^3) / 4

    the ti-89 comes up with another answer when this integration is performed, which is long and bizarre.
    You are almost on the right track.


    \int \frac{0.8(3 + 2 \ln(u))^3}{u} du

    Let m = 3 + 2 \ln(u) \,\, \therefore dm = \frac{2}{u}\,\,du

    So plugging in m and dm we get:

    \begin{align*}\frac{0.8(3 + 2 \ln(u))^3}{u} du =& \int \frac{0.8m^3}{2}\,\,dm\\=& \frac{0.8m^4}{8} + C\\=& \frac{0.8(3 + 2 \ln(u))^4}{8} + C....\text{[by plugging back the value of } m\text{]}\end{align*}

    So, \int \frac{0.8(3 + 2 \ln(u))^3}{u} du = \frac{0.8(3 + 2 \ln(u))^4}{8} + C....\text{[Answer]}

    You can check the answer by taking the derivative of \frac{0.8(3 + 2 \ln(u))^4}{8} + C which
    is equal to \frac{0.8(3 + 2 \ln(u))^3}{u}

    derivative 0.8((3 + 2 ln(u))^4}/8) + C - Wolfram|Alpha

    Hope it answers your question.
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  4. #4
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    Re: Integration: General Power Formula

    Hi how did 0.8m^3 / 2 turn into 0.8m^4 / 8?

    how did the division by 2 turn into a division by 8?
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  5. #5
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    Re: Integration: General Power Formula

    Surely you know that \displaystyle \int{ x^n \,dx} = \frac{x^{n+1}}{n+1} + C...
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  6. #6
    Senior Member x3bnm's Avatar
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    Re: Integration: General Power Formula

    Quote Originally Posted by togo View Post
    Hi how did 0.8m^3 / 2 turn into 0.8m^4 / 8?

    how did the division by 2 turn into a division by 8?
    Integration is the opposite of differentiation hence the other name anti-derivative.

    So by integrating what you'll get is when differentiated will be what you're integrating.

    So:

    \int \frac{0.8m^3}{2}\,\,dm you're integrating and you'll get \frac{0.8m^4}{8} + C

    When you differentiate \frac{0.8m^4}{8} + C you'll get what you're integrating. So when you integrate you've to rearrange variables and constants in such a way that when you differentiate the rearrangement you'll end up with what you're integrating.

    So if we differentiate \frac{0.8m^4}{8} + C we get:
    Let f(x) = \frac{0.8m^4}{8} + C\,\, \therefore \frac{df}{dx} = \frac{0.8* 4* m^{(4-1)}}{8} + 0= \frac{0.8m^3}{2} which is what you're integrating.

    And that's why \int \frac{0.8m^3}{2}\,\,dm = \frac{0.8m^4}{8} + C
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  7. #7
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    Re: Integration: General Power Formula

    Quote Originally Posted by Prove It View Post
    Surely you know that \displaystyle \int{ x^n \,dx} = \frac{x^{n+1}}{n+1} + C...
    what confuses me is according to the general power rule, you add 1 to the exponent found on the integral, which is 3, so ends up being 4 on the bottom. While I understand it differentiates properly (post #6), seems like something other than general power rule was used to come up with a denominator of 8.
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  8. #8
    Senior Member x3bnm's Avatar
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    Re: Integration: General Power Formula

    Quote Originally Posted by togo View Post
    what confuses me is according to the general power rule, you add 1 to the exponent found on the integral, which is 3, so ends up being 4 on the bottom. While I understand it differentiates properly (post #6), seems like something other than general power rule was used to come up with a denominator of 8.
    You first separate the constants from variable. So it will be:

    \begin{align*}\int \frac{0.8m^3}{2}\,\,dm =& \frac{0.8}{2} \int {m^3}\,\,dm \\=& \frac{0.8}{2}\left( \frac{m^{(3+1)}}{3+1}\right) + C\\=& \frac{0.8}{2}\left(\frac{m^4}{4}\right)+C\\=& \frac{0.8m^4}{8}+ C\end{align*}
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