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Math Help - Did I do this correct, integration of parts

  1. #1
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    Did I do this correct, integration of parts

    S x(x+1)^.5 dx

    u = 1x v= 2/3(x+1)^3/2
    du = 1 dv (x+1)^.5

    answer

    2/3x(x+1)^3/2 -2(x+1)^3 + c

    Let me know thanks.

    CORRECTED MY du, I copied it down wrong, it should be 1
    Last edited by bradycat; August 14th 2011 at 02:26 PM.
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  2. #2
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    Re: Did I do this correct, integration of parts

    u' = 1. Not sure how you found your du.

    Your dv and v is fine.

    \dfrac{2}{3}x(x+1)^{3/2} - \int \dfrac{2}{3}(x+1)^{3/2} \cdot 1 \ dx

    If you integrate using the power rule you end up with \dfrac{2}{3}x(x+1)^{3/2} - \dfrac{4}{15}(x+1)^{5/2} + C = \dfrac{2}{15}(x+1)^{3/2} \left(5x - 2(x+1)\right)

     = \dfrac{2}{15}(x+1)^{3/2} (3x-2) + C
    Last edited by e^(i*pi); August 14th 2011 at 03:28 PM.
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    Re: Did I do this correct, integration of parts

    OK, I got the problem with your help, but the (5-2(x+1)) shouldn't the 5 have a 5x, as the 2/3x has an x with it?? Or did I miss something?
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    Re: Did I do this correct, integration of parts

    hi bradycat

    i think you are right.but anyway besides that everything else should be correct
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    Re: Did I do this correct, integration of parts

    Ok just making sure, thanks again for your help. always appreciated with the help always.
    Joanne
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    Re: Did I do this correct, integration of parts

    Quote Originally Posted by bradycat View Post
    OK, I got the problem with your help, but the (5-2(x+1)) shouldn't the 5 have a 5x, as the 2/3x has an x with it?? Or did I miss something?
    Yes, there should be. Good spot!
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  7. #7
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    Re: Did I do this correct, integration of parts

    Quote Originally Posted by bradycat View Post
    S x(x+1)^.5 dx

    u = 1x v= 2/3(x+1)^3/2
    du = 1 dv (x+1)^.5

    answer

    2/3x(x+1)^3/2 -2(x+1)^3 + c

    Let me know thanks.

    CORRECTED MY du, I copied it down wrong, it should be 1
    I would make the substitution \displaystyle u = x + 1 \implies x = u - 1 and \displaystyle du = dx.

    Then

    \displaystyle \begin{align*} \int{x(x + 1)^{\frac{1}{2}}\,dx} &= \int{(u - 1)u^{\frac{1}{2}}\,du} \\ &= \int{u^{\frac{3}{2}} - u^{\frac{1}{2}}\,du} \\ &= \frac{u^{\frac{5}{2}}}{\frac{5}{2}} - \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C \\ &= \frac{2(x + 1)^{\frac{5}{2}}}{5} - \frac{2(x + 1)^{\frac{3}{2}}}{3} + C \end{align*}
    Last edited by Prove It; August 14th 2011 at 08:53 PM.
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  8. #8
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    Re: Did I do this correct, integration of parts

    I have another question. Tried this question again.
    2/3(x)(x+1)^3/2 -2/3 int sign (x+1)^3/2 * 1 dx
    So I get
    2/3(x)(x+1)^3/2 - 2/5x(x+1)^5/2 +c


    The question is the 1 above which is du, so the integral is x so I just pop it infront of the 2/5???
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    Re: Did I do this correct, integration of parts

    Quote Originally Posted by bradycat View Post
    I have another question. Tried this question again.
    2/3(x)(x+1)^3/2 -2/3 int sign (x+1)^3/2 * 1 dx
    So I get
    2/3(x)(x+1)^3/2 - 2/5x(x+1)^5/2 +c


    The question is the 1 above which is du, so the integral is x so I just pop it infront of the 2/5???
    I edited my post above which has the easiest method, using a u substitution, now that the LaTeX is fixed you should be able to see it.

    If you have to use integration by parts:

    \displaystyle u = x \implies du = dx, dv = (x + 1)^{\frac{1}{2}}\,dx \implies v = \frac{2}{3}(x + 1)^{\frac{3}{2}} , so

    \displaystyle \begin{align*} \int{x(x + 1)^{\frac{1}{2}}\,dx} &= \frac{2}{3}x(x + 1)^{\frac{3}{2}} - \int{\frac{2}{3}(x + 1)^{\frac{3}{2}}\,dx} \\ &= \frac{2}{3}x(x + 1)^{\frac{3}{2}} - \frac{2}{3}\int{(x + 1)^{\frac{3}{2}}\,dx} \\ &= \frac{2}{3}x(x + 1)^{\frac{3}{2}} - \frac{2}{3}\cdot \frac{2}{5}(x + 1)^{\frac{5}{2}} + C \\ &= \frac{2}{3}x(x + 1)^{\frac{3}{2}} - \frac{4}{15}(x + 1)^{\frac{5}{2}} + C \end{align*}
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  10. #10
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    Re: Did I do this correct, integration of parts

    if u = x and du = 1.
    well when you put the first line after the negative integral sign, it's 2/3(x+1)^3/2 * 1 dx. as du is 1?

    When you take the integral of the 1 wouldn't it be x then the other becomes 2/5(x+1)^5/2????
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    Re: Did I do this correct, integration of parts

    Quote Originally Posted by bradycat View Post
    if u = x and du = 1.
    You keep saying this even though the first response to your post told you it was wrong. If u= x, then du= dx, not 1.

    well when you put the first line after the negative integral sign, it's 2/3(x+1)^3/2 * 1 dx. as du is 1?

    When you take the integral of the 1 wouldn't it be x then the other becomes 2/5(x+1)^5/2????
    The "1" doesn't become anything- \int (2/3)(x+ 1)^{3/2}dx
    = (2/3)\int u^{3/2}du= (2/3)(2/5)u^{5/2}= (4/15)(x+1)^{5/2}+ C
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  12. #12
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    Re: Did I do this correct, integration of parts

    Quote Originally Posted by bradycat View Post
    if u = x and du = 1.
    well when you put the first line after the negative integral sign, it's 2/3(x+1)^3/2 * 1 dx. as du is 1?

    When you take the integral of the 1 wouldn't it be x then the other becomes 2/5(x+1)^5/2????
    \displaystyle du does NOT equal \displaystyle 1, \frac{du}{dx} = 1 and \displaystyle du = dx.
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