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Math Help - Is This Correct? Looks like a simple Intergral Problem...

  1. #1
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    Is This Correct? Looks like a simple Intergral Problem...

    Evaluate:

    Intergral of X Times squareroot 5X+1 DX
    (see attached pic of problem)


    So I used substitution,

    U= 5X+1
    Du= 5Dx
    Dx= 1/5Du

    Next I found Intergral UDU

    U^1/2 Times 1/5DU =

    2/3U^3/2 Times 1/5DU

    my final answer:

    2/15U^3/2 + C

    Simplified to:

    2/15(5x+1)^3/2 + C



    Is that correct?
    Attached Thumbnails Attached Thumbnails Is This Correct? Looks like a simple Intergral Problem...-calcproblemagain.bmp  
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  2. #2
    Senior Member MacstersUndead's Avatar
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    EDIT: wait... no, forgot the chain rule.

    EDIT EDIT:
    Take u = 5x + 1, so that x = (u - 1)/5 and du = 5 dx

    => the integral of (1/5) * (1/5) [(u-1)(sqrt(u)] du (substituting **x**, u and du)
    => the integral of (1/25) (u^3/2 - u^1/2) du
    (1/25) * (2/5)u^5/2 - (2/3)u^3/2
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  3. #3
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    Take u = 5x + 1, so that x = (u - 1)/5 and du = 5 dx

    => the integral of (1/5) * (1/5) [(u-1)(sqrt(u)] du (substituting **x**, u and du)
    => the integral of (1/25) (u^3/2 - u^1/2) du
    (1/25) * (2/5)u^5/2 - (2/3)u^3/2


    ...yes this is correct
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  4. #4
    Moo
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    Hello,

    Make t=\sqrt{5x+1}
    dt=\frac 52 \frac{dx}{\sqrt{5x+1}} \Rightarrow dx=\frac 25 ~ t ~dt
    And t^2=5x+1 \Rightarrow x=\frac{t^2-1}{5}

    So the integral is now :
    \frac 25 \int t \left(\frac{t^2-1}{5}\right) ~dt=\frac{2}{25} \int t^3-t ~dt
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