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Thread: Integral of (x-7)(3-2x)^(3/2)

  1. #1
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    Integral of (x-7)(3-2x)^(3/2)

    Hey guys, I have this problem I have been trying to solve for the longest time.
    Not looking for a complete solution, maybe part of one would be nice.

    I'm trying to integrate $\displaystyle \int(x-7)\sqrt{3-2x}^3$

    So far I have tried 2 ways of going at it.

    First I completely expanded the integral

    $\displaystyle \int(x-7)\sqrt{3-2x}^3=\int(-2x^2)(\sqrt{3-2x})+17x\sqrt{3-2x}-21\sqrt{3-2x}$

    I have no problem integrating
    $\displaystyle -21\sqrt{3-2x}$



    I just simply use the substitution technique.
    $\displaystyle u=3-2x$


    The next part gets tricky, I'm stumped. Any of the functions with x variables I simply can't seem to do.
    I tried substituting by parts by either making $\displaystyle u=x$ $\displaystyle dv=3-2x$ or $\displaystyle dv=\sqrt{3-2x}$ so on and so forth
    I went to wolframalpha.com and tried to understand what they did step by step

    For example, I tried to integrate $\displaystyle \int -21x^2\sqrt{3-2x}$ just like the site did. They say to substitute $\displaystyle u=\sqrt{3-2x}$ and $\displaystyle du=1/\sqrt{3-2x}$ to get

    $\displaystyle 2\int(1/4)u^2(3-2u^2)^2du$

    Either I am feeling really dumb right now and the answer is right in from of me or the answer is completely wrong. How did wolframalpha manage to get
    $\displaystyle \int -21x^2\sqrt{3-2x}$=$\displaystyle 2\int(1/4)u^2(3-2u^2)^2du$

    All help is greatly appreciated
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  2. #2
    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by emonimous View Post
    I'm trying to integrate $\displaystyle \int(x-7)\sqrt{3-2x}^3 \,dx$
    From that point use $\displaystyle s = 3 - 2x$, then $\displaystyle ds = -2dx$.
    I'll let you crunch through the integral as it should be pretty straightforward. But you should get $\displaystyle \frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C$, where C is a constant.
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  3. #3
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    Oh wow! Thanks a bunch!
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  4. #4
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    Quote Originally Posted by utopiaNow View Post
    From that point use $\displaystyle s = 3 - 2x$, then $\displaystyle ds = -2dx$.
    I'll let you crunch through the integral as it should be pretty straightforward. But you should get $\displaystyle \frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C$, where C is a constant.
    Actually when I try to find the derivative of $\displaystyle \frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C$ I get
    $\displaystyle (12x-18)(3 - 2x)^{5/2}$
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  5. #5
    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by emonimous View Post
    Actually when I try to find the derivative of $\displaystyle \frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C$ I get
    $\displaystyle (12x-18)(3 - 2x)^{5/2}$
    Well that's incorrect, when I calculate the derivative of $\displaystyle \frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C$ I get $\displaystyle (x - 7)(3 - 2x)^{3/2} $, which was precisely the original function whose integral we calculated. Perhaps you should try posting your steps here so I can spot where you went wrong.
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    Quote Originally Posted by utopiaNow View Post
    Well that's incorrect, when I calculate the derivative of $\displaystyle \frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C$ I get $\displaystyle (x - 7)(3 - 2x)^{3/2} $, which was precisely the original function whose integral we calculated. Perhaps you should try posting your steps here so I can spot where you went wrong.
    Oh don't get me wrong, I was trying to follow your instructions but I think I failed miserably. So I tried to back trace what you did by finding it's derivative.
    First I seperated the function in 2 functions so I can derive each
    $\displaystyle d/dx \frac{1}{14}(3 - 2x)^{7/2} =\frac{1}{14}[\frac{-7}{2}(3 - 2x)^{5/2}*-2]=\frac{-1}{2}(3 - 2x)^{5/2}$

    $\displaystyle
    d/dx \frac{11}{10}(3 - 2x)^{5/2}=\frac{-11}{2}[(3 - 2x)^{3/2}]
    $
    Combining them I get
    $\displaystyle \frac{-1}{2}(3 - 2x)^{5/2}+\frac{-11}{2}[(3 - 2x)^{3/2}$
    $\displaystyle
    = (3 - 2x)^{3/2}(\frac{-11}{2}-\frac{1}{2}(3-2x))
    =(3 - 2x)^{3/2}(-6(3-2x))
    =(3 - 2x)^{3/2}(-18+12x)
    $
    Last edited by emonimous; Oct 27th 2009 at 02:48 PM. Reason: need to add more formulas
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  7. #7
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    Combining them I get
    $\displaystyle \frac{-1}{2}(3 - 2x)^{5/2}+\frac{-11}{2}[(3 - 2x)^{3/2}= (3 - 2x)^{3/2}(\frac{-11}{2}-\frac{1}{2}(3-2x))=(3 - 2x)^{3/2}(-6(3-2x))
    =(3 - 2x)^{3/2}(-18+12x)

    $
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  8. #8
    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by emonimous View Post
    Oh don't get me wrong, I was trying to follow your instructions but I think I failed miserably. So I tried to back trace what you did by finding it's derivative.
    First I seperated the function in 2 functions so I can derive each
    $\displaystyle d/dx \frac{1}{14}(3 - 2x)^{7/2} =\frac{1}{14}[\frac{-7}{2}(3 - 2x)^{5/2}*-2]=\frac{-1}{2}(3 - 2x)^{5/2}$

    $\displaystyle
    d/dx \frac{11}{10}(3 - 2x)^{5/2}=\frac{-11}{2}[(3 - 2x)^{3/2}]
    $
    Combining them I get
    $\displaystyle \frac{-1}{2}(3 - 2x)^{5/2}+\frac{-11}{2}[(3 - 2x)^{3/2}$
    $\displaystyle
    = (3 - 2x)^{3/2}(\frac{-11}{2}-\frac{1}{2}(3-2x))
    $
    $\displaystyle
    =\color{red}(3 - 2x)^{3/2}(-6(3-2x))
    =\color{black}(3 - 2x)^{3/2}(-18+12x)
    $
    The problem is in the red step and obviously the step that follows from it. You see you did $\displaystyle \frac{-11}{2} - \frac{1}{2} = -6$. Which would be fine if that's what it was, but you see the $\displaystyle \frac{-1}{2}$ factor is attached to a $\displaystyle (3 - 2x)$ term, hence what you did does not make any sense. If you further factor out the $\displaystyle (-1/2)$ and then do $\displaystyle (11 + (3 - 2x))$ and carry forwards you will get the correct answer.
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  9. #9
    Junior Member utopiaNow's Avatar
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    I'll do a step-by-step solution if you're having difficulty with the integral. Or just to give you something to compare your work to.

    Let $\displaystyle I = \int(x-7)(3 - 2x)^{3/2} \,dx$

    Then, let $\displaystyle s = 3 - 2x$, hence $\displaystyle ds = -2dx$ ie. $\displaystyle \left(\frac{-1}{2}\right)ds = dx$.

    Then $\displaystyle x = \frac{3 - s}{2}$ Hence $\displaystyle x - 7 = \frac{-11 - s}{2}$

    Therefore $\displaystyle I = \frac{-1}{2}\int(11 + s)s^{3/2}\left(\frac{-1}{2}\right)ds$

    So by multiplying out we get $\displaystyle \frac{1}{4}\int 11s^{3/2} + s^{5/2} = \frac{1}{4}\left(\frac{22}{5}s^{5/2} + \frac{2}{7}s^{7/2}\right) + C$, since $\displaystyle s = 3 - 2x$ we get

    $\displaystyle I = \frac{11}{10}(3 - 2x)^{5/2} + \frac{1}{14}(3 - 2x)^{7/2} + C$, which we can actually further factor to give $\displaystyle \frac{-1}{35}(3 - 2x)^{5/2}(5x - 46) + C$ where $\displaystyle C \in \mathbb{R}$ of course.
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