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Math Help - 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 \int(x-7)\sqrt{3-2x}^3

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

    First I completely expanded the integral

    \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
    -21\sqrt{3-2x}



    I just simply use the substitution technique.
    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 u=x dv=3-2x or 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 \int -21x^2\sqrt{3-2x} just like the site did. They say to substitute u=\sqrt{3-2x} and du=1/\sqrt{3-2x} to get

    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
    \int -21x^2\sqrt{3-2x}= 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 \int(x-7)\sqrt{3-2x}^3 \,dx
    From that point use s = 3 - 2x, then ds = -2dx.
    I'll let you crunch through the integral as it should be pretty straightforward. But you should get \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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    Quote Originally Posted by utopiaNow View Post
    From that point use s = 3 - 2x, then ds = -2dx.
    I'll let you crunch through the integral as it should be pretty straightforward. But you should get \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 \frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C I get
    (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 \frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C I get
    (12x-18)(3 - 2x)^{5/2}
    Well that's incorrect, when I calculate the derivative of \frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C I get (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 \frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C I get (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
    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}

    <br />
d/dx \frac{11}{10}(3 - 2x)^{5/2}=\frac{-11}{2}[(3 - 2x)^{3/2}]<br />
    Combining them I get
    \frac{-1}{2}(3 - 2x)^{5/2}+\frac{-11}{2}[(3 - 2x)^{3/2}
    <br />
= (3 - 2x)^{3/2}(\frac{-11}{2}-\frac{1}{2}(3-2x))<br />
=(3 - 2x)^{3/2}(-6(3-2x))<br />
=(3 - 2x)^{3/2}(-18+12x)<br />
    Last edited by emonimous; October 27th 2009 at 03:48 PM. Reason: need to add more formulas
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  7. #7
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    Combining them I get
    \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))<br />
=(3 - 2x)^{3/2}(-18+12x)<br /> <br />
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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
    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}

    <br />
d/dx \frac{11}{10}(3 - 2x)^{5/2}=\frac{-11}{2}[(3 - 2x)^{3/2}]<br />
    Combining them I get
    \frac{-1}{2}(3 - 2x)^{5/2}+\frac{-11}{2}[(3 - 2x)^{3/2}
    <br />
= (3 - 2x)^{3/2}(\frac{-11}{2}-\frac{1}{2}(3-2x))<br />
    <br />
=\color{red}(3 - 2x)^{3/2}(-6(3-2x))<br />
=\color{black}(3 - 2x)^{3/2}(-18+12x)<br />
    The problem is in the red step and obviously the step that follows from it. You see you did \frac{-11}{2} - \frac{1}{2} = -6. Which would be fine if that's what it was, but you see the \frac{-1}{2} factor is attached to a (3 - 2x) term, hence what you did does not make any sense. If you further factor out the (-1/2) and then do (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 I = \int(x-7)(3 - 2x)^{3/2} \,dx

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

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

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

    So by multiplying out we get \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 s = 3 - 2x we get

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