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Math Help - Integration with u-substitution question

  1. #1
    Member nautica17's Avatar
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    Integration with u-substitution question

    I'm trying to solve this:

    integral of (y^2) / sqrt(4-3y)

    And so I see that I need to do a u-substitution. So I set u = 4-3y, and du = -3.
    So at this point I have:

    integral of ( (y^2) / sqrt(u) ) du/-3
    = (-1/3) integral of ( (y^2) / sqrt(u) ) du

    Now for the y^2 up top, I figure I'll need to do 3y = 4-u, but I'm not sure how to implement this. When I check my answers with wolfram alpha, they do the same thing up this point, but they put a 9 in front of the sqrt(u).
    Something like this:
    (-1/3) integral of ( ((4-u)^2) / 9sqrt(u) ) du

    And so I am sort of stuck right now, and I'm not sure how to get past this step. Can someone tell me where the 9 is from? Perhaps I'm overlooking some crucial algebraic step.
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  2. #2
    Member nautica17's Avatar
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    Nevermind... I found the answer to my question. >.<
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  3. #3
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    Hello, nautica17!

    You solved it yourself . . . Good for you!


    \displaystyle \int \dfrac{y^2\,dy}{\sqrt{4-3y}}

    Here's a tip that might make life a bit easier . . .

    If the expression under the radical is linear, let \,u = the entire radical.


    \text{Let: }\:u \:=\:\sqrt{4-3y} \quad\Rightarrow\quad y \:=\:\dfrac{4-u^2}{3} \quad\Rightarrow\quad dy \:=\:-\frac{2}{3}u\,du


    \displaystyle \text{Substitute: }\;\int\frac{(\frac{4-u^2}{3})^2}{u}\left(-\tfrac{2}{3}u\,du\right) \;=\;-\frac{2}{27}\int\left(16 - 8u^2 + u^4)\,du

    . . . =\;-\frac{2}{27}\left(16u - \frac{8}{3}u^3 + \frac{1}{5}u^5\right) + C \;=\;-\frac{2}{405}u\left(240-40u^2 + 3u^4\right)+C


    \text{Back-substitute: }\;-\frac{2}{405}\sqrt{4-3y}\bigg[240 - 40(4-3y) + 3(4-3y)^2\bigg] + C

    . . . =\;-\frac{2}{405}\sqrt{4-3y}\,\bigg[240 - 160 + 120y + 48 = 72y + 27y^4\bigg]


    . . . =\; -\frac{2}{405}\sqrt{4-3y}\,\left(27y^4 + 48y + 128\right) + C

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  4. #4
    Member nautica17's Avatar
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    I'll have to keep that trick in mind next time. Thank-you!
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  5. #5
    MHF Contributor

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    By the way:
    1) You do NOT want to integrate \int\frac{y^2}{\sqrt{4- 3y}}- you want to integrate \int\frac{y^2}{\sqrt{4- 3y}}dy. That "dy" is important.

    2) Related to that- if u= 4- 3y, then du is NOT -3, du= -3 dy.
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