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Math Help - Integral Refresher Question

  1. #1
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    Integral Refresher Question

    Hi, I'm having a little bit of trouble with a integration question, here is the text from the question.

    Find the antiderivitive of the given functions:

     F(x) = x(1-x^2)^7

    Basically I just have no idea how to expand  (1 - x^2)^7
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  2. #2
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    Quote Originally Posted by Crell View Post
    Hi, I'm having a little bit of trouble with a integration question, here is the text from the question.

    Find the antiderivitive of the given functions:

     F(x) = x(1-x^2)^7

    Basically I just have no idea how to expand  (1 - x^2)^7
    Dont expand!

    Use substition.

    u=1-x^2
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  3. #3
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    Well, you could expand it using the "binomial theorem": (a+ b)^n= \sum_{i=0}^n \begin{pmatrix}n \\ i\end{pmatrix}a^ib^{n- i}
    For (1- x^2)^7= 1- 7x^2+ 21x^4+ \cdot\cdot\cdot- 7x^{12}+ x^{14}

    But you don't want to do that. Instead, make the substituition u= 1- x^2 so that (1- x^2)^7= u^7 and du= -2 xdx.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    du= -2 xdx.
    Thank you much, this is the part I was missing.

    <br /> <br />
\int x(1-x^2)^7dx<br /> <br />

    Aside:

    Let  u = 1-x^2

    Derive this equation

    \frac{du}{dx} = -2x

    Solve for dx

     dx = \frac{du}{-2x}


    Manipulate the first equation to make it appear like  -2x(u)^7dx so that when you substitute dx, the x is eliminated.

    Therefore:

     -\frac{1}{2} \int -2x(u)^7dx

    Substitute dx = \frac{du}{-2x} into first equation
     -\frac{1}{2} \int \frac {-2x(u)^7du}{-2x}

     -\frac{1}{2} \int {(u)^7du}

    Integrate, sub in u and simplfy.


    - \frac{1}{2} [\frac {(u)^8}{8}]

     -\frac{1}{2} [\frac {(1-x^2)^8}{8}]


     -\frac {(1-x^2)^8}{16}
    Last edited by Crell; January 18th 2011 at 02:35 PM.
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  5. #5
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    Did you differentiate that answer to cheek it for yourself?
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  6. #6
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    Have now, thanks.
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  7. #7
    Super Member Quacky's Avatar
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    This is a perfect place to use my favourite rule of integration:

    \displaystyle\int{f'(x)f(x)^n} dx=\frac{1}{n+1}f(x)^{n+1}+c

    So let f(x)=(1-x^2)
    f'(x)=-2x
    n=7

    f'(x)f(x)^n= -2x(1-x^2)^7
    You'll notice that it isn't quite a match to what you are trying to integrate. However, if you divide by -2, it is exactly what you are trying to integrate. Therefore, we can apply the rule as normal, but we must remember to divide our final answer by -2.

    \displaystyle\int x(1-x^2)^7 dx

    =\displaystyle\frac{-1}{2}\times\frac{(1-x^2)^8}{8} + C

    =\displaystyle -\frac{(1-x^2)^8}{16} + C

    I think in this case, this method is much nicer than substitution - if you spot the rule, it will take a few lines of working to obtain the full solution with no messy work required at all.
    Last edited by Quacky; January 18th 2011 at 03:45 PM. Reason: Typo removal
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  8. #8
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    In speaking of the binomial theorem, you can find the anti-derivative
    without performing an expansion, but in somewhat undesirable forms:

     \displaystyle \sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+2}}{2k+2}\binom{7}{k} or  \displaystyle \sum_{k=0}^{7}\frac{(-1)^k(x)^{2k+2}}{2k+2}\binom{7}{k}.

    Plus the constant of integration. The u-sub does the job far better.

    Spoiler:
    1.

    \displaystyle x(1-x^2)^7 = x\left(\sum_{k=0}^{\infty}\binom{7}{k}(-1)^kx^{2k}\right) = \sum_{k=0}^{\infty}\binom{7}{k}(-1)^{k}x^{2k+1}.


     \begin{aligned}\displaystyle \therefore \int x(1-x^2)^7\;{dx} & = \int \left(\sum_{k=0}^{\infty}\binom{7}{k}(-1)^{k}x^{2k+1}\right)\;{dx} \\& = \sum_{k=0}^{\infty} \left(\int \binom{7}{k}(-1)^{k}x^{2k+1}\;{dx}\right) \\& = \sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+2}}{2k+2}\binom{7}{k}+C.\end{aligned}

    2.

    \displaystyle \begin{aligned}x(1-x^2)^7 & = x\left(\sum_{k=0}^{7}\binom{7}{k}(-1)^kx^{2k}\right) \\& = \sum_{k=0}^{7}\binom{7}{k}(-1)^kx^{2k+1}.\end{aligned}

    \displaystyle \begin{aligned}  \therefore  \int x(1-x^2)^7\;{dx} & = \int \left(\sum_{k=0}^{7}\binom{7}{k}(-1)^kx^{2k+1}\right)\;{dx} \\& = \sum_{k=0}^{7} \left(\int \binom{7}{k}(-1)^kx^{2k+1}\;{dx}\right) \\& = \sum_{k=0}^{7}\frac{(-1)^k(x)^{2k+2}}{2k+2}\binom{7}{k}+C.\end{aligned}
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