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Math Help - Integration problem

  1. #1
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    Integration problem

    Evaluate \int_{0}^{1}\frac{x^4(1- x)^4}{1 + x^2} dx. Deduce that π does not equal 22/7.

    I thought about expanding the numerator and then performing long division, but this way got really messy for me, so I am wondering what the correct method would be.
    Last edited by SyNtHeSiS; August 26th 2010 at 08:01 AM.
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  2. #2
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    Intermediate simplify step

    (1 + x^2)(1 - x^2) = (1 - x^4). Can you figure out the rest?
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  3. #3
    Senior Member eumyang's Avatar
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    That may be true if the original problem is
    \int_{0}^{1}\frac{x^4(1 - x^4)}{1 + x^2} dx

    Granted, we are not 100% sure what the problem is, because there is an extra ")", so it could be what I typed above or

    \int_{0}^{1}\frac{x^4(1- x)^4}{1 + x^2} dx

    OP: please fix.
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  4. #4
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    Quote Originally Posted by wonderboy1953 View Post
    (1 + x^2)(1 - x^2) = (1 - x^4). Can you figure out the rest?
    (1-x)^4 = (1-x)^2 (1-x)^2 \ne (1+x^2)(1-x^2).<br />
    Quote Originally Posted by eumyang View Post
    \int_{0}^{1}\frac{x^4(1- x)^4}{1 + x^2} dx
    It's this one (a fairly well-known integral).
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  5. #5
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    To the OP, let x = \tan{\theta}.
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  6. #6
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by SyNtHeSiS View Post
    Evaluate \int_{0}^{1}\frac{x^4(1- x)^4)}{1 + x^2} dx. Deduce that π does not equal 22/7.

    I thought about expanding the numerator and then performing long division, but this way got really messy for me, so I am wondering what the correct method would be.


     \displaystyle \int _0 ^1 \frac {x^4(1-x)^4}{1+x^2} \; dx = \int _0 ^1 (x^6-4x^5+5x^4-4x^2-\frac {4}{x^2+1} +4 ) \; dx = \frac {22}{7} -\pi
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  7. #7
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    There is an elegant solution using Vandermonde's substitution.

    And it isn't really messy if you properly expand (1-x)^4:

    (1-x)^4 = (1-x)^2(1-x)^2 = (1-2x+x^2)(1-2x+x^2) = x^4-4 x^3+6 x^2-4 x+1.

    \Rightarrow x^4(1-x)^4 = x^8-4 x^7+6 x^6-4 x^5+x^4

    \displaystyle \therefore \frac{x^4(1-x)^4}{x^2+1} = \frac{x^8-4 x^7+6 x^6-4 x^5+x^4}{x^2+1} = x^6-4 x^5+5 x^4-4 x^2-\frac{4}{x^2+1}+4

    \displaystyle \therefore \int_{0}^{1}\frac{x^4(1-x)^4}{x^2+1}\;{dx} = \int_{0}^{1}{x^6-4 x^5+5 x^4-4 x^2+4}\;{dx}-\int_{0}^{1}\frac{4}{x^2+1}\;{dx}

    Now, \displaystyle \int_{0}^{1}{x^6-4 x^5+5 x^4-4 x^2+4}\;{dx} = \bigg[\frac{1}{7}x^7-\frac{2}{3}x^6+x^5-\frac{4}{3}x^3+4x\bigg]_{0}^{1} = \frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4 = \frac{22}{7}

    And since \displaystyle \int\frac{x}{x^2+a^2}\;{dx} = \frac{1}{a}\arctan\frac{x}{a}+k (see here), we have

    \displaystyle \int_{0}^{1}\frac{4}{x^2+1}\;{dx}= 4\bigg[\arctan{x}\bigg]_{0}^{1} = 4\left(\frac{\pi}{4}\right) = \pi.

    Thus \displaystyle \int_{0}^{1}\frac{x^4(1-x)^4}{x^2+1}\;{dx} = \frac{22}{7}-\pi.

    Since \displaystyle \int_{0}^{1}\frac{x^4(1-x)^4}{x^2+1}\;{dx} >0 for 0\le x \le 1, we conclude that \pi \ne \frac{22}{7}

    (In fact, \pi > \frac{22}{7}, since supposing otherwise would imply that the integral is negative).
    Last edited by TheCoffeeMachine; August 11th 2010 at 03:03 PM.
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  8. #8
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    Quote Originally Posted by Vandermonde View Post
    (1-x)^4 = (1-x)^2 (1-x)^2 \ne (1+x^2)(1-x^2).<br />
    Why cant you say: (1-x)^{4} = (1^{4} - x^{4}) = 1 - x^{4} = (1 + x^{2})(1-x^{2})

    I cant see how you can deduce that \pi doesnt equal \frac{22}{7} from the answer \frac{22}{7}-\pi.
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  9. #9
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    Quote Originally Posted by SyNtHeSiS View Post
    Why cant you say: (1-x)^{4} = (1^{4} - x^{4}) = 1 - x^{4} = (1 + x^{2})(1-x^{2})
    Because (1-x)^4 = (1-x)(1-x)(1-x)(1-x) \neq 1^4 - x^4... It's for the same reason (1-x)^2 \neq 1^2 - x^2.

    I cant see how you can deduce that \pi doesnt equal \frac{22}{7} from the answer \frac{22}{7}-\pi.
    The integral is I = \frac{22}{7} - \pi. The integrand is strictly positive in 0 \le x \le 1. The integral of a strictly positive function is always positive, and hence not zero. The conclusion follows.
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