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Math Help - Challenge Integral

  1. #1
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    Challenge Integral

    Calculate \int_{0}^{1}\frac{x^4+1}{2 x^4-4 x^3+6 x^2-4 x+3}\;{dx}.
    Last edited by CaptainBlack; May 24th 2011 at 06:55 PM. Reason: Approved
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  2. #2
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    Hint:

    Spoiler:
    \begin{aligned} 2 x^4-4 x^3+6 x^2-4 x+3} & = x^4+(x^4-4x^3+6x^2-4x+1)+2 \\& = x^4+(1-x)^4+2. \end{aligned}
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  3. #3
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    Okay, following from the hint, here is the solution:

    \displaystyle \begin{aligned} I_{n} & = \int_{0}^{a}\frac{x^n+b}{x^n+(a-x)^n+2b}\;{dx}  \\& = \int_{0}^{a}\frac{(a-x)^n+b}{(a-x)^n+x^n+2b}\;{dx} \\& = \frac{1}{2}\int_{0}^{a}\frac{x^n+(a-x)^n+2b}{x^n+(a-x)^n+2b}\;{dx} \\& = \frac{1}{2}~a.\end{aligned}

    In general, we have the following nice result:

    \displaystyle \begin{aligned} J & = \int_{0}^{a} \frac{f(x)}{f(x)+f(a-x)}\;{dx} \\& = \int_{0}^{a}\frac{f(a-x)}{f(a-x)+f(x)}\;{dx} \\&= \frac{1}{2}\int_{0}^{a} \frac{f(x)+f(a-x)}{f(x)+f(a-x)}\;{dx} \\& = \frac{1}{2}~a. \end{aligned}.
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  4. #4
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    It's funny how you phrase the title, so methinks that it's actually an algebra problem rather than a calculus problem haha, and of course, having in mind the old symmetry trick.
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  5. #5
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    Quote Originally Posted by TheCoffeeMachine View Post
    Okay, following from the hint, here is the solution:

    \displaystyle \begin{aligned} I_{n} & = \int_{0}^{a}\frac{x^n+b}{x^n+(a-x)^n+2b}\;{dx}  \\& = \int_{0}^{a}\frac{(a-x)^n+b}{(a-x)^n+x^n+2b}\;{dx} \\& = \frac{1}{2}\int_{0}^{a}\frac{x^n+(a-x)^n+2b}{x^n+(a-x)^n+2b}\;{dx} \\& = \frac{1}{2}~a.\end{aligned}

    In general, we have the following nice result:

    \displaystyle \begin{aligned} J & = \int_{0}^{a} \frac{f(x)}{f(x)+f(a-x)}\;{dx} \\& = \int_{0}^{a}\frac{f(a-x)}{f(a-x)+f(x)}\;{dx} \\&= \frac{1}{2}\int_{0}^{a} \frac{f(x)+f(a-x)}{f(x)+f(a-x)}\;{dx} \\& = \frac{1}{2}~a. \end{aligned}.
    Dear TheCoffeeMachine,

    Thanks for posting this challenge integral. I enjoyed solving it. Alternatively we could also get the answer if we use use the substitution, x=1-y

    \int_{0}^{1}\frac{x^4+1}{2 x^4-4 x^3+6 x^2-4 x+3}\;{dx}

    =-\int_{1}^{0}\frac{(y-1)^4+1}{y^4+1+(y-1)^4+1}dy

    =\int_{0}^{1}\frac{(y-1)^4+1}{y^4+1+(y-1)^4+1}dy

    Hence, \int_{0}^{1}\frac{x^4+1}{x^4+1+(x-1)^4+1}dx=\int_{0}^{1}\frac{(x-1)^4+1}{x^4+1+(x-1)^4+1}dx-----------(1)

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

    By (1) and (2),

    \int_{0}^{1}\frac{x^4+1}{2 x^4-4 x^3+6 x^2-4 x+3}\;{dx}=0.5
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