Challenge Integral

**TheCoffeeMachine** · May 24th 2011, 02:42 PM

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

**TheCoffeeMachine** · May 26th 2011, 06:34 AM

Hint:

Spoiler:

**TheCoffeeMachine** · May 27th 2011, 08:42 AM

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:

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

**Krizalid** · May 28th 2011, 06:06 PM

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.

**Sudharaka** · June 5th 2011, 04:39 AM

Originally Posted by TheCoffeeMachine

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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