# Challenge Integral

• May 24th 2011, 03:42 PM
TheCoffeeMachine
Challenge Integral
Calculate $\int_{0}^{1}\frac{x^4+1}{2 x^4-4 x^3+6 x^2-4 x+3}\;{dx}.$
• May 26th 2011, 07:34 AM
TheCoffeeMachine
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}
• May 27th 2011, 09:42 AM
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}.
• May 28th 2011, 07:06 PM
Krizalid
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.
• Jun 5th 2011, 05:39 AM
Sudharaka
Quote:

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),

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