Integration problem

**SyNtHeSiS** · August 11th 2010, 07:54 AM

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.

**wonderboy1953** · August 11th 2010, 10:29 AM

$(1 + x^2)(1 - x^2) = (1 - x^4)$ . Can you figure out the rest?

**eumyang** · August 11th 2010, 10:40 AM

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.

**Vandermonde** · August 11th 2010, 11:08 AM

Originally Posted by wonderboy1953

$(1 + x^2)(1 - x^2) = (1 - x^4)$ . Can you figure out the rest?

$(1-x)^4 = (1-x)^2 (1-x)^2 \ne (1+x^2)(1-x^2).<br />$

Originally Posted by eumyang

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

It's this one (a fairly well-known integral).

**Vandermonde** · August 11th 2010, 11:11 AM

To the OP, let $x = \tan{\theta}$ .

**yeKciM** · August 11th 2010, 01:04 PM

Originally Posted by SyNtHeSiS

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.

$\displaystyle \int _0 ^1 \frac {x^4(1-x)^4}{1+x^2} \; dx = \int _0 ^1 (x^6-4x^5+5x^4-4x^2-\frac {4}{x^2+1} +4 ) \; dx = \frac {22}{7} -\pi$

**TheCoffeeMachine** · August 11th 2010, 01:29 PM

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).

**SyNtHeSiS** · August 26th 2010, 07:12 AM

Originally Posted by Vandermonde

$(1-x)^4 = (1-x)^2 (1-x)^2 \ne (1+x^2)(1-x^2).<br />$

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 $\frac{22}{7}-\pi$ .

**Defunkt** · August 26th 2010, 09:09 AM

Originally Posted by SyNtHeSiS

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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