Approximating PI: Int. of x^4(1-x)^4/(1+x^2)= 22/7 - Pi

[Zadey]

Approximating PI Show that $$\int$$$$\stackrel{1}{0}$$$$\frac{x^{4}(1-x)^{4}}{1+x^{2}}$$dx=$$\frac{22}{7}$$-$$\Pi$$ Why does this imply that $$\Pi$$$$\triangleleft$$$$\frac{22}{7}$$


I have no clue where to begin with this, I'm at a loss, this is one of the questions for in my university project, first year. Any help is appreciated.

 

[aPhilosopher]

Calculating the integral would probably be a good start.

What's the numerical value of the integral? Is it a big number or a small number? Is it positive or negative?

 

[Billy Bob]

I have no clue where to begin with this, I'm at a loss, this is one of the questions for in my university project, first year. Any help is appreciated.

Welcome to PF, Zadey.


To evaluate $$\int_0^1 \frac{x^{4}(1-x)^{4}}{1+x^{2}}\,dx$$

multiply out the numerator, then use long division, then integrate from 0 to 1.

 

[HallsofIvy]

Once you have done the integral and derived the result shown, if $\pi$ were greater than 22/7, the integral would be negative.

 

[Zadey]

Thanks, I got it. Don't know why I didin't see it earlier.
Now if I had to find the maximum of the numerator, how would I go about using it to show that $$\frac{22}{7}$$<$$\frac{\pi}{1024}$$<$$\frac{1}{100}$$ and how does it imply that the approximation $$\frac{22}{7}$$is accurate to 2 decimal places? I know that the $$\frac{1}{100}$$ would be used to imply that its accurate to 2 decimal places but how it does I'm not sure.