Maths - How do we show by integrating [x^4(1+x)^4/(1+x^2) between 0 and 1 ,that π≠22/7?

Responses (2)

Your presentation is a mess.

Votes: +0 / -0

f(x) = [x^4 * (x + 1)^4] / (x^2 + 1) ;;; I think, that's what you wanted to say?

Expand the Nominator:
f(x) = (x^8 + 4x^7 + 6x^6 + 4x^5 +x^4) / (x^2 + 1) ;;; Divide

f(x) = x^6 + 4x^5 + 5x^4 - 4x^2 + 4 + (-4 / (x^2 + 1))

Integrate
∫ f(x) dx = F(x) = -4 arctan(x) + ((3x^7 + 14x^6 + 21x^5 - 28x^3 + 84x) / 21) + C

between 0 and 1, well, for x=0 the whole lot is 0, thus

∫ ,0 to 1, f(x) dx = 94/21 - π
If π = 22/7, then the answer would be: 4/3.

Votes: +0 / -0

I also managed to come this far...But can I know how does that prove π≠22/7..We can tell that the value of the integral≠22/7 ,but...

94/21 - π = 1.334597823...
whereas, when π = 22/7, the result could be expressed as a rational number, a nice fraction: 4/3. Which is certainly not a transcendental number ...

But frankly, I don't see, what this question is trying to get my attention focused upon.
Sry. -- Let me know, when your teacher shares their wisdom ... ;)