A definite integral which shows us how close our approximation is to a transcendental number.

$\begin{aligned} I & = \int_0^1 \frac {x^4(1-x)^4}{1+x^2} dx \\ & = \int_0^1 \frac {x^4(x^4-4x^3+6x^2-4x+1)}{1+x^2} dx \\ & = \int_0^1 \frac {x^8-4x^7+6x^6-4x^5+x^4}{x^2+1} dx \\ & = \int_0^1 \left(x^6-4x^5+5x^4-4x^2+4- \frac 4{x^2+1} \right) dx \\ & = \frac {x^7}7- \frac {4x^6}6 + \frac {5x^5}5 - \frac {4x^3}3 + 4x - 4\tan^{-1}x \ \bigg|_0^1 \\ & = \frac 17- \frac 23 + 1 - \frac 43 + 4 - \pi \\ & = \frac {22}7 - \pi \approx 0.0012645 \end{aligned}$

$\implies \lceil 10000A \rceil = \lceil 12.645 \rceil = \boxed{13}$