Verified 30 May 2026.
If $\displaystyle\int_0^1 \frac{x^4(1-x)^4}{1+x^2} \, dx = \frac{22}{7} - \pi$, find the value of $7\displaystyle\int_0^1 \frac{x^4(1-x)^4}{1+x^2} \, dx + 7\pi$ rounded to the nearest integer.
Given: $\displaystyle\int_0^1 \frac{x^4(1-x)^4}{1+x^2} \, dx = \frac{22}{7} - \pi$
$$7\int_0^1 \frac{x^4(1-x)^4}{1+x^2} \, dx + 7\pi = 7\left(\frac{22}{7} - \pi\right) + 7\pi$$
$$= 22 - 7\pi + 7\pi = \boxed{22}$$
This is the famous integral that proves $\frac{22}{7} > \pi$ since the integrand is strictly positive on $(0,1)$.
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