Is there an integral that proves $\pi > 333/106$? The following integral,
$$ \int_0^1 \frac{x^4(1-x)^4}{x^2 + 1} \mathrm{d}x = \frac{22}{7} - \pi $$
is clearly positive, which proves that $\pi < 22/7$.
Is there a similar integral which proves $\pi > 333/106$?

Another solution is given by the integral
$$
0 < \int_0^1 \frac{x^4(1-x)^8}{4(1+x^2)}dx = \pi -\frac{2419}{770} = \pi - \frac{333}{106}-\frac{1}{20405}
$$
This proves the stricter condition
$$
\pi > \frac{333}{106}+\frac{1}{20405}
$$
which implies
$$
\pi > \frac{333}{106}
$$
Similarly, for the fourth convergent (formula (6) http://www.math.ucla.edu/~vsv/resource/general/Lucas.pdf)
$$0<\int_0^1 \frac{x^{10}(1-x)^8}{4(1+x^2)}dx=\frac{3849155}{1225224}-\pi=\frac{355}{113}-\frac{5}{138450312}-\pi$$
Therefore
$$\pi<\frac{355}{113}-\frac{5}{138450312}$$
and
$$\pi<\frac{355}{113}$$
Even for the first convergent
$$0<2\int_0^1 \frac{x(1-x)^2}{(1+x^2)}dx=\pi-3$$
so
$$\pi>3$$
(See https://math.stackexchange.com/a/1618454/134791 for a proof for $3<\pi<4$ that uses this integral)
A series proof that $\pi>\frac{333}{106}$ is given by

$$\frac{48}{371} \sum_{k=0}^\infty \frac{118720 k^2+762311 k+1409424}{(4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21) (4 k+23)} \\=\pi-\frac{333}{106}$$

