Computing ∫₀¹ sin(x²) dx with a series

Series. $\sin(x^2)=\sum_{k\ge0}\frac{(-1)^k}{(2k+1)!}x^{4k+2}$, and integrating on $[0,1]$,
$$I=\sum_{k\ge0}\frac{(-1)^k}{(2k+1)!\,(4k+3)}=\frac13-\frac1{42}+\frac1{1320}-\frac1{75\,600}+\cdots$$

Alternating series bound. The terms decrease in absolute value to $0$, so the error is less than the first omitted term. The term for $k=3$ is $\frac1{75\,600}\approx1.3\cdot10^{-5}<10^{-4}$, so stopping at $k=2$ suffices:
$$I\approx\frac13-\frac1{42}+\frac1{1320}=\frac{2867}{9240}\approx0.31028.$$

Check. The true value is $0.310268\dots$, with error $1.3\cdot10^{-5}$ ✓.

Remark. This is a Fresnel-type integral. Integrating a power series term by term is legitimate on the interval where the series converges uniformly (here, all of $\mathbb{R}$).