Solutions - 0250-03

Write the series of the integrand:

$$\begin{array}{c}\sin \left(x^2\right)=\sum _{n=0}^{\infty }{(-1)}^n\frac{{\left(x^2\right)}^{2n+1}}{(2n+1)!}\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{x^{4n+2}}{(2n+1)!}\end{array}$$

Integrate:

$$\begin{array}{c}{\int }_0^1\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{4n+2}}{(2n+1)!}dx\\ \\ \gg \gg {\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{4n+3}}{(2n+1)!(4n+3)}|}_0^1\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{(2n+1)!(4n+3)}\\ \\ \gg \gg \frac{1}{3}-\frac{1}{3!(7)}+\frac{1}{5!(11)}-\cdots \end{array}$$

Now apply the “Next Term Bound” and look for the first term below ${10}^{-3}$:

$$\frac{1}{3!(7)}=2.4\times {10}^{-2},\frac{1}{5!(11)}=7.6\times {10}^{-4}$$

So we simply add the first two terms:

$$\frac{1}{3}-\frac{1}{3!(7)}\approx 0.3095$$