1. ## Squeezing integrals

Use the fact that
$\displaystyle x-\frac{x^3}{6}\leq{sin{}x}\leq{x-\frac{x^3}{6}+\frac{x^5}{120}}$

to prove that $\displaystyle \int_0^{\frac{1}{2}} sin(x^2)=\frac{223}{5376}$ with error no greater than $\displaystyle \frac{1}{2703360}$

I attempted to use the inequality by replacing $\displaystyle x$ with $\displaystyle x^2$ and integrating both sides to get bounds on $\displaystyle sin (x^2)$, but I got nothing close to the actual answer

2. If $\displaystyle \displaystyle X - \frac{X^3}{6} \leq \sin{X} \leq X - \frac{X^3}{6} + \frac{X^5}{120}$

then by letting $\displaystyle \displaystyle X = x^2$ we have

$\displaystyle \displaystyle x^2 - \frac{x^6}{6} \leq \sin{(x^2)} \leq x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}$

$\displaystyle \displaystyle \int_0^{\frac{1}{2}}{x^2-\frac{x^6}{6}\,dx} \leq \int_0^{\frac{1}{2}}{\sin{(x^2)}\,dx} \leq \int_0^{\frac{1}{2}}{x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}\,dx}$

$\displaystyle \displaystyle \left[\frac{x^3}{3} - \frac{x^7}{42}\right]_0^{\frac{1}{2}} \leq \int_0^{\frac{1}{2}}{\sin{(x^2)}\,dx} \leq \left[\frac{x^3}{3} - \frac{x^7}{42} + \frac{x^{11}}{1320}\right]_0^{\frac{1}{2}}$

$\displaystyle \displaystyle \frac{223}{5376} \leq \int_0^{\frac{1}{2}}{\sin{(x^2)}} \leq \frac{784967}{18923520}$.

The difference between the endpoints is $\displaystyle \displaystyle \frac{1}{2703360}$.

Therefore $\displaystyle \displaystyle \int_0^{\frac{1}{2}}{\sin{(x^2)}\,dx} = \frac{223}{5376}$ with an error no greater than $\displaystyle \displaystyle \frac{1}{2703360}$.

3. I integrated wrong... Shame on me.

Thanks