$\displaystyle \frac{b-a}{2n}\bigg[f(x_0)+\sum_{m=1}^{n-1}2f(x_m)+f(x_n)\bigg]$hello i have a big problem i can solve this problem pleae help

Use the trapezoidal rule with n-8 to approximate the definite integral of sin(x^2)dx when a=0 and b=2

So we have for this case

$\displaystyle \Delta{x}=\frac{2-0}{8}=\frac{1}{4}$

$\displaystyle \frac{2-0}{16}\bigg[\sin(0^2)+2\sin\bigg(\frac{1}{16}\bigg)+2\sin\bigg (\frac{1}{4}\bigg)+$$\displaystyle 2\sin\bigg(\frac{9}{16}\bigg)+2\sin(1)+\sin\bigg(\ frac{25}{16}\bigg)+2\sin\bigg(\frac{9}{4}\bigg)+2\ sin\bigg(\frac{49}{16}\bigg)+\sin(4)\bigg]\approx\int_0^{2}\sin(x^2)dx$

I dont know how high you are in but if you are interested

$\displaystyle \int\sin(x^2)dx=\sum_{n=0}^{\infty}\frac{(-1)^nx^{4n+3}}{(4n+3)(2n+1)!}$