Answer
$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\sin(x)dx=\frac{1}{2}$
Work Step by Step
$$\lim\limits_{N \to \infty}\frac{\pi}{6N}\sum_{j=1}^{N}\sin(\frac{\pi}{3}+\frac{\pi j}{6N})$$
$$\lim\limits_{N \to \infty}\sum_{j=1}^{N}\sin(\frac{\pi}{3}+\frac{\pi j}{6N})\cdot \frac{\pi}{6N}$$
$$\lim\limits_{N \to \infty}\sum_{j=1}^{N}\sin(\frac{\pi}{3}+j\frac{\frac{\pi}{2}-\frac{\pi}{3}}{N})\cdot \frac{\frac{\pi}{2}-\frac{\pi}{3}}{N}$$
where $x_j=\frac{\pi}{3}+j\frac{\frac{\pi}{2}-\frac{\pi}{3}}{N}$ ,$\Delta x=\frac{\frac{\pi}{2}-\frac{\pi}{3}}{N}$,$a=\frac{\pi}{3}$,$b=\frac{\pi}{2}, $and $f(x)=\sin(x)$
so:
$$\lim\limits_{N \to \infty}\sum_{j=1}^{N}\sin(\frac{\pi}{3}+j\frac{\frac{\pi}{2}-\frac{\pi}{3}}{N})\cdot \frac{\frac{\pi}{2}-\frac{\pi}{3}}{N}=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\sin(x)dx$$
Using the $FTC I$ it follows:
$$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\sin(x)dx=[-\cos(x)]_{\frac{\pi}{3}}^{\frac{\pi}{2}}=-\cos(\frac{\pi}{2})-(-\cos(\frac{\pi}{3}))=\frac{1}{2}$$
