Hello i have this calculus problem but i cannot solve it as i don't understand very much about riemann sums.
Use regular partitions to find the aproximate value of the integral : $$ \int_{\pi/12}^{\pi/6} \frac12sin(3x)dx $$
for the integration interval use circunscrit rectangles and 3 partitions
... i don't know if i need to make n = 3 as the problem says to use 3 partitions and i don't know how to solve a sum with a sine function
Generally you would obtain for your integral a Riemann-sum as follows:
$$ \int_{\frac{\pi}{12}}^\frac{\pi}{6}\frac{\sin (3x)}{2}\,\mathrm dx = \lim_{n\to\infty}\sum_{k=1}^n\frac{\pi}{24n}\sin\left(\frac{\pi}{4}+\frac{k\pi}{4 n}\right) $$ Now you're asked explicitly to compute the sum whenever $n=3$:
$$ \sum_{k=1}^3\frac{\pi}{72}\sin\left(\frac{\pi}{4}+\frac{k\pi}{12}\right) $$
Note that $$\begin{aligned}\sin\left(\frac{\pi}{4}+\frac{\pi}{12}\right)&=\frac{\sqrt{3}}{2}\\ \sin\left(\frac{\pi}{4}+\frac{\pi}{6}\right)&=\frac{1+\sqrt{3}}{2\sqrt 2}\\ \sin\left(\frac{\pi}{4}+\frac{\pi}{4}\right)&=1\end{aligned}$$ Now you have all the ingredients to compute the sum.