Answer
$A=\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{(2\pi+\pi \sin^2\frac{\pi i}{n})}{n}$
Work Step by Step
Using the Definition of 2,
$A=\lim\limits_{n \to \infty}\sum_{i=1}^nf(x_i)\Delta x$ where $\Delta x=\frac{\text{upper bound - lower bound}}{n}$
Find $\Delta x$ and simplify:
$\Delta x=\frac{\pi-0}{n}=\frac{\pi}{n}$
Then,
$A=\lim\limits_{n \to \infty}\sum_{i=1}^n(2+\sin^2\frac{\pi i}{n})\cdot \frac{\pi}{n}$
$A=\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{(2\pi+\pi \sin^2\frac{\pi i}{n})}{n}$
