Answer
$\lim\limits_{n \to \infty}\sum_{i=1}^{n}sin(\frac{5\pi~i}{n})\cdot~\frac{\pi}{n} = \frac{2}{5}$
Work Step by Step
We can use the definition of the integral in Theorem 4 to evaluate the integral:
$\int_{a}^{b}f(x)~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i)\Delta x$
$\Delta x = \frac{b-a}{n} = \frac{\pi-0}{n} = \frac{\pi}{n}$
$x_i = \frac{\pi~i}{n}$
$\int_{0}^{\pi}sin~5x~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i)\Delta x$
$= \lim\limits_{n \to \infty}\sum_{i=1}^{n}sin(\frac{5\pi~i}{n})\cdot \frac{\pi}{n} = \frac{2}{5}$
