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