Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 328: 45

$$2$$

Given $$y=\sin (x), \ \ 0\leqslant x \leqslant \pi$$ From the graph of the region, we can observe that area of the bounded region approximately equal to $\frac{1}{2}$ area of the rectangle with width 1 and length $\pi$ $$\text{Area} \approx \frac{1}{2} (1)(\pi)=\frac{\pi}{2} $$ Now we use integration \begin{aligned} \text{Area}&= \int_0^{\pi}\sin{x}dx\\ &= -\cos(x)\bigg|_{0}^{\pi} \\ &=-[-1-1]=2\end{aligned}