Answer
$$\int_{0}^{\pi} f(x) d x =0$$
Work Step by Step
Given
$$\int_{0}^{\pi} f(x) d x \quad \text { where } f(x)= \begin{cases}\sin x & \text { if } 0 \leqslant x\lt\pi / 2 \\ \cos x & \text { if } \pi / 2 \leqslant x \leqslant \pi\end{cases}$$
Since
\begin{aligned}
\int_{0}^{\pi} f(x) d x &=\int_{0}^{\pi / 2} \sin x d x+\int_{\pi / 2}^{\pi} \cos x d x\\
&=(-\cos x)\bigg|_{0}^{\pi / 2}+(\sin x)\bigg|_{\pi / 2}^{\pi}\\
&=-\cos( \frac{\pi}{2})+\cos( 0)+\sin (\pi)-\sin( \frac{\pi}{2}) \\
&=-0+1+0-1\\
&=0
\end{aligned}
