5.3 The Fundamental Theorem of Calculus/41

$\int _{0}^{\pi }f(x)\,dx\quad {\text{where}}\;f(x)={\begin{cases}\sin(x)&0\leq x<{\frac {\pi }{2}}\\\cos(x)&{\frac {\pi }{2}}\leq x\leq \pi \end{cases}}$

${\begin{aligned}\int _{0}^{\pi }f(x)\,dx&=\int _{0}^{\frac {\pi }{2}}f(x)\,dx+\int _{\frac {\pi }{2}}^{\pi }f(x)\,dx=\int _{0}^{\frac {\pi }{2}}\sin(x)\,dx+\int _{\frac {\pi }{2}}^{\pi }\cos(x)\,dx\\[2ex]&=-\cos(x){\bigg |}_{0}^{\frac {\pi }{2}}+\sin(x){\bigg |}_{\frac {\pi }{2}}^{\pi }\\[2ex]&=\left[-\cos \left({\frac {\pi }{2}}\right)+\cos(0)\right]+\left[\sin(\pi )-\sin \left({\frac {\pi }{2}}\right)\right]\\[2ex]&=[0+1]+[0-1]\\[2ex]&=0\end{aligned}}$