5.3 The Fundamental Theorem of Calculus/41

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

$=\int _{0}^{\frac {\pi }{2}}f(x)dx+\int _{\frac {\pi }{2}}^{\pi }f(x)dx=\int _{0}^{\frac {\pi }{2}}\sin(x)dx+\int _{\frac {\pi }{2}}^{\pi }\cos(x)dx$

$=-\cos(x){\bigg |}_{0}^{\frac {\pi }{2}}+\sin(x){\bigg |}_{\frac {\pi }{2}}^{\pi }=\left[-\cos({\frac {\pi }{2}})+\cos(0)\right]+\left[\sin(\pi )-\sin({\frac {\pi }{2}})\right]$

$=0+1+0-1=0$