Evaluate ∫016π/3 |sinx|dx
212
0
192
172
∫016π/3 |sinx|dx=∫05π |sinx|dx+∫5π5π+π/3 |sinx|dx =5∫0π |sinx|dx+∫0π/3 |sinx|dx[∵|sinx| is periodic with period π=5∫0π sinxdx+∫0π/3 sinxdx =5[−cosx]0π+[−cosx]0π3 =5×2+−12+1=212