What is the area of the portion of the curve $$y = \sin \,x,$$   lying between $$x = 0,\,y = 0$$    and $$x = 2\pi \,?$$

Solution :
Required area
$$\eqalign{ & = \int\limits_0^{2\pi } {\sin \,x\,dx} \cr & = \left. { - \cos \,x} \right|_0^{2\pi } \cr & = - \cos \,2\pi - \left( { - \cos \,0} \right) \cr & = - \cos \left( {\pi + \pi } \right) + 1 \cr & = - \left[ { - \cos \,\pi } \right] + 1 \cr & = + \cos \left( {\frac{\pi }{2} + \frac{\pi }{2}} \right) + 1 \cr & = \sin \frac{\pi }{2} + 1 \cr & = 1 + 1 \cr & = 2{\text{ sq}}{\text{. units}} \cr} $$