Discussion Forum

The area (in square units) of the region bounded by the curve y=|sin 2x|  and X-axis  in [0, $2 \pi$] is 

Answer:

Explanation:

 Required area   $A=\int_{0}^{2 \pi} | \sin 2x|dx$

 $=4\int_{0}^{ \pi /2}  \sin 2xdx=4\left[ -\frac{\cos 2x}{2}\right]^{\pi/2}_{0}$

  = 2[-(-1-1)]=4

 Hence, option (d) is correct