Question

Area under the curve y = sin 2x + cos 2x between x=0 and $x=\frac{\pi }{4}$is

• A. 2 sq. unit
• B. 1 sq. unit
• C. 3 sq. unit
• D. 4 sq. unit

Answer 1

The correct option is B

1 sq. unit

Required area $={\int }_0^{\frac{\pi }{4}}(sin2x+cos2x)dx$
$={[-\frac{cos2x}{2}+\frac{sin2x}{2}]}_0^{\frac{\pi }{4}}$
$=\frac{1}{2}[-cos\frac{\pi }{2}+sin\frac{\pi }{2}+cos0-sin0]=1sq.unit$