Question

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

• A. $2sq.$ units
• B. $1sq.$ units
• C. $3sq.$ units
• D. $4sq.$ units

Answer 1

Answer: (B)

$1sq.$ units

$y=\sin 2x+\cos 2x$

$=\sqrt{2}\sin (2x+\frac{\pi }{4})$

Area under the curve between

$x=0,x=\frac{\pi }{4}$ is,

${\int }_0^{\pi /4}\sqrt{2}\sin (2x+\frac{\pi }{4})dx$

$\sqrt{2}{[-\frac{\cos (2x+\frac{\pi }{4})}{2}]}_o^{\frac{\pi }{4}}$

$\sqrt{2}[-\frac{\left(\frac{-1}{\sqrt{2}}\right)}{2}-\left(\frac{-1/\sqrt{2}}{2}\right)]$

$\sqrt{2}\times [\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}]$

$1$ sq. unit.