A) 2 sq. unit
B) 1 sq. unit
C) 3 sq. unit
D) 4 sq. unit
Correct Answer: B
Solution :
Required area \[=\int_{0}^{\pi /4}{(\sin 2x+\cos 2x)dx}\] \[=\left[ -\frac{\cos 2x}{2}+\frac{\sin 2x}{2} \right]_{0}^{\pi /4}\] \[=\frac{1}{2}\left[ -\cos \frac{\pi }{2}+\sin \frac{\pi }{2}+\cos 0-\sin 0 \right]=1\,sq.\]unit.
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