Definite Integration Question 279
Question: Area under the curve $ y=\sin 2x+\cos 2x $ between $ x=0 $ and $ x=\frac{\pi }{4}, $ is
[AI CBSE 1979]
Options:
A) 2 sq. unit
B) 1 sq. unit
C) 3 sq. unit
D) 4 sq. unit
Show Answer
Answer:
Correct Answer: B
Solution:
Required area $ =\int_0^{\pi /4}{(\sin 2x+\cos 2x)dx} $
$ =[ -\frac{\cos 2x}{2}+\frac{\sin 2x}{2} ]_0^{\pi /4} $
$ =\frac{1}{2}[ -\cos \frac{\pi }{2}+\sin \frac{\pi }{2}+\cos 0-\sin 0 ]=1sq. $ unit.
BETA
