Definite Integration Question 486
Question: The area between the curve $ y={{\sin }^{2}}x, $ $ x- $ axis and the ordinates $ x=0 $ and $ x=\frac{\pi }{2} $ is
[RPET 1996]
Options:
A) $ \frac{\pi }{2} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{8} $
D) $ \pi $
Show Answer
Answer:
Correct Answer: B
Solution:
Required area $ A=\int_0^{\pi /2}{{{\sin }^{2}}x.dx}=\int_0^{\pi /2}{( \frac{1-\cos 2x}{2} )}dxa $
$ =\frac{1}{2}[x]_0^{\pi /2}-\frac{1}{4}[\sin 2x]_0^{\pi /2}=\frac{\pi }{4} $ .
BETA
