Definite Integration Question 471
Question: $ \int _{0}^{\pi /2}{{x-[\sin x]}dx} $ is equal to
[AMU 1999]
Options:
A) $ \frac{{{\pi }^{2}}}{8} $
B) $ \frac{{{\pi }^{2}}}{8}-1 $
C) $ \frac{{{\pi }^{2}}}{8}-2 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_0^{\pi /2}{{x-[\sin x]}dx=\int_0^{\pi /2}{xdx-\int _{0}^{\pi /2}{[\sin x]dx}}} $
$ =( \frac{x^{2}}{2} )_0^{\pi /2} $
$ =\frac{{{\pi }^{2}}}{8} $ , $ [\because \int _{0}^{\pi /2}{[\sin x]dx=0}] $ .
BETA
