Definite Integration Question 414
Question: If for a real number $ y,
[y] $ is the greatest integer less than or equal to $ y, $ then the value of the integral $ \int\limits _{\pi /2}^{3\pi /2}{[2\sin x]dx} $ is [IIT 1999]
Options:
A) $ -\pi $
B) 0
C) $ -\frac{\pi }{2} $
D) $ \frac{\pi }{2} $
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Answer:
Correct Answer: C
Solution:
We know $ -1\le \sin x\le 1\Rightarrow -2\le 2\sin x\le 2 $
$ I=\int _{\frac{\pi }{2}}^{\frac{3\pi }{2}}{[2\sin x]dx} $
$ =\int _{\frac{\pi }{2}}^{\frac{5\pi }{6}}{[2\sin x]dx+\int _{\frac{5\pi }{6}}^{\pi }{[2\sin x]dx}} $
$ +\int _{\pi }^{\frac{7\pi }{6}}{[2\sin x]dx+\int _{\frac{7\pi }{6}}^{\frac{3\pi }{2}}{[2\sin x]dx}} $
$ =\int _{\frac{\pi }{2}}^{\frac{5\pi }{6}}{(1)dx+\int _{\frac{5\pi }{6}}^{\pi }{(0)dx+\int _{\pi }^{\frac{7\pi }{6}}{(-1)dx+\int _{\frac{7\pi }{6}}^{\frac{3\pi }{2}}{(-2)}}dx}} $
$ =( \frac{5\pi }{6}-\frac{\pi }{2} )+0-( \frac{7\pi }{6}-\pi )-2( \frac{3\pi }{2}-\frac{7\pi }{6} ) $
$ =\frac{2\pi }{6}-\frac{\pi }{6}-\frac{4\pi }{6}=-\frac{\pi }{2} $ .
BETA
