Definite Integration Question 159
What is the area of the portion of the curve $ y=\sin x $ , lying between $ x=0 $ and $ x=2\pi $ ?
Options:
A) 1 square unit
B) 2 square units
C) 4 square units
D) 8 square units
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Required area $ =\int\limits_0^{2\pi }{|\sin x|dx} $
$ =-\cos . x |_0^{2\pi }=-\cos 2\pi -(-\cos 0) $
$ =-\cos (\pi +\pi )+1=-[-\cos \pi ]+1 $
$ =+\cos ( \frac{\pi }{2}+\frac{\pi }{2} )+1 $
$ =\sin \frac{\pi }{2}+1=1+1=2 $ sq. units.
BETA
