Definite Integration Question 57
Question: Area bounded by the curve $ y=\sin x $ between $ x=0 $ and $ x=2\pi $ is
Options:
A) 2 sq. unit
B) 4 sq. unit
C) 8 sq. unit
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ y=\sin x $ $ x $ 0 $ \pi /6 $
$ \pi /2 $
$ \pi $
$ 3\pi /2 $
$ 2\pi $ $ y $ 0 0.5 1 0 -1 0 Join these points with a free hand to obtain a rough sketch Required area = (area of $ OAB $ ) + (area of $ BCD) $
= $ \int_0^{\pi }{ydx+\int _{\pi }^{2\pi }{(-y)dx}} $ , ( $ \because $ Area $ BCD $ is below $ x- $ axis) = $ \int_0^{\pi }{\sin xdx-\int _{\pi }^{2\pi }{\sin xdx=4}} $ sq. unit.
BETA
