Definite Integration Question 459
Question: Area bounded by curve $ y=k\sin x $ between $ x=\pi $ and $ x=2\pi , $ is
Options:
A) $ 2k $ sq. unit
B) 0
C) $ \frac{k^{2}}{2} $ sq. unit
D) $ k $ sq. unit
Show Answer
Answer:
Correct Answer: A
Solution:
Required area $ =k\int _{\pi }^{2\pi }{\sin xdx=k}[-\cos x] _{\pi }^{2\pi }=-2k $
Hence, area = 2k sq. unit.
BETA
