Definite Integration Question 93
Question: For $ 0\le x\le \pi , $ the area bounded by $ y=x $ and $ y=x+\sin x, $ is
[Roorkee Qualifying 1998]
Options:
A) 2
B) 4
C) $ 2\pi $
D) $ 4\pi $
Show Answer
Answer:
Correct Answer: A
Solution:
The curves $ y=x $ and $ y=x $ intersect at (0, 0) and $ (\pi ,\pi ) $ .
Hence area bounded by the two curves $ =\int\limits_0^{\pi }{(x+\sin x)dx-\int\limits_0^{\pi }{xdx}=\int\limits_0^{\pi }{\sin xdx}} $
$ =[-\cos x]_0^{\pi }=-\cos \pi +\cos 0=-(-1)+(1)=2 $ .
BETA
