Definite Integration Question 442
Question: The area enclosed by $ y=x^{2}+\cos x $ and its normal at $ x=\frac{\pi }{2} $ in the first quadrant is
Options:
A) $ 4(\sqrt{2}-1) $
B) $ 2\sqrt{2}(\sqrt{2}-1) $
C) $ 2(\sqrt{2}+1) $
D) $ 2\sqrt{2}(\sqrt{2}+1) $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Since $ \sin x $ and $ \cos x>0 $ for $ x\in [0,\pi /2] $ , the graph of $ y=\sin x+\cos x $ always lies above the graph of $ y=| \cos x-\sin x | $ Also $ \cos x>\sin x $ for $ x\in [0,\pi /4] $ and $ \sin x>\cos x $ for $ x\in [\pi /4,\pi /2] $
$ \Rightarrow $ Area $ =\int\limits_0^{\pi /4}{((sinx+cosx)-(cosx-sinx))dx} $
$ +\int\limits _{\pi /4}^{\pi /2}{((sinx+cosx)-(sinx-cosx))dx} $
$ =4-2\sqrt{2} $
BETA
