Definite Integration Question 296
Question: The area bounded by curves $ y=\cos x $ and $ y=\sin x $ and ordinates $ x=0 $ and $ x=\frac{\pi }{4} $ is
[Karnataka CET 2002]
Options:
A) $ \sqrt{2} $
B) $ \sqrt{2}+1 $
C) $ \sqrt{2}-1 $
D) $ \sqrt{2}(\sqrt{2}-1) $
Show Answer
Answer:
Correct Answer: C
Solution:
Given equations of curves $ y=\cos x $ and $ y=\sin x $ and ordinates $ x=0 $ to $ x=\frac{\pi }{4}. $ We know that area bounded by the curves $ =\int_x_1^{x_2}{ydx=\int_0^{\pi /4}{\cos xdx-\int_0^{\pi /4}{\sin xdx}}} $
$ =[\sin x]_0^{\pi /4}-[-\cos x]_0^{\pi /4} $
$ =( \sin \frac{\pi }{4}-\sin 0 )+( \cos \frac{\pi }{4}-\cos 0 )=( \frac{1}{\sqrt{2}}-0 )+( \frac{1}{\sqrt{2}}-1 ) $
$ =\sqrt{2}-1 $ .
BETA
