Definite Integration Question 30
Question: The area formed by triangular shaped region bounded by the curves $ y=\sin x,y=\cos x $ and $ x=0 $ is
[MP PET 2000]
Options:
A) $ \sqrt{2}-1 $
B) 1
C) $ \sqrt{2} $
D) $ 1+\sqrt{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
Given required area has been shown in the figure. $ x=\frac{\pi }{4} $ is the point of intersection of both curve
$ \therefore $ Required area = $ \int_0^{\pi /4}{(\cos x-\sin x)dx} $
$ =[\sin x+\cos x]_0^{\pi /4} $
$ =[ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1 ] $
= $ \frac{2}{\sqrt{2}}-1=\sqrt{2}-1 $ .
BETA
