A) \[x={{y}^{2}}\]
B) 1
C) \[\sqrt{2}\]
D) \[1+\sqrt{2}\]
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 = \[\int_{0}^{\pi /4}{(\cos x-\sin x)\,dx}\] \[=[\sin x+\cos x]_{0}^{\pi /4}\]\[=\left[ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1 \right]\] =\[\frac{2}{\sqrt{2}}-1=\sqrt{2}-1\].You need to login to perform this action.
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