A) \[\sqrt{2}\]
B) \[\sqrt{2}+1\]
C) \[\sqrt{2}-1\]
D) \[\sqrt{2}(\sqrt{2}-1)\]
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 \[=\int_{{{x}_{1}}}^{{{x}_{2}}}{ydx=\int_{0}^{\pi /4}{\cos xdx-\int_{0}^{\pi /4}{\,\,\,\,\,\,\sin x\,dx}}}\] \[=[\sin x]_{0}^{\pi /4}-[-\cos x]_{0}^{\pi /4}\] \[=\left( \sin \frac{\pi }{4}-\sin 0 \right)+\left( \cos \frac{\pi }{4}-\cos 0 \right)=\left( \frac{1}{\sqrt{2}}-0 \right)+\left( \frac{1}{\sqrt{2}}-1 \right)\] \[=\sqrt{2}-1\].
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