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Let the area of the region enclosed by the curve \(\mathrm{y}=\min \{\sin \mathrm{x}, \cos \mathrm{x}\}\) and the \(\mathrm{x}\)-axis between \(\mathrm{x}=-\pi\) to \(\mathrm{x}=\pi\) be \(\mathrm{A}\). Then \(\mathrm{A}^{2}\) is equal to ______.

1 Answer

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Best answer

Correct answer is : 16 

\(\mathrm{y}=\min \{\sin \mathrm{x}, \cos \mathrm{x}\}\)

\(\mathrm{x}\) - axis 

\(\mathrm{x}-\pi \) 

\( \mathrm{x}=\pi\)

Let the area of the region

\(\int_{0}^{\pi / 4} \sin \mathrm{x}=(\cos \mathrm{x})_{\pi / 4}^{0}=1-\frac{1}{\sqrt{2}}\)

\(\int_{-\pi}^{-3 \pi / 4}(\sin \mathrm{x}-\cos \mathrm{x})=(-\cos \mathrm{x}-\sin \mathrm{x})_{-\pi}^{-3 \pi / 4}\)

\(=(\cos \mathrm{x}+\sin \mathrm{x})_{-3 \pi / 4}^{-\pi}\)

\(=(-1+0)-\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)\)

\(=-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\)

\(\int_{\pi / 4}^{\pi / 2} \cos x d x=(\sin x)_{\pi / 4}^{\pi / 2}=1-\frac{1}{\sqrt{2}}\)

\(\mathrm{A}=4\)

\(\mathrm{A}^{2}=16\)

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