JEE MAIN - Mathematics (2024 - 8th April Morning Shift - No. 29)
Let the area of the region enclosed by the curve $$y=\min \{\sin x, \cos x\}$$ and the $$x$$ axis between $$x=-\pi$$ to $$x=\pi$$ be $$A$$. Then $$A^2$$ is equal to __________.
Answer
16
Explanation
$$y=f(x)=\min \{\sin x, \cos x\}$$
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$$\begin{aligned} & A=-\int_\limits{-\pi}^{\frac{-3 \pi}{4}} \cos x d x-\int_\limits{\frac{-3 \pi}{4}}^0 \sin x d x+\int_\limits0^{\frac{\pi}{4}} \sin x d x+\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x d x \\ & -\int_\limits{\frac{\pi}{2}}^\pi \cos x d x \\ & A=4 \\ & A^2=16 \end{aligned}$$
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