QJEEM24Mac0de: JEE Main 2024 Mathematics - Class 12 JEEM24Mac0de Solution: Let the area of the region enclosed by the curve y = min { sin…

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QJEEM24Mac0de: JEE Main 2024 Mathematics - Class 12 JEEM24Mac0de Solution: Let the area of the region enclosed by the curve y = min { sin…

Let the area of the region enclosed by the curve $y = min {\sin x, \cos x}$ and the $x$ axis between $x = - π$ to $x = π$ be $A$ . Then $A^{2}$ is equal to __________.

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Accepted Answer

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

JEE Main 2024 (Online) 8th April Morning Shift Mathematics - Area Under The Curves Question 15 English Explanation

$\begin{aligned} A = - \int_{- π}^{\frac{- 3 π}{4}} \cos x d x - \int_{\frac{- 3 π}{4}}^{0} \sin x d x + \int_{0}^{\frac{π}{4}} \sin x d x + \int_{\frac{π}{4}}^{\frac{π}{2}} \cos x d x \\ - \int_{\frac{π}{2}}^{π} \cos x d x \\ A = 4 \\ A^{2} = 16 \end{aligned}$

QJEEM24Mac0de: JEE Main 2024 Mathematics - Class 12 JEEM24Mac0de Solution: Let the area of the region enclosed by the curve y = min { sin…

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