Let the area of the region enclosed by the curve y = min { sin x , cos x } and the x -axis between be A . Then A 2 is equal to [2024]

Question

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 ______ . [2024]

Smart Achievers · Accepted Answer

(16)

Required area, $A = \int_{- π}^{\frac{- 3 π}{4}} - \cos x d x + \int_{\frac{- 3 π}{4}}^{0} - \sin x d x + \int_{0}^{\frac{π}{4}} \sin x d x + \int_{\frac{π}{4}}^{\frac{π}{2}} \cos x d x + \int_{\frac{π}{2}}^{π} - \cos x d x$

$= {[- \sin x]}_{- π}^{- \frac{3 π}{4}} + {[\cos x]}_{- \frac{3 π}{4}}^{0} - {[\cos x]}_{0}^{\frac{π}{4}} + {[\sin x]}_{\frac{π}{4}}^{\frac{π}{2}} - {[\sin x]}_{\frac{π}{2}}^{π}$

$= \frac{1}{\sqrt{2}} + 1 + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + 1 + 1 - \frac{1}{\sqrt{2}} + 1 = 4$

$∴$ $A^{2} = 16$

Solution

Q.
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 ______ . [2024]

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Solution

Q. Let the area of the region enclosed by the curve y=min{sinx,cosx} and the x-axis between x=-π to x=π be A. Then A2 is equal to ______ . [2024]

Ans. (16) Required area, A=∫-π-3π4-cosxdx+∫-3π40-sinxdx+∫0π4sinxdx+∫π4π2cosxdx+∫π2π-cosxdx =[-sinx]-π-3π4+[cosx]-3π40-[cosx]0π4+[sinx]π4π2-[sinx]π2π =12+1+12-12+1+1-12+1=4 ∴ A2=16

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Q.
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 ______ . [2024]