Let the area of the region enclosed by the curve y= \ x, x\ and the x axis between x=-π to x=π be A . Then A^2 is equal to ___________

Correct answer: 0 aligned & y= \ x, x\ & x- axis x-π x=π aligned aligned & ∫_0^ π / 4 x=( x)_ π / 4 ^0=1- 1 2 & ∫_ -π ^ -3 π / 4 ( x- x)=(- x- x)_ -π ^ -3 π / 4 & =( x+ x)_ -3 π / 4 ^ -π & =(-1+0)-(- 1 2 - 1 2 ) & =-1+ 1 2 + 1 2 & ∫_ π / 4 ^ π / 2 x d x=( x)_ π / 4 ^ π / 2 =1- 1 2 & A=4 & A^2=16 aligned

JEE Main 20248 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Area Under Curves

Let the area of the region enclosed by the curve y= \ x, x\ and the...

Let the area of the region enclosed by the curve

y=\min \{\sin x, \cos x\}

and the

x

axis between

x=-\pi

x=\pi

A

. Then

A^2

is equal to ___________

Correct answer

Step-by-step solution

\begin{aligned} & y=\min \{\sin x, \cos x\} \\ & x-\text{ axis } \quad x-\pi \quad x=\pi\end{aligned}

\begin{aligned} & \int_0^{\pi / 4} \sin x=(\cos x)_{\pi / 4}^0=1-\frac{1}{\sqrt{2}} \\ & \int_{-\pi}^{-3 \pi / 4}(\sin x-\cos x)=(-\cos x-\sin x)_{-\pi}^{-3 \pi / 4} \\ & =(\cos x+\sin 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}} \\ & A=4 \\ & A^2=16\end{aligned}

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