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

Mathematics · Area under the Curves

JEE Main 2024 — 8 April, Shift 1 — Question 29

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

Answer: 16

Numerical answer — enter this value.

Step-by-step solution

$y=\min \{\sin x, \cos x\}$ x -axis $\quad \mathrm{x}-\pi \quad \mathrm{x}=\pi$

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

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

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

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

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

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

$A=4$

$A^{2}=16$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Mathematics
Chapter: Area under the Curves
Topic: Area under the Curves