Let the area of the region bounded by the curve y=max \ sin x, cos x\ , lines x=0, x=3 π/2 , and the x -axis be A . Then, A + A 2 is equal to

Mathematics · Area under the Curves

JEE Main 2026 — 23 January, Morning Shift — Question 22

Let the area of the region bounded by the curve $y=\max \{\sin x, \cos x\}$ , lines $x=0, x=\frac{3 \pi}{2}$ , and the x -axis be A . Then, $\mathrm{A}+\mathrm{A}^{2}$ is equal to

Answer: 12

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{A}=\int_{0}^{\pi / 4} \cos \mathrm{xdx}+\int_{\pi / 4}^{\pi} \sin \mathrm{xdx}+\int_{\pi}^{5 \pi / 4}-\sin \mathrm{xdx}+\int_{5 \pi / 4}^{3 \pi / 2}-\cos \mathrm{xdx}$

$\mathrm{A}=(\sin \mathrm{x})_{0}^{\pi / 4}+(\cos \mathrm{x})_{\pi}^{\pi / 4}+(\cos \mathrm{x})_{\pi}^{5 \pi / 4}+(\sin \mathrm{x})_{3 \pi / 2}^{5 \pi / 4}$

$\mathrm{A}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+1+1-\frac{1}{\sqrt{2}}+1-\frac{1}{\sqrt{2}}=3$

$\mathrm{A}^{2}+\mathrm{A}=12$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Mathematics
Chapter: Area under the Curves
Topic: Area under the Curves