Exam

JEE Main

Year

2026

Date

23 January

Shift

Morning Shift

Subject

Mathematics

Chapter

Area Under Curves

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Area Under Curves

Let the area of the region bounded by the curve y= \ x, x\ , lines...

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 $\_\_\_\_$.

Step-by-step solution

$A = \int_{0}^{\pi/4} \cos x \, dx + \int_{\pi/4}^{\pi} \sin x \, dx + \int_{\pi}^{5\pi/4} -\sin x \, dx + \int_{5\pi/4}^{3\pi/2} -\cos x \, dx$
$A = (\sin x)_{0}^{\pi/4} + (-\cos x)_{\pi/4}^{\pi} + (\cos x)_{\pi}^{5\pi/4} + (\sin x)_{5\pi/4}^{3\pi/2}$
$A = \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} + 1 + 1 - \dfrac{1}{\sqrt{2}} + 1 - \dfrac{1}{\sqrt{2}} = 3$
$A^2 + A = 12$