Application of Integrals: Find the area of the region R=\(x,y):xyle8,\ 1le yle x^2,\ xge0\ is : | JEE Main 2026

JEE Main 2026 (Online) 28th January Morning ShiftQuestion 1

Find the area of the region $R=\{(x,y):xy\le8,\ 1\le y\le x^2,\ x\ge0\}$ is :

Application of Integrals

$\frac23\left(20\log_e(2)+9\right)$$\frac13\left(40\log_e(2)+27\right)$$\frac23\left(24\log_e(2)-7\right)$✓ Correct answer$\frac13\left(49\log_e(2)-15\right)$

Solution

$y\ge1$ forces $x\ge1$; curves $y=x^2$ and $y=8/x$ meet at $x=2$. Area $=\int_1^2(x^2-1)dx+\int_2^8\left(\frac8x-1\right)dx=\frac43+16\ln2-6=\frac23(24\ln2-7)$.

Application of Integrals - JEE Mathematics question (JEE Main 2026 (Online) 28th January Morning Shift)

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