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Mathematics · Area under the Curves

JEE Main 2026 — 28 January, Morning Shift — Question 12

The area of the region R={(x,y):xy≤8,1≤y≤x2,x≥0}R = \{(x,y) : xy \leq 8, 1 \leq y \leq x^2, x \geq 0\} is

  1. Option A:

    13(49log⁡e2−15)\dfrac{1}{3} (49 \log_e 2 - 15)

  2. Option B:

    23(20log⁡e2+9)\dfrac{2}{3} (20 \log_e 2 + 9)

  3. Option C:

    23(24log⁡e2−7)\dfrac{2}{3} (24 \log_e 2 - 7)

    Correct
  4. Option D:

    13(40log⁡e2+27)\dfrac{1}{3} (40 \log_e 2 + 27)

Answer: C

Step-by-step solution

A=∫12(x2−1) dx+∫28(8x−1) dxA = \int_{1}^{2} (x^2 - 1) \, dx + \int_{2}^{8} \left( \frac{8}{x} - 1 \right) \, dx A=8log⁡4−143=16log⁡2−143A = 8 \log 4 - \frac{14}{3} = 16 \log 2 - \frac{14}{3} =23(24log⁡2−7)= \frac{2}{3} (24 \log 2 - 7)
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Morning Shift
Subject
Mathematics
Chapter
Area under the Curves
Topic
Area under the Curves
The area of the region R = \ (x,y) : xy leq 8, 1 leq y leq x 2, x geq… | JEE Main 2026 PYQ with Solution · DhiX AI