Application of IntegralsmediumPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Area of Region xy≤27, 1≤y≤x²: 54ln3 - 52/3 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
The area of the region R={(x,y):xy≤27, 1≤y≤x2}R=\{(x,y):xy\le27,\ 1\le y\le x^2\} is
A78log⁡e3−52378\log_e 3-\dfrac{52}{3}
B54log⁡e3−52354\log_e 3-\dfrac{52}{3}correct
C54log⁡e3−26354\log_e 3-\dfrac{26}{3}
D54log⁡e3+26354\log_e 3+\dfrac{26}{3}
Solution
Step 1: x2=27x⇒x3=27⇒x=3x^2=\dfrac{27}{x}\Rightarrow x^3=27\Rightarrow x=3. For 1≤x≤31\le x\le3 top is x2x^2; for 3≤x≤273\le x\le27 top is 27x\dfrac{27}{x}; floor y=1y=1:
Area=∫13(x2−1) dx+∫327(27x−1)dx.\text{Area}=\int_1^3\big(x^2-1\big)\,dx+\int_3^{27}\left(\frac{27}{x}-1\right)dx.
 Step 2: ∫13(x2−1) dx=[x33−x]13=(9−3)−(13−1)=6+23=203.\displaystyle\int_1^3(x^2-1)\,dx=\left[\frac{x^3}{3}-x\right]_1^3=(9-3)-\left(\frac13-1\right)=6+\frac23=\frac{20}{3}. Step 3: ∫327(27x−1)dx=[27ln⁡x−x]327=(27ln⁡27−27)−(27ln⁡3−3)=27ln⁡9−24\displaystyle\int_3^{27}\left(\frac{27}{x}-1\right)dx=\big[27\ln x-x\big]_3^{27}=(27\ln27-27)-(27\ln3-3)=27\ln9-24, using ln⁡27−ln⁡3=ln⁡9\ln27-\ln3=\ln9. Step 4: ln⁡9=2ln⁡3\ln9=2\ln3:
Area=203+54ln⁡3−24=54ln⁡3+20−723=54ln⁡3−523.\text{Area}=\frac{20}{3}+54\ln3-24=54\ln3+\frac{20-72}{3}=54\ln3-\frac{52}{3}.
 Correct answer: (2)
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