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Maths Question 16 – JEE-MAIN 2026

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The area of the region R={(x,y)|xy≤27,1≤y≤x2} is equal to:

Sketch the given inequalities to visualize the region whose area needs to be calculated.

● Video Walkthrough
Step 1: Define the Region and Find Intersection Points✦ Active

The region R is defined by 1≤y≤x2 and y≤27x. This means 1≤y≤min(x2,27x). We need to find the intersection points of the boundary curves:

y=x2 and y=1⟹x2=1⟹x=1(since x>0)
y=27x and y=1⟹27x=1⟹x=27
y=x2 and y=27x⟹x2=27x⟹x3=27⟹x=3

Comparing x2 and 27x: for 1≤x≤3, x2≤27x, so y=x2 is the upper bound. For 3≤x≤27, x2≥27x, so y=27x is the upper bound. The lower bound is y=1 throughout.

Step 2: Set up the Definite Integral for Area○ Expand

The total area A can be calculated by splitting the integral into two parts based on the upper boundary function:

A=∫13(x2−1)dx+∫327(27x−1)dx
💡 Teacher's Secret Hint

Ensure the correct upper and lower bounds are used for each interval.

Step 3: Evaluate the Integrals○ Expand

Evaluate the first integral:

∫13(x2−1)dx=[x33−x]13=(333−3)−(133−1)=(9−3)−(13−1)=6−(−23)=6+23=203

Evaluate the second integral:

∫327(27x−1)dx=[27ln⁡|x|−x]327=(27ln⁡27−27)−(27ln⁡3−3)

Using ln⁡27=ln⁡(33)=3ln⁡3:

=(27⋅3ln⁡3−27)−(27ln⁡3−3)=81ln⁡3−27−27ln⁡3+3=54ln⁡3−24

Add the results of both integrals to find the total area:

A=203+54ln⁡3−24=54ln⁡3+203−723=54ln⁡3−523
💡 Teacher's Secret Hint

Remember that ln⁡x is the natural logarithm, often written as loge⁡x.

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