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JEE Mains Maths · Application of Integrals

Unknown Parameters and Area Ratios

The area is given, compared or optimised, and the question asks for a constant in a curve: find the area in terms of the constant, then solve.

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Why this matters

Twelve PYQs, five of them multiple choice, and three from 2026. Eight give an area, or set two areas equal, and ask for a constant in a curve; four divide an area in a ratio, or choose a constant that makes an area largest or smallest. Two ideas cover the page.

Concept 1 of 2: Area in terms of the constant, then solve

Keep the constant as a letter. Find the meeting points in terms of it, integrate, and set the result equal to the given area. Parabola-and-line areas have closed forms, so most of these reduce to one equation such as m3=216m^3=216.

Definition

  • y=ax2y=ax^2 and y=mxy=mx meet at x=max=\frac ma; the area between them is m36a2\frac{m^3}{6a^2}.
  • y2=4axy^2=4ax and y=mxy=mx: area 8a23m3\frac{8a^2}{3m^3}.
  • y=ax2y=ax^2 and y=cy=c: area 43cca\frac43c\sqrt{\frac ca}.

Parabola and a line through its vertex

y=ax2, y=mx:A=m36a2y=ax^2,\ y=mx:\quad A=\frac{m^3}{6a^2}

Worked example

The area between y=x2y=x^2 and y=mxy=mx (m>0m>0) is 36. Find m.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q61Moderate

Example 1 · Application of Integrals · Unknown Parameters and Area Ratios

Let P1:y=4x2P_{1}:y= 4x^{2} and P2:y=x2+27P_{2}:y=x^{2}+ 27 be two parabolas. If the area of the bounded region enclosed between P1P_{1} and P2P_{2} is six times the area of the bounded region enclosed between the line y=αx,α>0y =\alpha x,\alpha> 0 and P1P_{1}, then α\alpha is equal to :

Keep the sign condition

A square root gives two values of the constant; the condition in the question (a>0a>0, α>0\alpha>0) keeps one. Check the kept value also makes the curves meet as the region needs.
Drill 7 more on area in terms of the constant, then solve

Concept 2 of 2: Dividing an area, and making it largest or smallest

For a line that bisects an area or divides it in a ratio, find the piece on one side as a function of the constant and set it equal to the right fraction of the total. For a largest or smallest area, write the area as a function of the constant and minimise or maximise it — often by completing the square, without calculus.

Definition

  • Bisect: the piece on one side is half the total.
  • Ratio m:nm:n: the first piece is mm+n\frac m{m+n} of the total.
  • Largest or smallest: set dAdk=0\frac{dA}{dk}=0, or complete the square, and check the endpoints.

A piece in a given ratio

A1=mm+n AtotalA_1=\frac{m}{m+n}\,A_{\text{total}}

Worked example

The line y=cy=c bisects the area between y=x2y=x^2 and y=4y=4. Find c.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q53Moderate

Example 2 · Application of Integrals · Unknown Parameters and Area Ratios

Let the line x=−1x = - 1 divide the area of the region {(x,y):1+x2≤y≤3−x}\left\{ (x,y):1 +x^{2}\leq y \leq 3 - x \right\} in the ratio m:n,gcd(m,n)=1m:n,gcd(m,n) = 1. Then m+nm + n is equal to

Which piece comes first

A ratio m:nm:n depends on which piece the question names first. Compute both pieces and match the order in the question before reading off m and n.
Drill 3 more on dividing an area, and making it largest or smallest

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Watch out for (2)

Test yourself on Application of Integrals

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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