Maths · Class 12 · Integral Calculus

Area Under Curves for JEE

Everything you need for this topic on one page: the ideas, the formulas, the traps, and free lectures that explain it well.

Last updated: 26 October 2026

Direct answer

Area bounded by curves and lines using definite integrals, including areas between two curves.

Area bounded by curves and lines using definite integrals, including areas between two curves.

Key ideas

Area Under Curves: what you must understand

  • Sketch the curves first
  • Area is always positive: integrate |f|
  • Between curves: ∫(upper − lower) dx
  • Sometimes integrating in y is easier
Formula sheet

Must-know formulas for Area Under Curves

  • A = ∫ₐᵇ |f(x)| dx
  • Between curves: A = ∫ₐᵇ (f − g) dx where f ≥ g
  • Area of ellipse = πab
  • Area between y² = 4ax and x² = 4ay is 16a²/3
Print

One-page formula sheet

Printable Area Under Curves formula sheet: formulas, key ideas and traps on one page. Save it as a PDF for offline revision.

Avoid

Common mistakes students make

  • Not finding intersection points
  • Negative area from below the x-axis
  • Choosing the wrong upper curve
For the exam

How this topic is asked in JEE

Appears frequently. Almost every shift has an area question. A quick sketch is the whole trick.

Prerequisites: Definite Integration.

NCERT map

Where Area Under Curves is in NCERT

Chapter names follow the current rationalised NCERT textbooks. Read these sections before solving PYQs.

ClassBookChapterSections JEE asks from
Class 12Mathematics Part IIApplication of IntegralsArea under Simple Curves; Area of a Region Bounded by a Curve and a Line; Area between Two Curves
JEE Advanced level

A JEE Advanced-level example on Area Under Curves

Question. Find the area of the region defined by {(x, y) : x² + y² ≤ 4, x + y ≥ 2}. This is the region inside the circle of radius 2 centered at the origin that lies above the line x + y = 2.

Show solution
  1. Find intersection points of circle x² + y² = 4 and line x + y = 2. Substitute y = 2 - x into circle equation: x² + (2-x)² = 4, which gives 2x² - 4x = 0, so x = 0 or x = 2. Points are (0, 2) and (2, 0).
  2. The chord from (0, 2) to (2, 0) cuts off a circular segment. The required area is this segment (the smaller piece above the line).
  3. Find the angle subtended at center. Point (2, 0) is at angle 0 and point (0, 2) is at angle π/2. So the sector angle θ = π/2.
  4. Area of circular sector = (1/2)r²θ = (1/2)(4)(π/2) = π.
  5. Area of triangle with vertices at origin, (0, 2), and (2, 0) = (1/2) × base × height = (1/2)(2)(2) = 2.
  6. Required area = Area of sector - Area of triangle = π - 2.

Answer: π - 2 square units

This example was solved twice independently and both solutions agree.

Free lectures

Best free lectures for Area Under Curves

Full-chapter free lectures picked from YouTube search results by views, from established JEE channels (mostly in Hinglish). Use the subtopic list to jump straight to the part you need (timestamps come from each video's own chapter list). Watch one, then solve PYQs the same day.

Practise

Where to practise for free

FAQs

Frequently asked questions

Is Area Under Curves important for JEE Main 2027?

Appears frequently. It is part of the official 'Integral Calculus' unit in the JEE Main 2027 syllabus.

What are the most important formulas in Area Under Curves?

A = ∫ₐᵇ |f(x)| dx; Between curves: A = ∫ₐᵇ (f − g) dx where f ≥ g; Area of ellipse = πab.

How should I study Area Under Curves for JEE?

Understand the key ideas from NCERT or one full lecture, learn the formula sheet, then solve 25–30 past JEE questions on the topic. Almost every shift has an area question. A quick sketch is the whole trick.

WA✎
Sitemap