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CBSE 13 min read21 October 2026

CBSE Class 12 Application of Integrals - Complete Guide with Area Under Curves & Between Curves

CBSE Class 12 Application of Integrals complete guide. Area under curves, area between two curves, area bounded by lines and parabolas, circles, ellipses — all formulas, solved examples, and board exam strategies for 5 marks.

In short: Application of Integrals (Chapter 8) carries 5 marks in the CBSE Class 12 Maths board exam. It focuses on one powerful application: finding the area under curves and between curves using definite integration. You will work with parabolas, circles, ellipses, and straight lines. This guide covers every formula, every standard curve type, and the exact approach needed to solve the predictable 5-mark question that appears from this chapter every year.

Introduction — A 5-Mark Chapter with Predictable Questions

Application of Integrals belongs to Unit III (Calculus) of the CBSE Class 12 Maths syllabus. The Calculus unit carries 35 marks out of 80, and Chapter 8 contributes approximately 5 marks — almost always as a single 5-mark long answer question. The beautiful thing about this chapter is its narrow scope: you only need to find the area of regions bounded by curves and lines. There are no other applications to learn. The question types are limited to 6-7 standard curve combinations, and the same patterns repeat year after year. If you memorize the standard curve equations and practise the integration, this is one of the most predictable 5-mark questions in the paper. For the full marks breakdown, see the CBSE Class 12 Maths chapter-wise marks weightage.

Core Concepts, Formulas & Memory Tricks

Basic Principle — Area Under a Curve

The area bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b is:

A = ∫ₐᵇ f(x) dx (when f(x) ≥ 0 in [a,b])

If f(x) ≤ 0 in [a,b], the area is:

A = |∫ₐᵇ f(x) dx| = -∫ₐᵇ f(x) dx

For curves expressed as x = g(y): A = ∫_c^d g(y) dy (area between the curve, the y-axis, and y = c and y = d)

Memory trick: Area is always positive. If the integral gives a negative value, take the absolute value. The integral itself can be negative (when the curve is below the x-axis), but area never is.

Area Between Two Curves

If f(x) ≥ g(x) on [a,b], the area between the curves y = f(x) and y = g(x) is:

A = ∫ₐᵇ [f(x) - g(x)] dx

Where f(x) is the upper curve and g(x) is the lower curve.

If the curves cross: Split the integral at the intersection point. In each sub-interval, determine which curve is on top.

Memory trick — "Top minus bottom": Always subtract the lower curve from the upper curve. If you integrate "bottom minus top," you get a negative area and lose marks.

Standard Curves You Must Know

CurveEquationShape
Parabola (opens right)y² = 4axOpens to the right
Parabola (opens up)x² = 4ayOpens upward
Circle (centred at origin)x² + y² = r²Circle of radius r
Ellipsex²/a² + y²/b² = 1Ellipse with semi-axes a and b
Liney = mx + cStraight line

From these curves, express y (or x) for integration:

  • y² = 4ax → y = 2√(ax) (upper half) or y = -2√(ax) (lower half)
  • x² + y² = r² → y = √(r²-x²) (upper half)
  • x²/a² + y²/b² = 1 → y = (b/a)√(a²-x²) (upper half)

Standard Integrals Used in This Chapter

IntegralResult
∫ √(a²-x²) dx(x/2)√(a²-x²) + (a²/2)sin⁻¹(x/a) + C
∫ xⁿ dxxⁿ⁺¹/(n+1) + C
∫ √(ax) dx(2/3)√(a) × x^(3/2) + C

The first integral above is crucial — it appears in every circle and ellipse area problem.

Memory trick: The integral of √(a²-x²) has two parts: one with the square root and one with sin⁻¹. Think of it as "half the rectangle + half the arc."

Standard Area Results (Know These)

Area of full circle x²+y²=r²: πr² (verify: 4 × ∫₀ʳ √(r²-x²) dx = πr²)

Area of full ellipse x²/a²+y²/b²=1: πab

Area under parabola y²=4ax from x=0 to x=a: (8/3)a²

Step-by-Step Approach for Area Problems

Step 1: Sketch the curves. Draw a rough sketch showing all the given curves and lines. This is worth 1 mark and helps you visualize the bounded region.

Step 2: Find intersection points. Set the equations equal (or substitute) to find where the curves meet. These give the limits of integration.

Step 3: Identify the upper and lower curves. In the bounded region, determine which curve is on top (for horizontal integration) or which is to the right (for vertical integration).

Step 4: Set up the integral. A = ∫ₐᵇ (upper curve - lower curve) dx

Step 5: Evaluate the integral. Use standard integrals. For √(a²-x²), use the standard result directly.

Step 6: Use symmetry when possible. If the region is symmetric about the x-axis or y-axis, compute the area of one part and multiply by 2 (or 4).

Board Exam Question Types

Type 1: Area Between a Line and a Parabola (5 marks)

Example: Find the area of the region bounded by y² = 4x and the line y = x.

Approach:

  1. Find intersection: y² = 4x and y = x → x² = 4x → x(x-4) = 0 → x = 0, 4. Points: (0,0) and (4,4).
  2. For 0 ≤ x ≤ 4: parabola gives y = 2√x (upper), line gives y = x (lower, in this region).
  3. A = ∫₀⁴ (2√x - x) dx = [2·(2/3)x^(3/2) - x²/2]₀⁴ = [(4/3)(8) - 8] = 32/3 - 8 = 8/3 sq units.

Type 2: Area Between a Line and a Circle (or Ellipse) (5 marks)

Example: Find the area of the smaller region bounded by x²/4 + y²/9 = 1 and the line x/2 + y/3 = 1.

Approach:

  1. Find intersections: from the line, y = 3(1-x/2). Substitute into ellipse equation to find x = 0 and x = 2.
  2. Upper curve = ellipse: y = (3/2)√(4-x²). Lower curve = line: y = 3(1-x/2).
  3. A = ∫₀² [(3/2)√(4-x²) - 3(1-x/2)] dx
  4. Evaluate using the standard integral for √(a²-x²) with a = 2.

Type 3: Area Between Two Parabolas (5 marks)

Example: Find the area enclosed between y² = 4x and x² = 4y.

Approach:

  1. From y² = 4x: y = 2√x. From x² = 4y: y = x²/4.
  2. Intersection: 2√x = x²/4 → 8√x = x² → 64x = x⁴ → x = 0, 4.
  3. For 0 ≤ x ≤ 4: parabola y = 2√x is above y = x²/4.
  4. A = ∫₀⁴ (2√x - x²/4) dx = [(4/3)x^(3/2) - x³/12]₀⁴ = 32/3 - 16/3 = 16/3 sq units.

Type 4: Area of a Circle/Ellipse Using Integration (3-5 marks)

Example: Using integration, find the area of the circle x²+y²=16.

Approach:

  1. Upper half: y = √(16-x²), area of full circle = 4 × ∫₀⁴ √(16-x²) dx (by symmetry)
  2. ∫₀⁴ √(16-x²) dx = [(x/2)√(16-x²) + 8sin⁻¹(x/4)]₀⁴ = 0 + 8(π/2) = 4π
  3. Total area = 4 × 4π = 16π sq units (which equals πr² with r = 4 ✓)

Type 5: Area Bounded by Lines (Triangle/Quadrilateral) (3-5 marks)

Example: Find the area of the triangle with vertices (1,0), (2,2), (3,1) using integration.

Approach:

  1. Find the equations of the three sides.
  2. Integrate along the x-axis, splitting into sub-intervals at each vertex.
  3. In each sub-interval, identify the upper and lower boundary lines.

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Common Mistakes That Cost Marks

1. Not drawing the sketch. The sketch is typically worth 1 mark. More importantly, without it you cannot determine which curve is on top and what the limits are. Always draw the sketch first, even if it is rough.

2. Getting the upper and lower curves reversed. If you integrate (lower - upper) instead of (upper - lower), you get a negative area. Take the absolute value as a safety net, but it is better to set up the integral correctly from the start by looking at the sketch.

3. Wrong limits of integration. The limits are the x-coordinates (or y-coordinates) of the intersection points. A calculation error here propagates through the entire solution. Verify your intersection points by substituting back into both equations.

4. Forgetting the standard integral for √(a²-x²). If you cannot recall ∫√(a²-x²)dx = (x/2)√(a²-x²) + (a²/2)sin⁻¹(x/a) + C, you cannot solve any circle or ellipse area problem. This single formula is the key to most 5-mark questions in this chapter.

5. Not using symmetry. Many regions are symmetric about the x-axis or y-axis. Using symmetry halves the computation and reduces errors. If the region is symmetric about the x-axis, compute the area of the upper half and multiply by 2.

6. Errors in evaluating definite integrals at the bounds. After integrating, students make arithmetic errors when substituting the upper and lower limits. Write out each substitution step clearly. Do not try to evaluate everything in your head.

4-Week Practice Plan

Week 1: Fundamentals and Standard Curves

  • Read NCERT Chapter 8 (Application of Integrals) — all sections
  • Revise the standard curves: parabola, circle, ellipse, and their equations
  • Practice sketching these curves quickly (within 1 minute each)
  • Memorize the standard integral ∫√(a²-x²)dx
  • Solve NCERT Examples 8.1 to 8.5

Week 2: Standard Problem Types

  • Solve all NCERT back exercises (Exercises 8.1 and 8.2, including Miscellaneous)
  • Practice area between line and parabola — at least 10 problems
  • Practice area between line and circle/ellipse — at least 8 problems
  • Practice area between two parabolas — at least 5 problems
  • Refer to the CBSE Class 12 Maths formulas list for integration formulas

Week 3: Board-Level Practice

  • Solve CBSE previous year questions from 2019-2026 (at least 20 questions from this chapter)
  • Time yourself: a 5-mark area problem should take 8-10 minutes
  • Practice problems that require splitting the integral (curves that cross within the region)
  • Focus on setting up the integral correctly — the integration itself uses standard results
  • Cross-reference with the CBSE Class 12 Maths important chapters

Week 4: Revision and Mock Tests

  • Revise the standard integral for √(a²-x²) daily
  • Solve 2-3 full Maths sample papers, focusing on the area question
  • Review common mistakes, especially sketch, limits, and upper/lower curve identification
  • Practice one area problem every day as warm-up
  • Check the Maths marks weightage for the overall strategy

Quick Revision Card for Exam Day

  • Area under curve: A = ∫ₐᵇ f(x) dx (f(x) ≥ 0)
  • Area between curves: A = ∫ₐᵇ (upper - lower) dx
  • For x = g(y): A = ∫_c^d g(y) dy
  • Standard integral: ∫√(a²-x²)dx = (x/2)√(a²-x²) + (a²/2)sin⁻¹(x/a) + C
  • Parabola y²=4ax: upper half y = 2√(ax)
  • Circle x²+y²=r²: upper half y = √(r²-x²); area = πr²
  • Ellipse x²/a²+y²/b²=1: upper half y = (b/a)√(a²-x²); area = πab
  • Use symmetry: double (or quadruple) when symmetric about axes
  • Steps: Sketch → Find intersections → Identify upper/lower → Integrate → Simplify
  • Always draw the diagram — it is worth 1 mark

Start Practicing on Padhaao

Application of Integrals is a focused chapter with a narrow scope but high-value questions. The 5-mark question from this chapter is one of the most predictable in the entire Maths paper — if you know your standard curves and can set up the integral, full marks are within reach. Padhaao offers visual curve-sketching tools, step-by-step integration walkthroughs, and board-style area problems. 10 learning modes, 3 free chapters daily, no download needed.

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Frequently Asked Questions

How many marks does Application of Integrals carry?

Chapter 8 carries approximately 5 marks out of 80 in the CBSE Class 12 Maths theory paper. This almost always comes as a single 5-mark long answer question asking you to find the area of a region bounded by specific curves and/or lines. Sometimes it appears as a 4-mark question with one additional 1-mark MCQ related to area.

What is the most common 5-mark question?

Area between a line and a conic section (parabola, circle, or ellipse) is the most frequently asked. In recent years, questions on the area between a line and a parabola (like y²=4ax with y=mx) and area between a line and an ellipse have appeared repeatedly. Area between two parabolas is another common variant.

Do I need to memorize the integral of √(a²-x²)?

Absolutely. This integral is the backbone of every circle and ellipse area problem: ∫√(a²-x²)dx = (x/2)√(a²-x²) + (a²/2)sin⁻¹(x/a) + C. Without it, you cannot solve the most common 5-mark question. Write it on your formula sheet and revise daily in the week before the exam.

How important is the sketch/diagram?

Very important. The sketch is worth about 1 mark in a 5-mark question. Beyond the marks, it helps you correctly identify the bounded region, determine which curve is upper/lower, and find the correct limits. A wrong sketch leads to a wrong integral setup, which means losing all 5 marks. Spend 1-2 minutes drawing a reasonably accurate sketch.

How does this chapter connect to the Integration chapter?

Chapter 7 (Integrals) teaches you the techniques of integration — substitution, partial fractions, by parts. Chapter 8 uses only definite integrals with standard results. In practice, most area problems require only ∫xⁿdx and ∫√(a²-x²)dx, so the integration itself is simpler than Chapter 7. The challenge in Chapter 8 is the geometric setup, not the integration.

Is this chapter important for JEE?

Yes. Area under curves is a common JEE Main topic. JEE problems are more complex — they may involve trigonometric curves, bounded regions with multiple intersections, or parametric curves. But the board-level foundation (setting up integrals for standard curves) is essential. Board preparation covers about 50-60% of JEE requirements for area problems.

Final Thoughts

Application of Integrals is the narrowest chapter in the CBSE Class 12 Maths syllabus — it covers exactly one application (area computation) — but it delivers a predictable 5 marks. The key is threefold: (1) know your standard curves and their equations, (2) memorize the integral of √(a²-x²), and (3) always draw the sketch first. With these three pillars in place, the 6-7 standard question types become routine. Combine this chapter with strong preparation in Chapters 5 (Continuity), 6 (Application of Derivatives), and 7 (Integrals) to dominate the entire Calculus unit — the 35-mark backbone of the Maths paper. For the complete strategy, check the CBSE Class 12 Maths chapter-wise marks weightage and the important chapters for 2027.

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