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CBSE Class 12 Mathematics★ 5-6 marks · CBSE Class 12 Mathematics Chapter 8

Application of Integrals (Area Under Curves)

The definite integral is the most PRECISE way to find the area under a curve — and between curves. CBSE Class 12 Mathematics Chapter 8.

⏱ 14 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Set up the definite integral to find the area bounded by a curve y = f(x), x-axis, and vertical lines x = a and x = b
2Find the area between two curves by integrating the difference [upper curve − lower curve]
3Identify the limits of integration by finding intersection points of the two curves
4Calculate area of standard regions: circle, parabola, ellipse using integration
5Draw a rough sketch of the region and identify which curve is 'above' in the integration interval

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Why this chapter matters

Application of Integrals is a focused chapter with one core skill: finding areas using definite integrals. The board exam reliably asks one 5-6 mark question on area between curves (parabola and line, circle and line). The diagram-drawing requirement is explicit and earns marks — students who skip the sketch lose a guaranteed 1-2 marks.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Application of Integrals

"The definite integral turns geometry into algebra. Area becomes a calculation."

1. Chapter Overview

The definite integral measures the AREA under a curve. This chapter applies integration to: finding the area bounded by a curve and the x-axis (or y-axis), and the area BETWEEN TWO CURVES.

2. Area Under a Simple Curve

Area bounded by y = f(x), x-axis, and lines x = a, x = b: ∫ₐᵇ f(x) dx
IF f(x) ≥ 0 on [a,b]. If f(x) dips BELOW the x-axis: take the absolute value of the negative portion (area is ALWAYS positive).

Horizontal Strips

Area bounded by x = g(y), y-axis, y = c, y = d: ∫ᶜᵈ g(y) dy

3. Area Between Two Curves

Area between y = f(x) (upper) and y = g(x) (lower) from x = a to x = b: ∫ₐᵇ [f(x) — g(x)] dx
'Upper minus lower — integrated over the interval of intersection.'
Important: find the points of intersection FIRST (solve f(x) = g(x)) — these are the limits of integration.

4. Exam Focus

Area under simple curve (y = f(x), x-axis).
Area between two curves — upper minus lower. Find intersection points first.
Sketching the region (essential for setting up the correct integral).

5. Conclusion

The definite integral answers: how much land lies between the curve and the axis?

SIMPLE AREA: ∫ f(x) dx. 'Area is always positive — watch for pieces below the axis.'
BETWEEN CURVES: ∫ [upper — lower]. 'Sketch first. Then integrate. The sketch prevents sign errors.'

'The area under a curve is not just geometry. It could be total distance, total revenue, total probability — integration turns accumulation into a single number.'

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Area Under a Curve

Area bounded by y = f(x), x-axis, x = a and x = b: A = ∫ₐᵇ |f(x)| dx. If f(x) ≥ 0 throughout [a,b]: A = ∫ₐᵇ f(x) dx. If f(x) ≤ 0 throughout [a,b]: A = −∫ₐᵇ f(x) dx = |∫ₐᵇ f(x) dx|. If f(x) changes sign at c ∈ (a,b): A = ∫ₐᶜ f(x) dx + |∫ᶜᵇ f(x) dx| (split at the zero). AREA IN TERMS OF y: if curve is x = g(y), area between y = c and y = d: A = ∫ᶜᵈ |g(y)| dy.

The absolute value is critical. If part of the curve dips below the x-axis, taking ∫ₐᵇ f(x) dx gives a net signed area (positive − negative), NOT the geometric area. Always check sign of f(x) in [a,b] and split if necessary. CBSE problems usually specify regions above the x-axis to avoid the splitting complication.

Area Between Two Curves

Area between y = f(x) (upper) and y = g(x) (lower) from x = a to x = b: A = ∫ₐᵇ [f(x) − g(x)] dx, where f(x) ≥ g(x) throughout [a,b]. PROCEDURE: (1) Find intersection points: solve f(x) = g(x) → these are the limits a and b. (2) Determine which curve is on top in [a,b] (substitute a test point). (3) Set up A = ∫ₐᵇ [upper − lower] dx. (4) Evaluate. If curves cross within [a,b], split the integral at the crossing point.

MOST COMMON EXAM TYPE: Find area between parabola y² = 4ax (or y = x²) and a line. Step 1 is ALWAYS to find where they intersect (solve simultaneously). Substituting one equation into the other gives the x-limits. Test which is 'upper' by checking y-values at a midpoint — this determines the integrand order.

Areas of Standard Regions

CIRCLE x² + y² = r²: Area of full circle = πr². By integration: A = 4∫₀ʳ √(r²−x²) dx = πr². Area of semicircle = πr²/2. PARABOLA y² = 4ax (opens right): Area from x=0 to x=a: A = ∫₀ᵃ 2√(4ax) dx = (8a²/3)... standard approach is to use symmetry. ELLIPSE x²/a² + y²/b² = 1: Area = πab. SECTOR of circle (angle θ): A = (1/2)r²θ. KEY RESULT: Area between y = x and y = x² from 0 to 1 = 1/2 − 1/3 = 1/6.

For a circle x²+y²=r²: upper half is y = √(r²−x²) ≥ 0. Use symmetry: compute area in first quadrant and multiply by 4. ∫₀ʳ √(r²−x²) dx = πr²/4 (quarter circle). For ellipse: by substitution x=a sinθ or direct formula πab. Memorise πab — it is tested as a direct-substitution MCQ.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Forgetting to find intersection points and instead using arbitrary limits

✓ For area between two curves, the limits of integration are ALWAYS the x-coordinates where the curves INTERSECT. Solve f(x) = g(x) algebraically to find these points. Common mistakes: using x=0 and x=1 without checking if these are actually intersection points. Always solve f(x)=g(x) first.

WATCH OUT

✗ Putting the lower curve's function first in the integrand

✓ The integrand must be [upper curve − lower curve]. If the upper curve has y = f(x) and lower y = g(x) in the interval, write ∫[f(x) − g(x)]dx. If you reverse them, you get a negative area (wrong sign). Verify by substituting a midpoint test value: whichever gives a larger y-value is the upper curve.

WATCH OUT

✗ Skipping the rough sketch

✓ CBSE requires a rough sketch for area problems. Without the sketch, you cannot identify which curve is 'above,' where intersections are, and whether the area spans multiple regions. A correct sketch earns 1-2 marks independently. At minimum, mark the intersection points, the axes, and label the curves — this takes 60 seconds and protects multiple marks.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s 8.1 and 8.2

Exercises 8.1 and 8.2

Area under parabolas and lines; area of circle regions; area between parabola and line; area between two parabolas; area of ellipse using integration; miscellaneous region problems

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· area-under-curve

Find the area of the region bounded by y = x², y = 0, x = 0 and x = 3.

▶ Show solution

The curve y = x² is a parabola opening upward. In [0, 3], y = x² ≥ 0. Area = ∫₀³ x² dx = [x³/3]₀³ = 27/3 − 0 = 9 square units.

Q2MEDIUM· area-between-curves

Find the area of the region enclosed by the parabola y = x² and the line y = x + 2.

▶ Show solution

FIND INTERSECTION POINTS: Set x² = x + 2 → x² − x − 2 = 0 → (x−2)(x+1) = 0 → x = −1 and x = 2. So limits are x = −1 to x = 2. CHECK WHICH IS UPPER: At x = 0: parabola y = 0, line y = 2. Line is ABOVE parabola in [−1, 2]. DRAW SKETCH: Parabola y = x² with vertex at origin; line y = x+2 intersecting at (−1, 1) and (2, 4). AREA: A = ∫₋₁² [(x+2) − x²] dx = [x²/2 + 2x − x³/3]₋₁². At x = 2: 4/2 + 4 − 8/3 = 2 + 4 − 8/3 = 6 − 8/3 = 10/3. At x = −1: 1/2 − 2 − (−1/3) = 1/2 − 2 + 1/3 = 3/6 − 12/6 + 2/6 = −7/6. A = 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2 square units.

Q3HARD· circle-area

Using integration, find the area of the region in the first quadrant enclosed by the circle x² + y² = 4 and the line x = √3·y.

▶ Show solution

FIND INTERSECTION: x = √3·y and x² + y² = 4. Substitute: 3y² + y² = 4 → 4y² = 4 → y = 1 (first quadrant). So x = √3. Intersection point: (√3, 1). The region in the first quadrant is bounded by the line (from origin to (√3,1)) and the circle arc (from (√3,1) to (2,0) on x-axis). SKETCH: Line from O to (√3,1); circle arc from (√3,1) to (2,0). Area = Area under circle from x=0 to x=√3 using y=x/√3 (line) + Area under circle from x=√3 to x=2 using y=√(4−x²) (circle). Wait — reframe: Area = ∫₀^√3 (x/√3) dx + ∫_√3^2 √(4−x²) dx. First integral: (1/√3)[x²/2]₀^√3 = (1/√3)(3/2) = √3/2. Second integral: ∫_√3^2 √(4−x²) dx = [x√(4−x²)/2 + 2 sin⁻¹(x/2)]_√3^2. At x=2: [0 + 2·sin⁻¹(1)] = 2·(π/2) = π. At x=√3: [√3·1/2 + 2·sin⁻¹(√3/2)] = √3/2 + 2·(π/3) = √3/2 + 2π/3. Second integral = π − √3/2 − 2π/3 = π/3 − √3/2. TOTAL AREA = √3/2 + π/3 − √3/2 = π/3 square units.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Area = ∫ₐᵇ [upper curve − lower curve] dx. Limits are intersection points.
•If curve goes below x-axis: A = |∫f(x) dx| — use absolute value or split at zero.
•Circle x²+y²=r²: area = πr². First quadrant area = πr²/4 = ∫₀ʳ √(r²−x²) dx.
•Ellipse x²/a² + y²/b² = 1: area = πab.
•Parabola y = x² and line y = x + 2: area = 9/2 (classic NCERT result).
•ALWAYS sketch: mark intersection points, axes, shade region, label curves.
•For area w.r.t. y-axis: A = ∫ᶜᵈ [right curve − left curve] dy (sometimes easier).
•Test which curve is upper: substitute any x in (a, b) and compare y-values.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 5-6 marks

Question type	Marks each	Typical count	What it tests
Area Between Two Curves	5-6	1	Parabola and line; circle and line; two parabolas; ellipse region; compute intersection points, sketch, integrate

Prep strategy

The 4-step process for every area problem: (1) INTERSECT — solve the two curve equations simultaneously to find x-limits. (2) SKETCH — draw a rough graph, mark intersection points, shade the region. (3) IDENTIFY UPPER — determine which function is on top via a test point. (4) INTEGRATE — ∫[upper − lower] dx over the correct limits.
Know these areas by integration derivation, not memorisation: circle area = πr² (via ∫₀ʳ √(r²−x²) dx with substitution x = r sin θ). Practice this derivation at least twice.
For line-and-parabola problems, always factorise f(x) − g(x) = 0 to find limits. Then integrate [line − parabola] or [parabola − line] depending on which is on top.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Computing Work and Energy in Physics

In physics, the work done by a variable force F(x) over displacement from a to b is W = ∫ₐᵇ F(x) dx — the area under the force-displacement graph. Engineers use this to compute the energy stored in springs (∫₀ˣ kx dx = kx²/2), the work done by a gas expanding against a piston, and the energy absorbed by brakes. Every energy calculation in mechanics, thermodynamics, and electromagnetism is a definite integral evaluated as an area.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

The sketch earns 1 mark explicitly in CBSE board marking. Draw it first, label the curves (y = x², y = x+2), mark the intersection points with coordinates, and shade the bounded region. A labelled sketch takes 60 seconds and can be the difference between full marks and losing 2.
For standard circle/parabola areas: use the formula ∫√(r²−x²)dx = (x/2)√(r²−x²) + (r²/2)sin⁻¹(x/r) + C without re-deriving it each time. Memorise this antiderivative — it appears in every circle-area problem.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Study CAVALIERI'S PRINCIPLE: if two solids have equal cross-sectional areas at every height, they have equal volumes. This is the 3D extension of area-between-curves — it's how Cavalieri showed that the volume of a sphere equals 2/3 the volume of its circumscribed cylinder (before calculus!)
Explore GREEN'S THEOREM: the area enclosed by a closed curve C is A = (1/2)∮(x dy − y dx) — the area becomes a LINE INTEGRAL around the boundary. This connects area (a 2D integral) to a 1D integral around the perimeter, unifying the concepts of area and integration at a deep level

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 Board (Mathematics)High

JEE Main (Area under Curves)High

CUET (Mathematics)Medium

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Integrate with respect to y when the region is naturally described by horizontal strips — typically when curves are written as x = f(y) (like the parabola y² = 4x). In this case, A = ∫ᶜᵈ [x_right − x_left] dy. It's also useful when integrating with respect to x would require splitting the interval (e.g., the region is above and below the curve at different x-values). Choose whichever variable makes the region a single integral without splitting.

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