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Mathematics Grade XII

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Chapter 8: Application of Integrals

Class 12 • Mathematics • Chapter 8

Application of Integrals

Using the definite integral to measure the exact area of regions bounded by curves.

CBSE / NCERT Aligned

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Chapter Roadmap

Area Under a Curve • Area Bounded by a Curve and the Axis • Area Between Two Curves • Horizontal Strips • Curves Meeting a Line • Using Symmetry

Measuring Curved Regions Exactly

Straight-sided shapes have simple area formulas, but the world is full of curved boundaries: a parabolic arch, the cross-section of a lens, the region a road sweeps out. The definite integral gives the exact area of such regions by adding up infinitely many vertical strips, each of height equal to the curve and width dx.

This short chapter is the geometric pay-off of integration. You will compute areas under a single curve, areas trapped between a curve and the x-axis, and areas enclosed between two curves. The key skills are sketching the region, choosing whether to slice vertically or horizontally, and setting the correct limits. Symmetry, where present, halves the work.

Foundation

The area under y = f(x) from x = a to x = b, where the curve is above the axis, is ∫ from a to b of f(x) dx. Between two curves the area is ∫ (top curve − bottom curve) dx over the overlap.

Key Terms You Must Know

Term

Meaning

Example

Area under a curve

Definite integral of the height f(x) between two x-values.

∫ from a to b of f dx

Bounded region

A finite region enclosed by given curves and lines.

between a parabola and a chord

Top and bottom curves

The upper and lower boundaries when slicing vertically.

subtract bottom from top

Vertical strip

A thin slice of width dx and height the curve value.

building block of the area

Horizontal strip

A thin slice of height dy; integrate x with respect to y.

useful when x is a function of y

Symmetry

Using mirror images to integrate half and double.

circle and ellipse areas

Core Concepts, Step by Step

1. Area Under a Single Curve

When a curve y = f(x) stays above the x-axis between a and b, the enclosed area is simply the definite integral of f from a to b. You sketch the curve, identify the limits, integrate and evaluate. This is the most direct application of the Fundamental Theorem of Calculus.

2. When the Curve Dips Below the Axis

A definite integral gives signed area, counting parts below the x-axis as negative. To find the true geometric area we split the interval at the points where the curve crosses the axis and take the absolute value of each piece before adding. This keeps every contribution positive, as area should be.

3. Area Between Two Curves

When two curves enclose a region, the area is the integral of the upper curve minus the lower curve across the overlap. First find where they meet by setting the two expressions equal; those intersection x-values become the limits. The subtraction automatically gives a positive height wherever the chosen top really is on top.

The area between y = x and y = x² from 0 to 1 equals one sixth

Shaded region enclosed between the line y equals x and the parabola y equals x squared from 0 to 1

4. Slicing with Horizontal Strips

Sometimes the boundary is easier to describe with x as a function of y. Then we slice horizontally, using strips of height dy, and integrate x with respect to y between the relevant y-limits. Choosing the slicing direction that matches the boundaries keeps the integral simple.

5. Regions Bounded by a Curve and a Line

A common exam region is the piece trapped between a parabola and a straight line, or a curve and one of the axes. The method is unchanged: find the intersection points, decide which boundary is on top, and integrate the difference. A clear sketch prevents sign mistakes.

6. Exploiting Symmetry

If a region is symmetric about an axis, integrate over the half that is easy and double the result. The area of a circle of radius r, for instance, can be found by integrating the upper semicircle and doubling, and an ellipse follows the same idea. Symmetry turns a long integral into a short one.

Key Results with Proofs

Result 1: Area Under a Curve

Statement. If f(x) ≥ 0 on [a, b], the area between the curve and the x-axis is ∫ from a to b of f(x) dx.

Proof

The area is built from infinitely many thin rectangles under the curve.

Divide [a, b] into thin strips of width dx. – slice the region

Each strip is almost a rectangle of height f(x), so its area is about f(x) dx. – approximate a strip

Adding all strips and letting the width shrink to 0 gives the integral. – sum and take the limit

Hence the total area is ∫ from a to b of f(x) dx. – the formula

If f dips below the axis, split at the crossings and take absolute values.

Result 2: Area Between Two Curves

Statement. If f(x) ≥ g(x) on [a, b], the area between them is ∫ from a to b of [f(x) − g(x)] dx.

Proof

Each strip’s height is the distance between the two curves.

At each x the vertical gap between the curves has height f(x) − g(x). – height of a strip

A strip of width dx has area [f(x) − g(x)] dx. – strip area

Summing the strips and taking the limit gives the integral of the difference. – integrate the gap

So the area is ∫ from a to b of [f(x) − g(x)] dx. – result

The limits a and b are the x-coordinates where the curves intersect.

Result 3: Area of a Circle by Integration

Statement. The area enclosed by x² + y² = r² is πr².

Proof

Integrate the top half of the circle and double it.

The upper half is y = √(r² − x²), from x = −r to x = r. – upper semicircle

Area = 2 ∫ from −r to r of √(r² − x²) dx, doubling for the lower half. – use symmetry

This standard integral evaluates to 2 × (πr²/2). – evaluate the integral

= πr². – the familiar formula

The same method with x²/a² + y²/b² = 1 gives an ellipse area of πab.

Worked Examples

Example 1

Question: Find the area under y = x² from x = 0 to x = 3.

▶ Show full working

Integrate f.

Area = ∫ from 0 to 3 of x² dx = [x³/3] from 0 to 3. – set up

= 27/3 = 9. – evaluate

Answer: 9 square units.

Example 2

Question: Find the area under y = x from x = 0 to x = 4.

▶ Show full working

Integrate.

∫ from 0 to 4 of x dx = [x²/2] = 16/2 = 8. – evaluate

Answer: 8 square units.

Example 3

Question: Find the area under y = sin x from 0 to π.

▶ Show full working

Integrate.

∫ from 0 to π of sin x dx = [−cos x] = −cos π + cos 0 = 1 + 1. – evaluate

Answer: 2 square units.

Example 4

Question: Find the area between y = x and y = x² from 0 to 1.

▶ Show full working

Top minus bottom.

On (0, 1) the line y = x is above the parabola y = x². – decide top

∫ from 0 to 1 of (x − x²) dx = [x²/2 − x³/3] = 1/2 − 1/3. – integrate

= 1/6. – simplify

Answer: 1/6 square unit.

Example 5

Question: Find the area enclosed by y = 4 − x² and the x-axis.

▶ Show full working

Integrate between the roots.

The curve meets the axis at x = −2 and x = 2. – find limits

∫ from −2 to 2 of (4 − x²) dx = [4x − x³/3] from −2 to 2. – integrate

= (8 − 8/3) − (−8 + 8/3) = 32/3. – evaluate

Answer: 32/3 square units.

Example 6

Question: Find the area enclosed by y² = 4x and the line x = 1.

▶ Show full working

Use symmetry about the x-axis.

For each x, y runs from −2√x to 2√x, a height of 4√x. – height of a strip

Area = ∫ from 0 to 1 of 4√x dx = [4 · (2/3) x^(3/2)] = 8/3. – integrate

Answer: 8/3 square units.

Example 7

Question: Find the area of a circle of radius 2 by integration.

▶ Show full working

Top half doubled.

Area = 2 ∫ from −2 to 2 of √(4 − x²) dx. – set up

The standard integral gives 2 × (π · 2²/2) = 4π. – evaluate

Answer: 4π square units.

Example 8

Question: Find the area between y = 2x and y = x² from 0 to 2.

▶ Show full working

Top minus bottom.

On (0, 2) the line y = 2x is above y = x². – decide top

∫ from 0 to 2 of (2x − x²) dx = [x² − x³/3] = 4 − 8/3. – integrate

= 4/3. – simplify

Answer: 4/3 square units.

Example 9

Question: Find the area under y = eₓ from 0 to 1.

▶ Show full working

Integrate.

∫ from 0 to 1 of eₓ dx = [eₓ] = e − 1. – evaluate

Answer: e − 1 square units.

Example 10

Question: Find the area under y = 1/x from x = 1 to x = e.

▶ Show full working

Integrate.

∫ from 1 to e of (1/x) dx = [ln x] = ln e − ln 1 = 1. – evaluate

Answer: 1 square unit.

Example 11

Question: Find the area between y = √x and y = x from 0 to 1.

▶ Show full working

Decide which is on top.

On (0, 1), √x ≥ x, so √x is the top curve. – decide top

∫ from 0 to 1 of (√x − x) dx = [(2/3) x^(3/2) − x²/2] = 2/3 − 1/2. – integrate

= 1/6. – simplify

Answer: 1/6 square unit.

Example 12

Question: Find the area of the ellipse x²/9 + y²/4 = 1.

▶ Show full working

Use the πab result.

Here a = 3 and b = 2. – read a and b

Area = πab = π × 3 × 2 = 6π. – apply the formula

Answer: 6π square units.

Where You Meet This in Real Life

Architecture

The area of arches, domes and curved facades is found by integrating their boundary curves.

Land surveying

Plots with curved boundaries, such as those along a river, are measured by integrating the boundary.

Manufacturing

The cross-sectional area of pipes, lenses and machined parts feeds directly into material and strength calculations.

Physics

The work done by a variable force is the area under its force-distance graph, a definite integral.

Statistics

The probability that a continuous variable falls in a range is the area under its density curve.

Practice Sets A–D

Practice Set A – Basics

A1. Find the area under y = x from 0 to 2.

▶ Reveal full working

Integrate.

[x²/2] = 2.

Answer: 2 square units.

A2. Find the area under y = 3 from x = 0 to x = 5.

▶ Reveal full working

Constant height.

3 × 5 = 15.

Answer: 15 square units.

A3. Find the area under y = x² from 0 to 1.

▶ Reveal full working

Integrate.

[x³/3] = 1/3.

Answer: 1/3 square unit.

A4. What does a negative definite integral indicate about the region?

▶ Reveal full working

Signed area.

The curve lies below the x-axis there.

Answer: The region is below the x-axis.

Practice Set B – Conceptual

B1. Why do we take absolute values when a curve dips below the axis?

▶ Reveal full working

Signed vs true area.

A definite integral counts area below the axis as negative. – signed area

Taking absolute values of each piece keeps the geometric area positive. – true area

Answer: Because the integral gives signed area, and area must be positive.

B2. How do you choose the limits for the area between two curves?

▶ Reveal full working

Where they meet.

Set the two curves equal and solve to find their intersection x-values. – intersections

Answer: Use the x-values where the curves intersect.

B3. When is a horizontal strip more convenient than a vertical one?

▶ Reveal full working

Boundary form.

When x is more easily written as a function of y, slice with dy strips.

Answer: When x is a clean function of y.

B4. How does symmetry shorten an area calculation?

▶ Reveal full working

Half then double.

Integrate over one symmetric half and double the result.

Answer: Integrate half the region and double it.

Practice Set C – Application / Numerical

C1. Find the area between y = x² and y = 4.

▶ Reveal full working

Horizontal line on top.

They meet at x = ±2; the line y = 4 is above the parabola. – limits and top

∫ from −2 to 2 of (4 − x²) dx = 32/3. – integrate

Answer: 32/3 square units.

C2. Find the area under y = cos x from 0 to π/2.

▶ Reveal full working

Integrate.

[sin x] from 0 to π/2 = 1 − 0 = 1. – evaluate

Answer: 1 square unit.

C3. Find the area enclosed by y = x² and y = x from 0 to 1.

▶ Reveal full working

Top minus bottom.

Line on top: ∫ from 0 to 1 of (x − x²) dx = 1/6. – integrate

Answer: 1/6 square unit.

C4. Find the area of the region bounded by y = x³, the x-axis and x = 2.

▶ Reveal full working

Integrate.

∫ from 0 to 2 of x³ dx = [x⁴/4] = 16/4 = 4. – evaluate

Answer: 4 square units.

Practice Set D – HOTS / Multi-step

D1. Find the area enclosed between y² = x and x = 4.

▶ Reveal full working

Symmetry about the x-axis.

For each x, height is from −√x to √x, that is 2√x. – strip height

∫ from 0 to 4 of 2√x dx = [2 · (2/3) x^(3/2)] = (4/3)(8) = 32/3. – integrate

Answer: 32/3 square units.

D2. Find the area between y = x² and y = 2x.

▶ Reveal full working

Find intersections first.

x² = 2x gives x = 0 and x = 2; the line is on top. – limits and top

∫ from 0 to 2 of (2x − x²) dx = 4/3. – integrate

Answer: 4/3 square units.

D3. Find the area of one loop region under y = sin x from 0 to π, then explain why ∫ from 0 to 2π of sin x dx = 0.

▶ Reveal full working

Compute one hump and reason about cancellation.

∫ from 0 to π of sin x dx = 2, the area of the positive hump. – positive hump

From π to 2π the curve is below the axis, contributing −2. – negative hump

The signed integral over 0 to 2π is 2 + (−2) = 0, though the true area is 4. – cancellation

Answer: Positive area 2; the full signed integral is 0 due to cancellation.

D4. Find the area of the ellipse x²/16 + y²/9 = 1.

▶ Reveal full working

Use πab.

a = 4, b = 3. – read off

Area = πab = 12π. – apply

Answer: 12π square units.

Chapter Summary

Everything in One Glance

Area Under a Curve

∫ from a to b of f(x) dx when f ≥ 0; split and take absolute values where f dips below.

Between Two Curves

∫ (top − bottom) dx across the overlap; limits are the intersection points.

Slicing Direction

Vertical strips use dx; horizontal strips use dy when x is a clean function of y.

Circle & Ellipse

Area of a circle is πr²; an ellipse is πab, both by integration with symmetry.

Method

Sketch the region, find intersections, choose top and bottom, then integrate.

Signed vs True

The integral gives signed area; geometric area needs absolute values.

Are You Exam-Ready?

8-Point Exam Quick-Check

Find the area under y = x² from 0 to 2.

Find the area under y = sin x from 0 to π.

Find the area between y = x and y = x² from 0 to 1.

What does a negative definite integral tell you?

Find the area under y = 1/x from 1 to e.

State the area of an ellipse with semi-axes a and b.

Find the area between y = 2x and y = x² from 0 to 2.

How do you find the limits for the area between two curves?

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Class 12 Mathematics Chapter 8: Application of Integrals, Complete Notes and Practice

These free Class 12 Maths Application of Integrals notes follow the NCERT 2026 to 27 syllabus and cover area under a curve, area bounded by a curve and the axis, area between two curves, horizontal strips and the use of symmetry, with twelve worked examples and sixteen graded practice questions. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.