# Application of Integrals, Class 12 Mathematics | SwaVid

Application of Integrals, NCERT Class 12 Mathematics: notes with key ideas, formulas, common mistakes and 3 quick checks with worked answers.

Canonical: https://www.swavid.com/cbse/class-12/maths/chapter/application-of-integrals

Source: https://www.swavid.com/cbse/class-12/maths/chapter/application-of-integrals

# Application of Integrals

## Formulas and definitions

## Key ideas

## Common mistakes

## Quick checks

## Keep going

## Questions parents ask

## Class 12 Mathematics on SwaVid

### Area under a curve

### Area along the y-axis

### Regions below the axis

### Symmetry in plane figures

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SwaVid › CBSE › Class 12 › Mathematics › Application of Integrals

Notes written by SwaVid from the NCERT textbook.

Class 12 Mathematics, Chapter 8. Key ideas, formulas and common mistakes.

CBSE does not print marks for a single chapter.

Short answer. This chapter deals with finding the area of bounded plane regions using definite integrals. A student learns to set up vertical or horizontal strips to calculate areas under standard curves such as lines, parabolas, circles, and ellipses.

Source: NCERT Mathematics Part II (Reprint 2026-27) , Chapter 8, Application of Integrals

The area bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b is given by the definite integral of f(x) from a to b. The region is thought of as made up of infinitely many thin vertical strips of width dx and height y.

The area bounded by the curve x = g(y), the y-axis, and the horizontal lines y = c and y = d is evaluated using horizontal strips. The formula for this area is the definite integral of g(y) with respect to y from c to d.

When a curve lies below the x-axis, the definite integral evaluates to a negative value. Because area is a non-negative quantity, the absolute value or numerical value of the integral must be taken.

Symmetrical curves like circles, ellipses, and standard parabolas can be divided into equal parts. Finding the area of one portion and multiplying by the number of symmetric sections simplifies the calculation.

Try each question first. Then tap it to see the answer.

Area = ∫ (from 1 to 3) x² dx

= [x³ / 3] (from 1 to 3)

= (3³ / 3) - (1³ / 3) = 9 - 1/3 = 26/3

26/3 sq units

Express x in terms of y: x = y² / 4

Area = ∫ (from 0 to 4) (y² / 4) dy

= (1 / 4) × [y³ / 3] (from 0 to 4)

= (1 / 12) × (64 - 0) = 64 / 12 = 16/3

16/3 sq units

In first quadrant, y = √(4 - x²)

Area = ∫ (from 0 to 2) √(4 - x²) dx

= [(x / 2) √(4 - x²) + (4 / 2) sin⁻¹(x / 2)] (from 0 to 2)

= [0 + 2 sin⁻¹(1)] - [0 + 2 sin⁻¹(0)] = 2 × (π / 2) = π

π sq units

Differential Equations .

Integrals .

The 13 chapters of Class 12 Mathematics .

CBSE does not give marks to one chapter. It prints marks for units or parts, and the chapter list page shows them all.

No. They are written by SwaVid from the NCERT textbook. The CBSE curriculum is the reference for what is assessed.

Differential Equations.

SwaVid is an AI tutor for Indian students in Classes 6 to 12. It finds the earlier ideas a child missed, then teaches today’s chapter. Start free to see how it works.

- Ch 8 NCERT book
- 80 marks, theory paper
- 3 quick checks
- Leaving a negative sign in the final calculated area . Area is always non-negative. If a curve lies below the x-axis, take the absolute value of the integral for that section.
- Mixing variables and integration limits . When integrating with respect to x using dx, limits must be x-values; when integrating with respect to y using dy, limits must be y-values.
- Forgetting to multiply symmetric sections . Check whether the question asks for the area in the first quadrant or the total area enclosed by the curve, and multiply by the appropriate factor like 2 or 4.

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