Class 12 Maths Chapter 8: Application of Integrals

Application of Integrals is Chapter 8 of NCERT Class 12 Maths. It uses definite integrals to find the area of regions bounded by simple curves such as lines, circles, parabolas and ellipses together with the coordinate axes or vertical and horizontal lines. The chapter is short, but a clean diagram and a correct set-up usually decide full marks in the CBSE board exam.

What you will learn

How a definite integral represents the area under a curve as a sum of thin strips
Finding area with vertical strips (integrating in x) and horizontal strips (integrating in y)
Areas bounded by lines, circles, parabolas and ellipses in standard form
Handling regions that lie below the x-axis or on both sides of an axis
Using symmetry to cut down the working

Key Concepts and Formulas

Area as a Limit of Thin Strips

Imagine the region under a curve y = f(x) cut into very thin vertical strips. Each strip is almost a rectangle of height y and width dx. Adding all such strips from x = a to x = b gives the definite integral, valid when f(x) ≥ 0 on [a, b].

Elementary strip: dA = y dx
Area between y = f(x), x-axis, x = a, x = b:
A = ∫ₐᵇ y dx = ∫ₐᵇ f(x) dx

Area under the curve y = x² from x = 1 to x = 3 using vertical strips, Class 12 Application of Integrals — Adding thin vertical strips gives 26/3 square units

Area is always taken as a positive quantity

Using Horizontal Strips

For curves that are easier to write as x = g(y), or for regions bounded by the y-axis, it is simpler to use horizontal strips of length x and width dy.

Area between x = g(y), y-axis, y = c, y = d:
A = ∫ x dy = ∫ g(y) dy   (from y = c to y = d)

Area between x² = 4y and the y-axis from y = 1 to y = 4 using horizontal strips, Class 12 Maths — Horizontal strips of length 2√y give 28/3 square units

Choose the strip direction that gives a single integral without splitting the region

Regions Below the Axis

If the curve lies below the x-axis, then f(x) < 0 and the integral comes out negative. The area is its absolute value. If the curve crosses the axis at x = c, split the region there.

If f(x) ≤ 0 on [a, b]:  A = |∫ₐᵇ f(x) dx|
Crossing at x = c:
A = |∫ f dx from a to c| + |∫ f dx from c to b|

Never add a positive part and a negative part directly, they cancel and give a wrong answer

Standard Curves to Remember

Most board questions use the following curves, so learn how each looks and where it cuts the axes. The parabola y² = 4ax opens to the right and is symmetric about the x-axis, while x² = 4ay opens upward and is symmetric about the y-axis. The ellipse cuts the axes at (±a, 0) and (0, ±b).

Circle x² + y² = a²:          Area = πa²
Ellipse x²/a² + y²/b² = 1:    Area = πab
∫₀ᵃ √(a² − x²) dx = πa²/4

Using Symmetry

When a region is symmetric about an axis, find the area of one part and multiply. This saves time and reduces sign errors.

Circle or ellipse at origin:
Total area = 4 × (first quadrant area)
y² = 4ax cut by a vertical line:
Total area = 2 × (area above x-axis)

Area of the ellipse x²/25 + y²/9 = 1 found from its first quadrant, Class 12 Application of Integrals — One quarter is 15π/4, so the ellipse has area 15π

Solved Examples

Example 1: Find the area bounded by y = x², the x-axis and the lines x = 1 and x = 3.

A = ∫₁³ x² dx
= [x³/3] from 1 to 3
= (27 − 1)/3
∴ A = 26/3 square units

Example 2: Find the area bounded by the line y = 2x + 1, the x-axis and the lines x = 0 and x = 3.

Line is above the x-axis on [0, 3]
A = ∫₀³ (2x + 1) dx
= [x² + x] from 0 to 3
= 9 + 3
∴ A = 12 square units

Example 3: Find the area of the region in the first quadrant enclosed by the circle x² + y² = 16.

A = ∫₀⁴ √(16 − x²) dx
= [(x/2)√(16 − x²) + 8 sin⁻¹(x/4)] from 0 to 4
= 0 + 8 × (π/2) − 0
∴ A = 4π square units

Example 4: Find the area enclosed by the ellipse x²/25 + y²/9 = 1.

First quadrant: y = (3/5)√(25 − x²)
A = 4 × (3/5) ∫₀⁵ √(25 − x²) dx
= (12/5) × (25π/4)
∴ A = 15π square units

Example 5: Find the area bounded by the parabola y² = 8x and the line x = 2.

By symmetry, A = 2 ∫₀² √(8x) dx
= 4√2 ∫₀² √x dx
= 4√2 × (2/3)(2√2)
= 4√2 × (4√2/3)
∴ A = 32/3 square units

Example 6: Find the area bounded by x² = 4y, the y-axis and the lines y = 1 and y = 4 in the first quadrant.

Horizontal strips: x = 2√y
A = ∫₁⁴ 2√y dy
= (4/3)[y^(3/2)] from 1 to 4
= (4/3)(8 − 1)
∴ A = 28/3 square units

Example 7: Find the area between y = x³, the x-axis and the lines x = −2 and x = 2.

Curve is below the x-axis for x < 0 (odd function)
A = 2 ∫₀² x³ dx
= 2 × [x⁴/4] from 0 to 2
= 2 × 4
∴ A = 8 square units (not 0)

Common Mistakes to Avoid

Always draw a rough sketch and shade the required region; examiners give marks for the figure and it prevents choosing wrong limits.
Do not report a negative or zero answer for an area; take the modulus of each part that lies below the axis.
Find the points of intersection of the curve and the line first, because they become your limits of integration.
Write the final answer with square units; it is a common step mark in CBSE marking schemes.

Quick Revision

Vertical strips:   A = ∫ₐᵇ y dx
Horizontal strips: A = ∫ x dy
Circle x² + y² = a²:        Area = πa²
Ellipse x²/a² + y²/b² = 1:  Area = πab
∫₀ᵃ √(a² − x²) dx = πa²/4

Area is always positive; split the region where the curve crosses the axis
Use symmetry: multiply the first quadrant area by 4 or by 2 as needed

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