Is it correct to say: Integrating straight across a point where the curve crosses the x-axis?

The negative part cancels part of the positive part, giving the net value. Find the crossing point, split the integral there, and add the absolute values.

Is it correct to say: Forgetting the factor 4 (or 2) when using symmetry?

If you integrate only over the first quadrant, multiply by the number of identical pieces: 4 for a circle or ellipse, 2 for y² = 4ax about the x-axis.

Is it correct to say: Using dx when the curve is given as x = g(y) and bounded by horizontal lines?

For a region against the y-axis between y = c and y = d, integrate x = g(y) with respect to y, using the limits c and d.

Is it correct to say: Writing the ellipse strip as (a/b)√(a² − x²)?

From x²/a² + y²/b² = 1 you get y = (b/a)√(a² − x²). As a check, the final area must be πab.

Is it correct to say: Leaving sin⁻¹(1) as 90 or using degrees?

Inside the integral the angle is in radians: sin⁻¹(1) = π/2 and sin⁻¹(1/2) = π/6.

Is it correct to say: Leaving out the unit?

Area answers are written in square units, for example 20π sq units.

📄 Formula sheet (PDF)

Application of Integrals

Area under simple curves using vertical and horizontal strips, regions below the x-axis, and areas of circles, ellipses and parabolic regions in standard form — NCERT Class 12 Maths Ch 8

📐 13 formulas✏️ 4 examples🎯 6 practice⚖️ Unit III: 35 marks (with Ch 5, 6, 7 and 9)🏫 CBSE📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Always draw a rough sketch and shade the region. The sketch gives the limits and shows whether vertical or horizontal strips are easier.
→Area is never negative. If the curve dips below the x-axis, split the integral where it crosses and add the absolute values.
→Use symmetry. For a circle or ellipse in standard form, find the first-quadrant area and multiply by 4.
→Keep ∫√(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) ready. Every circle and ellipse question uses it.
→If the region is bounded by the y-axis and horizontal lines y = c, y = d, write x in terms of y and integrate with respect to y.
→Write the final answer in square units.

📐 Formulas(13)

Elementary Strip

d A = y d x (vertical strip), d A = x d y (horizontal strip)

Area Under a Curve (about the x-axis)★ Board fav

A = \int_{a}^{b} y d x = \int_{a}^{b} f (x) d x, f (x) \geq 0 on [a, b]

Symbol	Meaning
$A$	Area of the region (square units)
$a, b$	Left and right ordinates (x = a, x = b), with b > a
$f (x)$	Curve y = f(x) bounding the region from above

Area About the y-axis

A = \int_{c}^{d} x d y = \int_{c}^{d} g (y) d y, g (y) \geq 0 on [c, d]

Symbol	Meaning
$c, d$	Lower and upper lines y = c and y = d, with d > c
$g (y)$	Curve x = g(y) bounding the region on the right

Curve Below the x-axis

A = \int_{a}^{b} f (x) d x, f (x) \leq 0 on [a, b]

Curve Crossing the x-axis★ Board fav

A = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x, f (c) = 0

Odd Function on a Symmetric Interval

f (- x) = - f (x) : \int_{- a}^{a} f (x) d x = 0 but A = 2 \int_{0}^{a} f (x) d x

Area Under y = kxⁿ

\int_{0}^{b} k x^{n} d x = \frac{k b ^{n + 1}}{n + 1}, n \geq 0

One Arch of the Sine Curve

\int_{0}^{π} sin x d x = 2

Standard Integral for Circle and Ellipse★ Board fav

\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a} + C

Quarter-circle Integral

\int_{0}^{a} a^{2} - x^{2} d x = \frac{π a ^{2}}{4}

Area of a Circle

x^{2} + y^{2} = a^{2} : A = 4 \int_{0}^{a} a^{2} - x^{2} d x = π a^{2}

Symbol	Meaning
$a$	Radius of the circle (units)

Area of an Ellipse★ Board fav

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 : A = 4 \cdot \frac{b}{a} \int_{0}^{a} a^{2} - x^{2} d x = π ab

Symbol	Meaning
$a$	Semi-axis along x (units)
$b$	Semi-axis along y (units)

Parabola y² = 4ax up to x = h

A = 2 \int_{0}^{h} 2 a x d x = \frac{8}{3} a h^{3/2}

Symbol	Meaning
$a$	Parabola parameter in y² = 4ax (a > 0)
$h$	Ordinate x = h closing the region (h > 0)

✏️ Solved Examples

1Solved Exampleeasy2 steps

Find the area of the region bounded by the curve y = x², the x-axis and the ordinates x = 1 and x = 3.

y = x² ≥ 0 on [1, 3], so the region lies above the x-axis. Use vertical strips.

A = \int_{1}^{3} x^{2} d x

2Solved Exampleboard4 steps

Using integration, find the area enclosed by the ellipse x²/25 + y²/16 = 1.

Here a = 5 and b = 4. In the first quadrant, take the positive root for y.

y = \frac{4}{5} 25 - x^{2}

3Solved Exampleboard4 steps

Find the area of the region bounded by the line y = 2x − 1, the x-axis and the ordinates x = 0 and x = 2.

The line crosses the x-axis where 2x − 1 = 0. It is below the axis on [0, 1/2] and above it on [1/2, 2].

x = \frac{1}{2}

4Solved ExampleHOTS3 steps

Find the area of the region in the first quadrant bounded by the circle x² + y² = 16, the x-axis and the ordinates x = 0 and x = 2.

In the first quadrant y = √(16 − x²). The region is a vertical slice of the quarter circle.

A = \int_{0}^{2} 16 - x^{2} d x

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Integrating straight across a point where the curve crosses the x-axis
✓The negative part cancels part of the positive part, giving the net value. Find the crossing point, split the integral there, and add the absolute values.
2
Forgetting the factor 4 (or 2) when using symmetry
✓If you integrate only over the first quadrant, multiply by the number of identical pieces: 4 for a circle or ellipse, 2 for y² = 4ax about the x-axis.
3
Using dx when the curve is given as x = g(y) and bounded by horizontal lines
✓For a region against the y-axis between y = c and y = d, integrate x = g(y) with respect to y, using the limits c and d.
4
Writing the ellipse strip as (a/b)√(a² − x²)
✓From x²/a² + y²/b² = 1 you get y = (b/a)√(a² − x²). As a check, the final area must be πab.
5
Leaving sin⁻¹(1) as 90 or using degrees
✓Inside the integral the angle is in radians: sin⁻¹(1) = π/2 and sin⁻¹(1/2) = π/6.
6
Leaving out the unit
✓Area answers are written in square units, for example 20π sq units.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the area bounded by y = x³, the x-axis and the ordinates x = 0 and x = 2.
Q2
Using integration, find the area of the circle x² + y² = 9.
Q3
Find the area of the region bounded by the curve x = 2y², the y-axis and the lines y = 0 and y = 3.
Q4
Find the area bounded by y = cos x and the x-axis between x = 0 and x = π.
Q5
Find the area bounded by the line y = x − 3, the x-axis and the ordinates x = 0 and x = 4.
Q6
Find the area of the region bounded by the parabola y² = 8x and the line x = 2.

📝 Notes

Application of Integrals

Chapter 8 uses the definite integral from Chapter 7 to find areas bounded by simple curves: lines, circles, parabolas and ellipses in standard form. A definite integral can be negative, but an area cannot, so the main skill is setting up the region correctly.

Four-step method

Sketch the curve and the bounding lines, and shade the region.
Choose strips. Use vertical strips (dx) when the region rests on the x-axis between x = a and x = b. Use horizontal strips (dy) when it rests on the y-axis between y = c and y = d.
Check the sign. If any part of the curve lies below the x-axis (or left of the y-axis), split the integral at the crossing point.
Integrate, substitute the limits, and add absolute values. Give the answer in square units.

Using symmetry

Circles and ellipses in standard form are symmetric about both axes, so integrate over the first quadrant only and multiply by 4. A parabola y² = 4ax is symmetric about the x-axis, so integrate y = 2√(ax) and double. This avoids negative square roots and keeps the limits simple.

The integral you need most

Every circle or ellipse problem reduces to ∫√(a² − x²) dx. On the limits 0 to a it gives πa²/4. For other limits, such as 0 to a/2, substitute carefully: both the algebraic term and the sin⁻¹ term usually survive.

Scope reminder

The current NCERT chapter covers areas under simple curves (about the x-axis or y-axis) and the standard circle, ellipse and parabola regions shown here. The section on the area between two curves is not part of the rationalised textbook.