Lesson 1 of 7 · 10 min

Area from thin strips

NCERT §8.1–8.2

The school garden has a flower bed whose back edge is a curved hedge. The gardener needs its area to order compost, and no formula from geometry fits a curved edge.

The story this chapter follows: The school garden

The school is replanting its garden: a flower bed with a curved hedge, a round fountain, an oval lawn, a drain below ground level and a gravel strip between two edgings. Each needs an area, and none has straight sides. The measurements are for illustration.

The lesson in notes

In short

Formulas from geometry measure figures with straight edges and circles. A region with a curved boundary needs integral calculus.

Take the region bounded by y = f(x) (with f ≥ 0), the x-axis and the vertical lines x = a and x = b. Slice it into thin vertical strips. A strip at position x has height y = f(x) and width dx, so its area is dA = y dx.

Adding all the strips from x = a to x = b and letting their width shrink gives the area: A = ∫ₐᵇ y dx = ∫ₐᵇ f(x) dx.

The strip dA = y dx is called the elementary area. It sits at an arbitrary x between a and b and stands for every strip at once.

Worked example: the region under y = x² from x = 0 to x = 3 has area ∫₀³ x² dx = [x³/3] from 0 to 3 = 9.

Worked example: the region under y = √x from x = 0 to x = 4 has area ∫₀⁴ x^(1/2) dx = (2/3) × 4^(3/2) = 16/3.

Worked example: under y = x² + 1 between x = 1 and x = 2 the area is [x³/3 + x] from 1 to 2 = 14/3 − 4/3 = 10/3.

With a finite number of strips the sum only approximates the area; more and thinner strips bring it closer. The integral is the exact value those sums approach.