Simulation · Maths · Class 12

Adding up thin strips

From the lesson Area from thin strips in Application of Integrals. Change the values and watch what happens.

The idea behind it

NCERT §8.1–8.2

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.

1 more