Area under a curve

A definite integral can measure area. Cut the region under a curve into very thin vertical strips. Each strip is almost a rectangle of height y and width dx, so its area is y dx. Adding all the strips from x = a to x = b gives:

Area = ∫ₐᵇ y dx

This is the area between the curve y = f(x), the x-axis, and the lines x = a and x = b.

Example: under y = x² from 0 to 3: ∫₀³ x² dx = [x³/3]₀³ = 9 square units.

If the curve is given as x in terms of y, use horizontal strips instead: Area = ∫ᶜᵈ x dy, the area between the curve and the y-axis. Example: x = y² from y = 0 to y = 2 gives ∫₀² y² dy = 8/3.

Always draw a rough sketch first. It shows which strips to use, where the region starts and ends, and whether symmetry can help.

Symmetry saves work. A circle or an ellipse centred at the origin has four equal quarter parts, so find the first-quadrant part and multiply by 4.

• Circle x² + y² = a²: y = √(a² – x²), and 4∫₀ᵃ √(a² – x²) dx = πa².
• Ellipse x²/a² + y²/b² = 1: area = πab.
• The region inside y² = 4ax up to the line x = a is symmetric about the x-axis: area = 2∫₀ᵃ 2√(ax) dx = 8a²/3.

For the √(a² – x²) integral, use the standard result from Chapter 7: (x/2)√(a² – x²) + (a²/2) sin⁻¹(x/a).

Write the answer with its unit: square units.