Chapter 8 Applications of Integrals

Notes

Applications of Integrals

1. Area Under the Curve

Integration can be used to find the area of a region bounded by a curve and given lines.

1.1 Area Bounded by y = f(x), x = a and x = b

Consider the curve:

y = f(x)

bounded between the vertical lines:

x = a and x = b.

The required area is:

A = ∫ₐᵇ y dx

Since y = f(x),

A = ∫ₐᵇ f(x) dx

1.2 Area Bounded by x = f(y), y = c and y = d

Consider the curve:

x = f(y)

bounded between the horizontal lines:

y = c and y = d.

The required area is:

A = ∫cᵈ x dy

Since x = f(y),

A = ∫cᵈ f(y) dy

2. Area Enclosed by a Circle

Find the area enclosed by the circle:

x² + y² = a²

The given equation represents a circle with centre (0, 0) and radius a.

Due to symmetry, the total area of the circle is four times the area in the first quadrant.

Therefore,

A = 4∫₀ᵃ y dx

From:

x² + y² = a²

we get:

y² = a² - x²

Therefore,

y = √(a² - x²)

Hence,

A = 4∫₀ᵃ √(a² - x²) dx

Using:

∫√(a² - x²)dx = (x/2)√(a² - x²) + (a²/2)sin⁻¹(x/a)

we get:

A = 4[(x/2)√(a² - x²) + (a²/2)sin⁻¹(x/a)]₀ᵃ

A = 4[a²π/4]

Therefore,

A = πa² square units

3. Area Enclosed by an Ellipse

Find the area enclosed by the ellipse:

x²/a² + y²/b² = 1,   a > b

From the equation:

x²/a² + y²/b² = 1

we get:

y²/b² = 1 - x²/a²

Therefore,

y = (b/a)√(a² - x²)

Due to symmetry, the total area of the ellipse is four times the area in the first quadrant.

Therefore,

A = 4∫₀ᵃ y dx

A = (4b/a)∫₀ᵃ √(a² - x²) dx

Using the standard integral:

∫√(a² - x²)dx = (x/2)√(a² - x²) + (a²/2)sin⁻¹(x/a)

we get:

A = (4b/a)[(x/2)√(a² - x²) + (a²/2)sin⁻¹(x/a)]₀ᵃ

A = (4b/a)(a²π/4)

Therefore,

A = πab square units

4. Area Bounded by a Parabola and a Line

Find the area of the region bounded by the curve: