Find the area of the region that is outside the curve r = 2 and inside the curve r = 4 cos θ. Your work must include the integral, but you may use your calculator to find the area to 3 decimal places.

4π/3 + 2√3 ≈ 7.653

Step-by-step explanation:

Area between two polar curves is:

A = ∫ₐᵇ ½ (R² − r²) dθ

First, graph the curves.  r = 2 is a circle with radius 2 and center (0,0).  r = 4 cos θ is a circle with radius 2 and center (2,0).

Find where the curves intersect:

2 = 4 cos θ

1/2 = cos θ

θ = ±π/3

Between θ = -π/3 and θ = π/3, 4 cos θ > 2.  So R = 4 cos θ and r = 2.

So the integral is:

A = ∫₋ᵖⁱ'³ ½ ((4 cos θ)² − 2²) dθ

A = ∫₋ᵖⁱ'³ ½ (16 cos²θ − 4) dθ

A = ∫₋ᵖⁱ'³ (8 cos²θ − 2) dθ

Using symmetry, we can write this as:

A = 2 ∫₀ᵖⁱ'³ (8 cos²θ − 2) dθ

A = ∫₀ᵖⁱ'³ (16 cos²θ − 4) dθ

Use power reduction formula:

A = ∫₀ᵖⁱ'³ (8 + 8 cos(2θ) − 4) dθ

A = ∫₀ᵖⁱ'³ (4 + 8 cos(2θ)) dθ

Integrate:

A = (4θ + 4 sin(2θ)) |₀ᵖⁱ'³

A = (4π/3 + 4 sin(2π/3)) − (0 + 0)

A = 4π/3 + 2√3

A ≈ 7.653

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3