Find the integral?

Question

Find the integral of sqrt(4-x^2)dx from 0 to 2.

 

Answer: pi

 

Please show all your work.

Answer 1

Find the integral of sqrt(4-x^2)dx from 0 to 2.

I = ∫{0,2}(sqrt(4-x^2))dx

Draw Right Triangle with sides 2, x, sqrt(4-x^2), and θ = arcsin(x / 2).

x = 2 sin(θ) and sqrt(4-x^2) = 2 cos(θ)

dx = 2 cos(θ) dθ

upper limit: θ = arcsin(2/2) = pi/2
lower limit: θ = arcsin(0/2) = 0

I = ∫{0,pi/2}(2 cos(θ))(2 cos(θ) dθ)

I = 4∫{0,pi/2} (cos^2(θ)) dθ

cos(2θ) = cos^2(θ) - sin^2(θ) = 2 cos^2(θ) - 1

cos^2(θ) = (1 + cos(2θ))/2

I = 4∫{0,pi/2} ((1 + cos(2θ))/2) dθ

I = 2∫{0,pi/2} (1 + cos(2θ)) dθ

I = 2 [θ + (sin(2θ))/2]{0,pi/2}

I = 2 [pi/2 + (sin(pi))/2 - 0 - (sin(0))/2]

I = pi

Answer 2

This is just the area of a quarter of a circle of radius 2 so you get (1/4)*π*2²=π

 

Or you can use trig substitution: x = 2 sin θ, dx=2 cos θ dθ.

 

∫₀²√(4 - x²) dx = ∫₀^π/2 √(4 - sin² θ) * 2 cos  θ dθ = ∫₀^π/2 4 cos² θ dθ = ∫₀^π/2 (2+2 cos 2θ) dθ

 

= (2θ + sin 2θ) |₀^π/2 = π