3 integration Qs

1) Rewriting the equation of the curve with y as the subject:

y = (2/3)√(x2 - 9) or - (2/3)√(x2 - 9)

So the area can be expressed as:

A = 2 ∫ (x = 3 → 4) (2/3)√(x2 - 9) dx

= (4/3) ∫ (x = 3 → 4) √(x2 - 9) dx

Sub x = 3 sec θ, dx = 3 sec θ tan θ dθ

When x = 3, θ = 0, when x = 4, θ = cos-1 (3/4)

A = (4/3) ∫ [θ = 0 → cos-1 (3/4)] (3 tan θ) 3 sec θ tan θ dθ

= 12 ∫ [θ = 0 → cos-1 (3/4)] sec θ tan2 θ dθ

= 12 ∫ [θ = 0 → cos-1 (3/4)] (sec3 θ - sec θ) dθ

= 12 ∫ [θ = 0 → cos-1 (3/4)] sec3 θ dθ - 12 ∫ [θ = 0 → cos-1 (3/4)] sec θ dθ

Now,

∫ [θ = 0 → cos-1 (3/4)] sec3 θ dθ = ∫ [θ = 0 → cos-1 (3/4)] sec θ d(tan θ)

= [sec θ tan θ] [θ = 0 → cos-1 (3/4)] - ∫ [θ = 0 → cos-1 (3/4)] tan θ d(sec θ)

= (4√7)/9 - ∫ [θ = 0 → cos-1 (3/4)] tan2 θ sec θ dθ

= (4√7)/9 - ∫ [θ = 0 → cos-1 (3/4)] (sec3 θ - sec θ) dθ

2 ∫ [θ = 0 → cos-1 (3/4)] sec3 θ dθ = (4√7)/9 + ∫ [θ = 0 → cos-1 (3/4)] sec θ dθ

Thus

A = 6 x (4√7)/9 + 6 ∫ [θ = 0 → cos-1 (3/4)] sec θ dθ - 12 ∫ [θ = 0 → cos-1 (3/4)] sec θ dθ

= (8√7)/3 - 6 ∫ [θ = 0 → cos-1 (3/4)] sec θ dθ

= (8√7)/3 - 6 [ln |sec θ + tan θ|] [θ = 0 → cos-1 (3/4)]

= (8√7)/3 - 6 ln [(4 + √7)/3]

2) Rewriting the equation of the curve with y as the subject:

y = x3/2 or -x3/2

So the area is:

A = 2 ∫ (x = 0 → 4) x3/2 dx

= [4x5/2/5] (x = 0 → 4)

= 128/5

3) First of all, finding the points of intersection:

x + 2/x = 3

x2 + 2 = 3x

x2 - 3x + 2 = 0

x = 1 or 2

So the area is:

A = ∫ (x = 1 → 2) (3 - x - 2/x) dx

= [3x - x2/2 - 2 ln |x|] (x = 1 → 2)

= 3/2 - 2 ln 2