How do you solve this difficult problem below?

Question

What is the arc length, in length units, of the curve y = (4 + x)^3/2 from the point on the curve where x= 3 to the point on the curve where x= 6 ?

these were the possible solutions in the back of the book...

a. 13.44

b. 16.32

c. 17.39

d. 18.40

e. 19.65

f. 21.05

g. 24.38

or is it none of these ?

Answer 1

L =  ∫√((dy/dx)²+ 1)dx
dy/dx = 3/2(1)(4+x)^(1/2)
dydx² = 9/4(4+x) = 9 + 9x/4
L = ∫√(10 + 9x/4)dx

let u = 10+9x/4 then du = 9/4 dx
L = 4/9 ∫√u du = 4/9 (2/3) u^(3/2)
L = 8/27*(10+9x/4)^(3/2) |x=3..6
L = 13.44

Answer 2

Length= Int sqrt(1+y´2)dx between 3 and 6
y´=3/2(4+x)1/2
so the integral is
Intsqrt(1+9/4(4+x) dx
call 1+9+9/4 x =z so 9/4 dx =dz
and you get
4/9Int sqrt z dz = 4/9 *2/3(z)^3/2 +C
=8/27(10+9/4x)^3/2 beween 3 and 6
8/27*(113.92-68.55)=13.44 (a)

Answer 3

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Answer 4

Seven!!!No?Well,that was my best guess...