Find the length of the curve

(1/3)(x^2+2)^1.5 from x=0 to x=3

Length is int(sqrt(1 + (dy/dx)^2)dx)|(0,3) where int represents an integral and sqrt is the square root of what is inside the parenthesis

y = (x^2 + 2)^1.5/3
dy/dx = x(x^2 + 2)^0.5

int(sqrt(1 + x^2(x^2 + 2))dx)|(0,3)
int(sqrt(x^4 + 2x^2 +1)dx)|(0,3)
int(x^2 + 1 dx)|(0,3)
x^3/3 + x |(0,3)
(9 + 3) - (0 + 0) = 12

I hope this helps!

The arclength of some curve y =f(x) on the interval [a,b] = ∫√(1 + (dy/dx)²) dx

In this problem:

dy/dx = x√(x² + 2)

Arc-length = ∫√(1 + x²(x² + 2)) dx from 0 to 3

= ∫√(1 + 2x² + x^4) dx from 0 to 3

= ∫(x² + 1) dx from 0 to 3 = x^3/3 + x eval. from 0 to 3 = 9 + 3 = 12