Find the exact length of the curve

Find the exact length of the curve y = 2 + 4x^(3/2), 0 ≤ x ≤ 1

y = 2 + 4 * x^(3/2)
dy/dx = 4 * (3/2) * x^(1/2) = 6 * x^(1/2)

sqrt(1 + (dy/dx)^2) * dx =>
sqrt(1 + 36 * x) * dx

u = 1 + 36x
du = 36 * dx

Now we have:

(1/36) * u^(1/2) * du

Integrate:

(1/36) * u^(3/2) * (2/3) + C =>
(1/54) * (1 + 36x)^(3/2)

From 0 to 1

(1/54) * (1 + 36)^(3/2) - (1/54) * (1 + 0)^(3/2) =>
(1/54) * (37 * sqrt(37) - 1)

The length of some curve y = f(x) on the interval [a,b] is defined as:

∫√(1 + (dy/dx)²) dx from a to b

For this problem:

dy/dx = 6√x

Arc length = ∫√(1 + 36x) dx from 0 to 1

u = 1 + 36x

du/36 = dx

1/36*∫√u du from 1 to 37

1/54*[(37)^(3/2) - 1]

formula:integral of sqrt(1+(f'(x))^2)


f'(x)=6sqrt(x)

f'(x)^2=36x


integral[sqrt(1+36x)]

u=1+36x
du=36dx

du/36=dx

1/36*integral[sqrt(u)*du] from 0 to 1

1/36*[2/3 * u^(3/2)]

1/36*[2/3(1+36x)^3/2]

plug in values

1/54[37^(3/2)-1] is the answer.