Find the arc length of y = x^2 from x = 0 to x = 1.

• A. (√5)/2 + (1/4) asinh(0)
• B. (√5)/2 + (1/4) asinh(1)
• C. (√5)/2 + (1/4) asinh(2)
• D. (√5)/2 + (1/4) asinh(3)

Answer: (C)

The arc length is found by integrating the slope into the length element: L = ∫ from 0 to 1 sqrt(1 + (dy/dx)^2) dx. Here y = x^2, so dy/dx = 2x, giving the integrand sqrt(1 + 4x^2).

Compute L = ∫_0^1 sqrt(4x^2 + 1) dx. Let u = 2x, so du = 2 dx and dx = du/2. The bounds go from u = 0 to u = 2, and

L = (1/2) ∫_0^2 sqrt(u^2 + 1) du.

Using the antiderivative ∫ sqrt(u^2 + 1) du = (u/2) sqrt(u^2 + 1) + (1/2) asinh(u) + C, we get

L = (1/2) * [ (u/2) sqrt(u^2 + 1) + (1/2) asinh(u) ]_0^2

= (1/4) [ u sqrt(u^2 + 1) + asinh(u) ]_0^2.

Evaluate at u = 2 and u = 0: u sqrt(u^2 + 1) = 2√5 and asinh(2) remains; at 0 everything is 0. Thus

L = (1/4) [ 2√5 + asinh(2) ] = (√5)/2 + (1/4) asinh(2).

So the arc length is (√5)/2 + (1/4) asinh(2).