Find the arc length of y = x^4 from the origin to (1, 1). Answer in length units, correct to two decimal places, please.

The integral required is:

I=∫sqrt(1+f'(x)^2)dx from 0 to 1
=∫sqrt(1+(4x^3)^2)dx from 0 to 1

An analytic result involves elliptic integrals. Numerical integration would therefore be recommended.

I=1.600... (to 3 decimal places)

To find the arc length of the curve y = x^4 from the origin to (1, 1), we can use the formula for arc length in Cartesian coordinates:

L = ∫(a to b) √(1 + (dy/dx)^2) dx

In this case, the curve is y = x^4, and we want to find the arc length from the origin (0, 0) to (1, 1). So our values for a and b are 0 and 1, respectively.

First, let's find dy/dx:

dy/dx = d/dx (x^4)
= 4x^3

Now, we can substitute the expression for dy/dx into the formula for arc length:

L = ∫(0 to 1) √(1 + (4x^3)^2) dx
= ∫(0 to 1) √(1 + 16x^6) dx

To evaluate this integral, we can use a numerical method, such as Simpson's rule or the Trapezoidal rule. Let's use the Trapezoidal rule for simplicity:

L ≈ Δx/2 [f(x0) + 2f(x1) + 2f(x2) + ... + 2f(xn-1) + f(xn)]

where Δx = (b - a)/n, and n is the number of subintervals.

Let's choose n = 100 for a reasonably accurate approximation. Therefore, Δx = (1 - 0)/100 = 0.01.

L ≈ 0.01/2 [f(0) + 2f(0.01) + 2f(0.02) + ... + 2f(0.99) + f(1)]

Now, we substitute these values into the formula and calculate the sum:

L ≈ 0.01/2 [f(0) + 2f(0.01) + 2f(0.02) + ... + 2f(0.99) + f(1)]
≈ 0.01/2 [0 + 2(√(1 + 16(0.01)^6) + 2(√(1 + 16(0.02)^6) + ... + 2(√(1 + 16(0.99)^6) + (√(1 + 16(1)^6)]

Using a numerical method or a computer program, we can calculate this sum and find the approximate value of L. Round to two decimal places as requested.