Consider the curve defined by the equation y = 4x^3 +3x. Set up an integral that represents the length of curve from the point (0,0) to the point (4,268).

∫[0,4] √(1+y'^2) dx

= ∫[0,4] √(1+(12x^2+3)^2) dx

To find the length of the curve from the point (0,0) to the point (4,268), we need to set up an integral using the arc length formula. The arc length formula for a curve defined by the equation y = f(x) is given by:

L = ∫[a,b] sqrt(1 + (f'(x))^2) dx

Where a and b are the x-coordinates of the starting and ending points.

In this case, the starting point is (0,0) and the ending point is (4,268). The equation of the curve is y = 4x^3 + 3x.

First, let's find the derivative of the function f(x) = 4x^3 + 3x:

f'(x) = 12x^2 + 3

Now, we can set up the integral:

L = ∫[0,4] sqrt(1 + (12x^2 + 3)^2) dx

Simplifying further:

L = ∫[0,4] sqrt(1 + 144x^4 + 72x^2 + 9) dx

Now we can integrate this expression to find the length of the curve from (0,0) to (4,268).

To find the length of a curve between two points, we can use the arc length formula:

L = ∫[a, b] √(1 + (dy/dx)^2) dx

In this case, we want to find the length of the curve from point (0,0) to (4,268). We need to find the interval [a, b] within which the curve lies.

Let's first find the x-values of the two points:
For point (0,0), x = 0
For point (4,268), x = 4

Now, we need to find the derivative dy/dx of the curve y = 4x^3 + 3x:
dy/dx = 12x^2 + 3

With the interval and the derivative, we can set up the definite integral for the length of the curve:

L = ∫[0, 4] √(1 + (12x^2 + 3)^2) dx