 In order to find arc length, we'll need to find a representative length and sum it. Suppose we have the curve y equals f of x. A representative length is that between the point x, y, and another point x plus a small bit of the x-axis, y plus a small bit of the y-axis, x plus dx, y plus dy. We can use the distance formula to find the length of the segment joining these two points. And even though it isn't a fraction, part of the reason why the dy dx notation is useful is that it reminds us of things that we can do. In this case, we can rewrite this expression. We'll factor out a dx squared. Or, since dy dx is the derivative of our function, we have our expression for the representative length of the arc. We can then sum this over the entirety of the curve. And this gives us our theorem. Let y equals f of x be a curve that is continuous and differentiable over the interval between a and b. Then the arc length of the curve is going to be the integral from a to b of square root 1 plus f prime of x squared. For example, let's find the length of the curve y equals x to power 3 halves over the interval between 0 and 4. So our arc length is going to be the sum of the representative length, square root 1 plus f prime of x squared, and our sum is going to go from the start of the curve at x equals 0 to the end of the curve at x equals 4. We need the derivative of our function, and we can do a little bit of algebraic cleanup. To evaluate this, we'll use a u substitution. We'll try u equals 1 plus 9 fourths x, and so du will be 9 fourths x. We need a 9 fourths dx, so we'll put it in as long as we take it out by multiplying by 4 ninths. Making our u substitutions, we can then find the integral. Putting everything back where we found it gives us the function we'll need to evaluate. Since everything is multiplied by 827s, we'll remove it as a common factor. So we need to evaluate our function at 4 and at 0. And while this is a perfectly good answer, we'll rewrite this without the fractional exponents because we can.