Calculate the arc length of the curve \( y = x^{2} \) from \( x = 1 \) to \( x = 3 \).

To find the arc length of the curve \( y = x^{2} \) from \( x = 1 \) to \( x = 3 \), you can use the arc length formula: \[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] First, we need to find \( \frac{dy}{dx} \) for \( y = x^2 \): \[ \frac{dy}{dx} = 2x \] Now, we square this derivative: \[ \left(\frac{dy}{dx}\right)^2 = (2x)^2 = 4x^2 \] Substituting this into the arc length formula gives us: \[ L = \int_{1}^{3} \sqrt{1 + 4x^2} \, dx \] Now, we need to evaluate the integral. The integral \( \int \sqrt{1 + 4x^2} \, dx \) can be solved using trigonometric substitution or other methods, but it can be complex. For this specific integral, we can utilize a calculator or integration techniques to find: \[ L = \left[ \frac{x}{2} \sqrt{1 + 4x^2} + \frac{1}{4} \ln \left| 2x + \sqrt{1 + 4x^2} \right| \right]_{1}^{3} \] Evaluating the definite integral at the bounds \( x = 1 \) and \( x = 3 \): 1. At \( x = 3 \): \[ L(3) = \frac{3}{2} \sqrt{1 + 4(3^2)} + \frac{1}{4} \ln |2(3) + \sqrt{1 + 4(3^2)}| \] \[ = \frac{3}{2} \sqrt{37} + \frac{1}{4} \ln (6 + \sqrt{37}) \] 2. At \( x = 1 \): \[ L(1) = \frac{1}{2} \sqrt{5} + \frac{1}{4} \ln (2 + \sqrt{5}) \] Now, calculating \( L = L(3) - L(1) \) will give the final arc length. After evaluating both parts, the arc length \( L \) from \( x = 1 \) to \( x = 3 \) is approximately: \[ L \approx 8.713 \] Thus, the arc length of the curve \( y = x^2 \) from \( x = 1 \) to \( x = 3 \) is \( \approx 8.713 \) units.