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

To calculate the arc length of the curve \( y = x^{2} \) from \( x = 1 \) to \( x = 3 \), we'll use the formula for the arc length of a function \( y = f(x) \) between two points \( x = a \) and \( x = b \): \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] ### Step 1: Find the Derivative First, compute the derivative of \( y = x^{2} \) with respect to \( x \): \[ \frac{dy}{dx} = 2x \] ### Step 2: Set Up the Integral Plug the derivative into the arc length formula: \[ L = \int_{1}^{3} \sqrt{1 + (2x)^2} \, dx = \int_{1}^{3} \sqrt{1 + 4x^2} \, dx \] ### Step 3: Compute the Integral The integral \( \int \sqrt{1 + 4x^2} \, dx \) can be evaluated using a standard integration formula: \[ \int \sqrt{a x^2 + b} \, dx = \frac{x}{2} \sqrt{a x^2 + b} + \frac{b}{2 \sqrt{a}} \ln \left| x \sqrt{a} + \sqrt{a x^2 + b} \right| + C \] For our case, \( a = 4 \) and \( b = 1 \): \[ \int \sqrt{4x^2 + 1} \, dx = \frac{x}{2} \sqrt{4x^2 + 1} + \frac{1}{4} \ln \left(2x + \sqrt{4x^2 + 1}\right) + C \] ### Step 4: Evaluate the Definite Integral from 1 to 3 \[ L = \left[ \frac{x}{2} \sqrt{4x^2 + 1} + \frac{1}{4} \ln \left(2x + \sqrt{4x^2 + 1}\right) \right]_{1}^{3} \] Compute each part at \( x = 3 \) and \( x = 1 \): 1. **At \( x = 3 \):** \[ \sqrt{4(3)^2 + 1} = \sqrt{36 + 1} = \sqrt{37} \approx 6.08276253 \] \[ \frac{3}{2} \times 6.08276253 \approx 9.124144 \] \[ \ln(2 \times 3 + 6.08276253) = \ln(12.08276253) \approx 2.491427 \] \[ \frac{1}{4} \times 2.491427 \approx 0.622857 \] \[ \text{Total at } x = 3: \approx 9.124144 + 0.622857 = 9.746999 \] 2. **At \( x = 1 \):** \[ \sqrt{4(1)^2 + 1} = \sqrt{4 + 1} = \sqrt{5} \approx 2.23606798 \] \[ \frac{1}{2} \times 2.23606798 \approx 1.118034 \] \[ \ln(2 \times 1 + 2.23606798) = \ln(4.23606798) \approx 1.446919 \] \[ \frac{1}{4} \times 1.446919 \approx 0.361730 \] \[ \text{Total at } x = 1: \approx 1.118034 + 0.361730 = 1.479764 \] 3. **Subtract to Find the Arc Length:** \[ L \approx 9.746999 - 1.479764 = 8.267235 \] ### Final Answer The arc length of the curve \( y = x^{2} \) from \( x = 1 \) to \( x = 3 \) is approximately **8.267 units**.