Find the arc length of the following curve on the given interval. \( y=\frac{1}{3} x^{3 / 2} \) on \( [0,42] \) The length of the curve is (Type an integer or decimal rounded to two decimal places as needed.)

We want to compute the arc length of the curve \[ y = \frac{1}{3} x^{\frac{3}{2}} \] on the interval \([0,42]\). ### Step 1. Formula for Arc Length The arc length \( L \) of a curve \( y = f(x) \) from \( x=a \) to \( x=b \) is given by: \[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\, dx. \] ### Step 2. Compute the Derivative Given \[ y = \frac{1}{3} x^{\frac{3}{2}}, \] differentiate with respect to \( x \): \[ \frac{dy}{dx} = \frac{1}{3} \cdot \frac{3}{2} x^{\frac{1}{2}} = \frac{1}{2} x^{\frac{1}{2}}. \] Then, square the derivative: \[ \left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{2} x^{\frac{1}{2}}\right)^2 = \frac{1}{4} x. \] ### Step 3. Set Up the Integral Substitute into the arc length formula: \[ L = \int_{0}^{42} \sqrt{1 + \frac{1}{4}x}\, dx. \] ### Step 4. Use a Substitution Let \[ u = 1 + \frac{x}{4}. \] Then the derivative of \( u \) with respect to \( x \) is: \[ \frac{du}{dx} = \frac{1}{4} \quad \Longrightarrow \quad dx = 4\, du. \] Determine the new limits: - When \( x = 0 \), \( u = 1 + \frac{0}{4} = 1 \). - When \( x = 42 \), \( u = 1 + \frac{42}{4} = 1 + 10.5 = 11.5 \). Now, the integral becomes: \[ L = \int_{u=1}^{u=11.5} \sqrt{u} \cdot 4\, du = 4 \int_{1}^{11.5} u^{\frac{1}{2}}\, du. \] ### Step 5. Evaluate the Integral Integrate: \[ \int u^{\frac{1}{2}}\, du = \frac{2}{3} u^{\frac{3}{2}}. \] Thus, we have: \[ L = 4 \cdot \frac{2}{3} \left[u^{\frac{3}{2}}\right]_{1}^{11.5} = \frac{8}{3} \left( 11.5^{\frac{3}{2}} - 1^{\frac{3}{2}} \right). \] Since \( 1^{\frac{3}{2}} = 1 \), this simplifies to: \[ L = \frac{8}{3} \left( 11.5^{\frac{3}{2}} - 1 \right). \] ### Step 6. Compute the Numerical Value First, calculate \( 11.5^{\frac{3}{2}} \): \[ 11.5^{\frac{3}{2}} = \left(\sqrt{11.5}\right)^3. \] Using a calculator, \[ \sqrt{11.5} \approx 3.39116, \] so, \[ 11.5^{\frac{3}{2}} \approx 3.39116^3 \approx 39.00 \quad (\text{rounded}). \] Now, plug back into the formula: \[ L \approx \frac{8}{3} (39.00 - 1) = \frac{8}{3} \times 38.00. \] Multiply: \[ \frac{8 \times 38.00}{3} = \frac{304.00}{3} \approx 101.33. \] ### Final Answer The length of the curve is approximately: \[ \boxed{101.33}. \]