Let's solve the problem step by step: Given function: \[ y = \frac{2}{3}x^{3/2} + 2 \] We are to **find the arc length** of the graph over the interval \( [0, 3] \). ### **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 \( \frac{dy}{dx} \)** \[ y = \frac{2}{3}x^{3/2} + 2 \] Differentiate: \[ \frac{dy}{dx} = \frac{2}{3} \cdot \frac{3}{2} x^{1/2} = x^{1/2} \] ### **Step 3: Substitute into Arc Length Formula** \[ L = \int_0^3 \sqrt{1 + (x^{1/2})^2} \, dx \] \[ = \int_0^3 \sqrt{1 + x} \, dx \] ### **Step 4: Evaluate the Integral** Let \( u = 1 + x \Rightarrow du = dx \). - When \( x = 0, u = 1 \) - When \( x = 3, u = 4 \) So, \[ L = \int_{1}^{4} \sqrt{u} \, du \] \[ = \int_{1}^{4} u^{1/2} du \] \[ = \left[ \frac{2}{3} u^{3/2} \right]_1^4 \] Calculate at the bounds: \[ \left[ \frac{2}{3} u^{3/2} \right]_1^4 = \frac{2}{3} (4^{3/2}) - \frac{2}{3} (1^{3/2}) \] - \( 4^{3/2} = (4^1)^{3/2} = (2^2)^{3/2} = 2^{3} = 8 \) - \( 1^{3/2} = 1 \) So, \[ = \frac{2}{3} (8 - 1) \] \[ = \frac{2}{3} \times 7 = \frac{14}{3} \] ### **Step 5: Final Answer (Rounded to Three Decimal Places)** \[ L = \frac{14}{3} \approx 4.667 \] --- ## **Final Answer** \[ \boxed{4.667} \] This is the arc length of the graph of the function from \( x = 0 \) to \( x = 3 \), rounded to three decimal places.