Consider the following. \[ y=\frac{2}{3} x^{3 / 2}+7 \] \( \square \) Find the arc length of the graph of the function over the indicated interval. (Round your answer to three decimal places.) \( \square \)
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To find the arc length of the graph of the function \( y = \frac{2}{3}x^{3/2} + 7 \) over a given interval, we use the arc length formula: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] First, we need to calculate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{2}{3}x^{3/2} + 7 \right) = \frac{2}{3} \cdot \frac{3}{2} x^{1/2} = x^{1/2} \] Next, we substitute this into the arc length formula: \[ L = \int_{a}^{b} \sqrt{1 + (x^{1/2})^2} \, dx \] \[ = \int_{a}^{b} \sqrt{1 + x} \, dx \] Let’s calculate the definite integral over a specified interval \([a, b]\). For the example, let’s say \( a = 0 \) and \( b = 4 \): \[ L = \int_{0}^{4} \sqrt{1 + x} \, dx \] To solve this integral, we can use the substitution \( u = 1 + x \), so that \( du = dx \) and when \( x = 0 \), \( u = 1 \) and when \( x = 4 \), \( u = 5 \). Thus, we have: \[ L = \int_{1}^{5} \sqrt{u} \, du = \int_{1}^{5} u^{1/2} \, du \] Now, integrating: \[ = \left[ \frac{u^{3/2}}{\frac{3}{2}} \right]_{1}^{5} = \frac{2}{3} [u^{3/2}]_{1}^{5} = \frac{2}{3} \left[ 5^{3/2} - 1^{3/2} \right] \] Calculating \( 5^{3/2} \) gives \( 5^{3/2} = 5 \sqrt{5} \). Now, calculating: \[ L = \frac{2}{3} \left( 5 \sqrt{5} - 1 \right) \] Evaluating \( 5 \sqrt{5} \) numerically, we know \( \sqrt{5} \approx 2.236 \), thus: \[ 5 \sqrt{5} \approx 5 \cdot 2.236 \approx 11.18 \] Then: \[ L \approx \frac{2}{3} (11.18 - 1) \approx \frac{2}{3} \cdot 10.18 \approx \frac{20.36}{3} \approx 6.79 \] Rounding to three decimal places, the arc length \( L \) is approximately: \[ \boxed{6.792} \]