1. **State the problem:** We need to find the length $L$ of the curve defined by $y = 2x^{3/2}$ over the interval $0 \leq x \leq 4$. 2. **Formula for arc length:** The length of a curve $y=f(x)$ from $x=a$ to $x=b$ is given by $$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$ 3. **Find the derivative:** Given $y = 2x^{3/2}$, $$\frac{dy}{dx} = 2 \cdot \frac{3}{2} x^{\frac{3}{2} - 1} = 3x^{1/2} = 3\sqrt{x}$$ 4. **Substitute into the formula:** $$L = \int_0^4 \sqrt{1 + (3\sqrt{x})^2} \, dx = \int_0^4 \sqrt{1 + 9x} \, dx$$ 5. **Evaluate the integral:** Let $u = 1 + 9x$, then $du = 9 dx$ or $dx = \frac{du}{9}$. When $x=0$, $u=1$; when $x=4$, $u=1+36=37$. So, $$L = \int_1^{37} \sqrt{u} \cdot \frac{1}{9} \, du = \frac{1}{9} \int_1^{37} u^{1/2} \, du$$ 6. **Integrate:** $$\int u^{1/2} \, du = \frac{2}{3} u^{3/2}$$ Therefore, $$L = \frac{1}{9} \cdot \frac{2}{3} \left[u^{3/2}\right]_1^{37} = \frac{2}{27} \left(37^{3/2} - 1^{3/2}\right)$$ 7. **Calculate values:** $$37^{3/2} = (\sqrt{37})^3 \approx (6.082)^3 \approx 225.6$$ So, $$L \approx \frac{2}{27} (225.6 - 1) = \frac{2}{27} \times 224.6 \approx 16.63$$ **Final answer:** $$L \approx 16.63$$ The length of the curve $y=2x^{3/2}$ from $x=0$ to $x=4$ is approximately 16.63 units.