1. **State the problem:** Find the length of the curve defined by $y = 3 \cdot x^{\frac{3}{2}}$ for $0 \leq x \leq 2$. 2. **Formula for arc length:** The length $L$ of a curve $y=f(x)$ from $x=a$ to $x=b$ is given by $$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$ 3. **Find the derivative:** $$y = 3x^{\frac{3}{2}}$$ Using the power rule, $$\frac{dy}{dx} = 3 \cdot \frac{3}{2} x^{\frac{3}{2} - 1} = \frac{9}{2} x^{\frac{1}{2}} = \frac{9}{2} \sqrt{x}$$ 4. **Square the derivative:** $$\left(\frac{dy}{dx}\right)^2 = \left(\frac{9}{2} \sqrt{x}\right)^2 = \frac{81}{4} x$$ 5. **Set up the integral:** $$L = \int_0^2 \sqrt{1 + \frac{81}{4} x} \, dx = \int_0^2 \sqrt{1 + \frac{81}{4} x} \, dx$$ 6. **Substitute:** Let $$u = 1 + \frac{81}{4} x \implies du = \frac{81}{4} dx \implies dx = \frac{4}{81} du$$ When $x=0$, $u=1$; when $x=2$, $$u = 1 + \frac{81}{4} \times 2 = 1 + \frac{162}{4} = 1 + 40.5 = 41.5$$ 7. **Rewrite the integral:** $$L = \int_1^{41.5} \sqrt{u} \cdot \frac{4}{81} du = \frac{4}{81} \int_1^{41.5} u^{\frac{1}{2}} du$$ 8. **Integrate:** $$\int u^{\frac{1}{2}} du = \frac{2}{3} u^{\frac{3}{2}}$$ 9. **Evaluate the definite integral:** $$L = \frac{4}{81} \times \frac{2}{3} \left[u^{\frac{3}{2}}\right]_1^{41.5} = \frac{8}{243} \left(41.5^{\frac{3}{2}} - 1^{\frac{3}{2}}\right)$$ 10. **Calculate powers:** $$41.5^{\frac{3}{2}} = (\sqrt{41.5})^3 \approx (6.442)^3 \approx 267.5$$ 11. **Final length:** $$L \approx \frac{8}{243} (267.5 - 1) = \frac{8}{243} \times 266.5 \approx 8.78$$ **Answer:** The length of the curve is approximately **8.78** units.