1. **State the problem:** Find the arc length of the curve defined by the function $y = x^{\frac{3}{2}}$ from the point $(0,0)$ to $(4,8)$. 2. **Formula for arc length:** The arc length $L$ of a curve $y=f(x)$ from $x=a$ to $x=b$ is given by: $$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$ 3. **Find the derivative:** Given $y = x^{\frac{3}{2}}$, use the power rule: $$\frac{dy}{dx} = \frac{3}{2} x^{\frac{3}{2} - 1} = \frac{3}{2} x^{\frac{1}{2}} = \frac{3}{2} \sqrt{x}$$ 4. **Substitute into the formula:** $$L = \int_0^4 \sqrt{1 + \left(\frac{3}{2} \sqrt{x}\right)^2} \, dx = \int_0^4 \sqrt{1 + \frac{9}{4} x} \, dx$$ 5. **Simplify the integrand:** $$\sqrt{1 + \frac{9}{4} x} = \sqrt{\frac{4}{4} + \frac{9}{4} x} = \sqrt{\frac{4 + 9x}{4}} = \frac{\sqrt{4 + 9x}}{2}$$ 6. **Rewrite the integral:** $$L = \int_0^4 \frac{\sqrt{4 + 9x}}{2} \, dx = \frac{1}{2} \int_0^4 \sqrt{4 + 9x} \, dx$$ 7. **Use substitution:** Let $$u = 4 + 9x \implies du = 9 \, dx \implies dx = \frac{du}{9}$$ When $x=0$, $u=4$; when $x=4$, $u=4 + 9 \times 4 = 40$. 8. **Change limits and integral:** $$L = \frac{1}{2} \int_4^{40} \sqrt{u} \cdot \frac{1}{9} \, du = \frac{1}{18} \int_4^{40} u^{\frac{1}{2}} \, du$$ 9. **Integrate:** $$\int u^{\frac{1}{2}} \, du = \frac{2}{3} u^{\frac{3}{2}}$$ So, $$L = \frac{1}{18} \times \frac{2}{3} \left[u^{\frac{3}{2}}\right]_4^{40} = \frac{1}{27} \left(40^{\frac{3}{2}} - 4^{\frac{3}{2}}\right)$$ 10. **Evaluate powers:** $$40^{\frac{3}{2}} = (\sqrt{40})^3 = (6.324555)^3 \approx 252.982$$ $$4^{\frac{3}{2}} = (\sqrt{4})^3 = 2^3 = 8$$ 11. **Calculate the difference:** $$252.982 - 8 = 244.982$$ 12. **Final arc length:** $$L = \frac{244.982}{27} \approx 9.074$$ **Answer:** The arc length of the curve from $(0,0)$ to $(4,8)$ is approximately **9.074** (rounded to the nearest thousandth).