1. **Problem statement:** Find the length $L$ of the curve defined by the function $$f(y) = 3 \cdot (y + 4)^2$$ for $$0 \leq y \leq 4$$. 2. **Formula for arc length:** The length of a curve $y \mapsto f(y)$ from $y=a$ to $y=b$ is given by $$L = \int_a^b \sqrt{1 + \left(\frac{df}{dy}\right)^2} \, dy$$ 3. **Calculate the derivative:** $$\frac{df}{dy} = 3 \cdot 2 (y+4) = 6(y+4)$$ 4. **Substitute into the formula:** $$L = \int_0^4 \sqrt{1 + (6(y+4))^2} \, dy = \int_0^4 \sqrt{1 + 36(y+4)^2} \, dy$$ 5. **Use substitution:** Let $$u = y + 4$$, then when $$y=0, u=4$$ and when $$y=4, u=8$$. 6. **Rewrite the integral:** $$L = \int_4^8 \sqrt{1 + 36u^2} \, du$$ 7. **Integral formula:** $$\int \sqrt{a^2 + u^2} \, du = \frac{u}{2} \sqrt{a^2 + u^2} + \frac{a^2}{2} \ln\left|u + \sqrt{a^2 + u^2}\right| + C$$ Here, $$a^2 = \frac{1}{36}$$ is not correct because inside the root is $$1 + 36u^2$$, so rewrite as $$\sqrt{1 + (6u)^2}$$. Better to use $$a=1$$ and $$b=6$$, integral of $$\sqrt{1 + (b u)^2}$$ is $$\frac{u}{2} \sqrt{1 + b^2 u^2} + \frac{\sinh^{-1}(b u)}{2b} + C$$ 8. **Apply the formula:** $$L = \left[ \frac{u}{2} \sqrt{1 + 36 u^2} + \frac{\sinh^{-1}(6 u)}{12} \right]_4^8$$ 9. **Calculate values:** - For $$u=8$$: $$\frac{8}{2} \sqrt{1 + 36 \cdot 64} + \frac{\sinh^{-1}(48)}{12} = 4 \sqrt{1 + 2304} + \frac{\sinh^{-1}(48)}{12} = 4 \sqrt{2305} + \frac{\sinh^{-1}(48)}{12}$$ - For $$u=4$$: $$\frac{4}{2} \sqrt{1 + 36 \cdot 16} + \frac{\sinh^{-1}(24)}{12} = 2 \sqrt{1 + 576} + \frac{\sinh^{-1}(24)}{12} = 2 \sqrt{577} + \frac{\sinh^{-1}(24)}{12}$$ 10. **Numerical approximations:** - $$\sqrt{2305} \approx 48.02$$ - $$\sqrt{577} \approx 24.02$$ - $$\sinh^{-1}(48) = \ln(48 + \sqrt{48^2 + 1}) \approx \ln(48 + 48.01) = \ln(96.01) \approx 4.56$$ - $$\sinh^{-1}(24) = \ln(24 + \sqrt{24^2 + 1}) \approx \ln(24 + 24.02) = \ln(48.02) \approx 3.87$$ 11. **Calculate final length:** $$L \approx (4 \times 48.02 + \frac{4.56}{12}) - (2 \times 24.02 + \frac{3.87}{12}) = (192.08 + 0.38) - (48.04 + 0.32) = 192.46 - 48.36 = 144.10$$ **Final answer:** $$L \approx 144.10$$ (correct to 2 decimal places)