1. **State the problem:** We need to find the length $L$ of the curve defined by the function $$f(y) = 4 \cdot (y + 1)^2$$ over the interval $$3 \leq y \leq 6$$. 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:** $$ \frac{df}{dy} = 4 \cdot 2 (y + 1) = 8(y + 1) $$ 4. **Substitute into the formula:** $$ L = \int_3^6 \sqrt{1 + (8(y + 1))^2} \, dy = \int_3^6 \sqrt{1 + 64(y + 1)^2} \, dy $$ 5. **Simplify the integral:** Let $u = y + 1$, then when $y=3$, $u=4$ and when $y=6$, $u=7$. $$ L = \int_4^7 \sqrt{1 + 64u^2} \, du $$ 6. **Use the standard 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, $a^2 = \frac{1}{64}$, but since we have $1 + 64u^2$, rewrite as $64u^2 + 1$. Rewrite the integral as: $$ L = \int_4^7 \sqrt{64u^2 + 1} \, du $$ 7. **Factor out 64 inside the square root:** $$ \sqrt{64u^2 + 1} = \sqrt{64\left(u^2 + \frac{1}{64}\right)} = 8 \sqrt{u^2 + \frac{1}{64}} $$ 8. **Rewrite the integral:** $$ L = \int_4^7 8 \sqrt{u^2 + \frac{1}{64}} \, du = 8 \int_4^7 \sqrt{u^2 + \left(\frac{1}{8}\right)^2} \, du $$ 9. **Apply the formula with $a = \frac{1}{8}$:** $$ \int \sqrt{u^2 + a^2} \, du = \frac{u}{2} \sqrt{u^2 + a^2} + \frac{a^2}{2} \ln\left|u + \sqrt{u^2 + a^2}\right| + C $$ 10. **Evaluate the definite integral:** $$ L = 8 \left[ \frac{u}{2} \sqrt{u^2 + \left(\frac{1}{8}\right)^2} + \frac{\left(\frac{1}{8}\right)^2}{2} \ln\left|u + \sqrt{u^2 + \left(\frac{1}{8}\right)^2}\right| \right]_4^7 $$ 11. **Calculate constants:** $$ \left(\frac{1}{8}\right)^2 = \frac{1}{64} $$ 12. **Plug in the limits:** $$ L = 8 \left[ \frac{7}{2} \sqrt{49 + \frac{1}{64}} + \frac{1}{128} \ln\left(7 + \sqrt{49 + \frac{1}{64}}\right) - \left( \frac{4}{2} \sqrt{16 + \frac{1}{64}} + \frac{1}{128} \ln\left(4 + \sqrt{16 + \frac{1}{64}}\right) \right) \right] $$ 13. **Simplify the square roots:** $$ \sqrt{49 + \frac{1}{64}} = \sqrt{\frac{3136}{64} + \frac{1}{64}} = \sqrt{\frac{3137}{64}} = \frac{\sqrt{3137}}{8} $$ $$ \sqrt{16 + \frac{1}{64}} = \sqrt{\frac{1024}{64} + \frac{1}{64}} = \sqrt{\frac{1025}{64}} = \frac{\sqrt{1025}}{8} $$ 14. **Substitute back:** $$ L = 8 \left[ \frac{7}{2} \cdot \frac{\sqrt{3137}}{8} + \frac{1}{128} \ln\left(7 + \frac{\sqrt{3137}}{8}\right) - \left( 2 \cdot \frac{\sqrt{1025}}{8} + \frac{1}{128} \ln\left(4 + \frac{\sqrt{1025}}{8}\right) \right) \right] $$ 15. **Simplify coefficients:** $$ \frac{7}{2} \cdot \frac{\sqrt{3137}}{8} = \frac{7 \sqrt{3137}}{16} $$ $$ 2 \cdot \frac{\sqrt{1025}}{8} = \frac{\sqrt{1025}}{4} $$ 16. **Final expression:** $$ L = 8 \left[ \frac{7 \sqrt{3137}}{16} + \frac{1}{128} \ln\left(7 + \frac{\sqrt{3137}}{8}\right) - \frac{\sqrt{1025}}{4} - \frac{1}{128} \ln\left(4 + \frac{\sqrt{1025}}{8}\right) \right] $$ 17. **Multiply through by 8:** $$ L = 8 \cdot \frac{7 \sqrt{3137}}{16} + 8 \cdot \frac{1}{128} \ln\left(7 + \frac{\sqrt{3137}}{8}\right) - 8 \cdot \frac{\sqrt{1025}}{4} - 8 \cdot \frac{1}{128} \ln\left(4 + \frac{\sqrt{1025}}{8}\right) $$ $$ = \frac{7 \sqrt{3137}}{2} + \frac{1}{16} \ln\left(7 + \frac{\sqrt{3137}}{8}\right) - 2 \sqrt{1025} - \frac{1}{16} \ln\left(4 + \frac{\sqrt{1025}}{8}\right) $$ 18. **This is the exact length of the curve.** **Answer:** $$ L = \frac{7 \sqrt{3137}}{2} - 2 \sqrt{1025} + \frac{1}{16} \ln\left(\frac{7 + \frac{\sqrt{3137}}{8}}{4 + \frac{\sqrt{1025}}{8}}\right) $$