1. **State the problem:** Calculate the approximate value of the integral $$\int_0^3 \frac{dx}{x^3 + 9}$$ using Simpson's Rule with $N=6$ subintervals. 2. **Formula for Simpson's Rule:** $$S_N = \frac{\Delta x}{3} \left[f(x_0) + 4 \sum_{i=1,3,5}^{N-1} f(x_i) + 2 \sum_{i=2,4}^{N-2} f(x_i) + f(x_N)\right]$$ where $\Delta x = \frac{b-a}{N}$ and $x_i = a + i\Delta x$. 3. **Calculate $\Delta x$:** $$\Delta x = \frac{3-0}{6} = 0.5$$ 4. **Calculate the $x_i$ values:** $$x_0=0, x_1=0.5, x_2=1, x_3=1.5, x_4=2, x_5=2.5, x_6=3$$ 5. **Evaluate $f(x) = \frac{1}{x^3 + 9}$ at each $x_i$:** $$f(0) = \frac{1}{0 + 9} = \frac{1}{9} \approx 0.11111$$ $$f(0.5) = \frac{1}{0.5^3 + 9} = \frac{1}{0.125 + 9} = \frac{1}{9.125} \approx 0.10959$$ $$f(1) = \frac{1}{1 + 9} = \frac{1}{10} = 0.1$$ $$f(1.5) = \frac{1}{3.375 + 9} = \frac{1}{12.375} \approx 0.08081$$ $$f(2) = \frac{1}{8 + 9} = \frac{1}{17} \approx 0.05882$$ $$f(2.5) = \frac{1}{15.625 + 9} = \frac{1}{24.625} \approx 0.04061$$ $$f(3) = \frac{1}{27 + 9} = \frac{1}{36} \approx 0.02778$$ 6. **Apply Simpson's Rule formula:** $$S_6 = \frac{0.5}{3} \left[f(0) + 4(f(0.5) + f(1.5) + f(2.5)) + 2(f(1) + f(2)) + f(3)\right]$$ 7. **Calculate sums:** $$4 \times (0.10959 + 0.08081 + 0.04061) = 4 \times 0.23101 = 0.92404$$ $$2 \times (0.1 + 0.05882) = 2 \times 0.15882 = 0.31764$$ 8. **Sum all terms inside brackets:** $$0.11111 + 0.92404 + 0.31764 + 0.02778 = 1.38057$$ 9. **Calculate final approximation:** $$S_6 = \frac{0.5}{3} \times 1.38057 = \frac{0.5}{3} \times 1.38057 = 0.23010$$ **Final answer:** $$\boxed{0.23010}$$ (rounded to five decimal places)