1. **State the problem:** We need to find the approximate area of the cross section of a river using Simpson's one-third rule given the depth $y$ at distances $x$ from one bank. 2. **Given data:** $$x = 0, 10, 20, 30, 40, 50, 60, 70, 80$$ $$y = 0, 4, 7, 9, 12, 15, 14, 8, 3$$ 3. **Formula for Simpson's one-third rule:** For an even number of intervals $n$, with equal spacing $h$, the area $A$ is approximated by: $$A = \frac{h}{3} \left[y_0 + 4(y_1 + y_3 + y_5 + \dots) + 2(y_2 + y_4 + y_6 + \dots) + y_n \right]$$ 4. **Check number of intervals:** There are 9 points, so $n=8$ intervals (which is even), and spacing $h = 10$ m. 5. **Apply the formula:** Identify terms: - $y_0 = 0$ - Odd indices sum: $y_1 + y_3 + y_5 + y_7 = 4 + 9 + 15 + 8 = 36$ - Even indices sum (excluding first and last): $y_2 + y_4 + y_6 = 7 + 12 + 14 = 33$ - $y_8 = 3$ 6. **Calculate area:** $$A = \frac{10}{3} \left[0 + 4(36) + 2(33) + 3 \right] = \frac{10}{3} \left[0 + 144 + 66 + 3 \right] = \frac{10}{3} \times 213$$ 7. **Simplify:** $$A = \frac{10}{3} \times 213 = \frac{\cancel{10}}{3} \times 213 = 10 \times 71 = 710$$ 8. **Final answer:** The approximate area of the cross section of the river is **710 square meters**.