1. **Problem statement:** We need to find the volume of material used to build a river bridge arch shaped like a downward-opening parabola. The bridge is 4 meters wide, with a span of 6 meters at the base and a maximum height of 4.5 meters at the center. The sides have different heights: 4.5 m on the left and 5.5 m on the right, with 1 m widths at the base on both sides. 2. **Modeling the parabola:** Let the horizontal axis $x$ run from $-3$ m to $3$ m (total span 6 m), with $x=0$ at the center. The parabola opens downward with vertex at $(0,4.5)$. 3. **Parabola equation:** The general form is $y = a x^2 + b x + c$. Since the parabola is symmetric about the center, $b=0$. The vertex form is $y = a x^2 + 4.5$. 4. **Find $a$ using the base points:** At $x=\pm 3$, $y=0$ (base of the arch). So, $$0 = a \cdot 3^2 + 4.5 \Rightarrow 0 = 9a + 4.5 \Rightarrow a = -\frac{4.5}{9} = -0.5$$ 5. **Parabola equation:** $$y = -0.5 x^2 + 4.5$$ 6. **Cross-sectional area:** The bridge width is 4 m, with 1 m on each side at the base, so the arch width is 2 m at the top (4 m total width minus 1 m on each side). The cross-section is the area under the parabola from $x=-1$ to $x=1$ (the 2 m width of the arch). 7. **Calculate the area under the parabola from $x=-1$ to $x=1$:** $$A = \int_{-1}^{1} (-0.5 x^2 + 4.5) dx$$ 8. **Integrate:** $$A = \left[-0.5 \frac{x^3}{3} + 4.5 x \right]_{-1}^{1} = \left(-\frac{0.5}{3} x^3 + 4.5 x\right)_{-1}^{1}$$ 9. **Evaluate at bounds:** $$A = \left(-\frac{0.5}{3} (1)^3 + 4.5 (1)\right) - \left(-\frac{0.5}{3} (-1)^3 + 4.5 (-1)\right)$$ $$= \left(-\frac{0.5}{3} + 4.5\right) - \left(\frac{0.5}{3} - 4.5\right) = \left(4.5 - \frac{0.5}{3}\right) - \left(-4.5 + \frac{0.5}{3}\right)$$ 10. **Simplify:** $$= 4.5 - \frac{0.5}{3} + 4.5 - \frac{0.5}{3} = 9 - \frac{1}{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3} \approx 8.67 \text{ m}^2$$ 11. **Volume calculation:** The length of the bridge (depth) is 4 m. Volume $V = A \times \text{length} = 8.67 \times 4 = 34.67$ m$^3$. **Final answer:** The volume of material used is approximately **34.67 m$^3$**.