1. **State the problem:** We want to approximate the area under the curve of the function $f(x) = \sqrt{x^3 + 1}$ from $x=0$ to $x=2$ using the Midpoint (Mid-ordinate) Rule with six ordinates. 2. **Formula for Midpoint Rule:** The Midpoint Rule for $n$ subintervals is given by: $$\text{Area} \approx \Delta x \sum_{i=1}^n f\left(x_i^*\right)$$ where $\Delta x = \frac{b-a}{n}$ and $x_i^*$ is the midpoint of the $i$th subinterval. 3. **Calculate $\Delta x$:** $$\Delta x = \frac{2-0}{6} = \frac{2}{6} = \frac{1}{3}$$ 4. **Find midpoints $x_i^*$:** The subintervals are $[0, \frac{1}{3}], [\frac{1}{3}, \frac{2}{3}], [\frac{2}{3}, 1], [1, \frac{4}{3}], [\frac{4}{3}, \frac{5}{3}], [\frac{5}{3}, 2]$. Midpoints are: $$x_1^* = \frac{0 + \frac{1}{3}}{2} = \frac{1}{6}$$ $$x_2^* = \frac{\frac{1}{3} + \frac{2}{3}}{2} = \frac{1}{2}$$ $$x_3^* = \frac{\frac{2}{3} + 1}{2} = \frac{5}{6}$$ $$x_4^* = \frac{1 + \frac{4}{3}}{2} = \frac{7}{6}$$ $$x_5^* = \frac{\frac{4}{3} + \frac{5}{3}}{2} = \frac{3}{2}$$ $$x_6^* = \frac{\frac{5}{3} + 2}{2} = \frac{11}{6}$$ 5. **Evaluate $f(x_i^*) = \sqrt{(x_i^*)^3 + 1}$:** $$f\left(\frac{1}{6}\right) = \sqrt{\left(\frac{1}{6}\right)^3 + 1} = \sqrt{\frac{1}{216} + 1} = \sqrt{\frac{217}{216}}$$ $$f\left(\frac{1}{2}\right) = \sqrt{\left(\frac{1}{2}\right)^3 + 1} = \sqrt{\frac{1}{8} + 1} = \sqrt{\frac{9}{8}}$$ $$f\left(\frac{5}{6}\right) = \sqrt{\left(\frac{5}{6}\right)^3 + 1} = \sqrt{\frac{125}{216} + 1} = \sqrt{\frac{341}{216}}$$ $$f\left(\frac{7}{6}\right) = \sqrt{\left(\frac{7}{6}\right)^3 + 1} = \sqrt{\frac{343}{216} + 1} = \sqrt{\frac{559}{216}}$$ $$f\left(\frac{3}{2}\right) = \sqrt{\left(\frac{3}{2}\right)^3 + 1} = \sqrt{\frac{27}{8} + 1} = \sqrt{\frac{35}{8}}$$ $$f\left(\frac{11}{6}\right) = \sqrt{\left(\frac{11}{6}\right)^3 + 1} = \sqrt{\frac{1331}{216} + 1} = \sqrt{\frac{1547}{216}}$$ 6. **Calculate the approximate area:** $$\text{Area} \approx \Delta x \sum_{i=1}^6 f\left(x_i^*\right) = \frac{1}{3} \left( \sqrt{\frac{217}{216}} + \sqrt{\frac{9}{8}} + \sqrt{\frac{341}{216}} + \sqrt{\frac{559}{216}} + \sqrt{\frac{35}{8}} + \sqrt{\frac{1547}{216}} \right)$$ 7. **Numerical approximation:** $$\sqrt{\frac{217}{216}} \approx 1.0023$$ $$\sqrt{\frac{9}{8}} \approx 1.0607$$ $$\sqrt{\frac{341}{216}} \approx 1.2550$$ $$\sqrt{\frac{559}{216}} \approx 1.6078$$ $$\sqrt{\frac{35}{8}} \approx 2.0917$$ $$\sqrt{\frac{1547}{216}} \approx 2.6787$$ Sum: $$1.0023 + 1.0607 + 1.2550 + 1.6078 + 2.0917 + 2.6787 = 9.6962$$ Area: $$\frac{1}{3} \times 9.6962 = 3.2321$$ **Final answer:** The approximate area bounded by $y=\sqrt{x^3+1}$ from $0$ to $2$ using the Midpoint Rule with six ordinates is approximately $3.232$.