1. The definite integral from 3 to 10 of $x$ dx is calculated using the formula for the integral of $x$, which is $\frac{x^2}{2}$. Evaluating from 3 to 10: $$\int_3^{10} x \, dx = \left[ \frac{x^2}{2} \right]_3^{10} = \frac{10^2}{2} - \frac{3^2}{2} = \frac{100}{2} - \frac{9}{2} = \frac{91}{2} = 45.5$$ 2. The summation from $i=2$ to $6$ of $i$ is: $$\sum_{i=2}^6 i = 2 + 3 + 4 + 5 + 6 = 20$$ 3. The logarithm base 3 of 23 is approximately: $$\log_3(23) \approx 2.85$$ 4. The factorial of 5 is: $$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$ 5. The fraction $\frac{13}{19}$ is approximately: $$\frac{13}{19} \approx 0.6842$$ 6. The value of $e^3$ is approximately: $$e^3 \approx 20.0855$$ 7. The infinity symbol $\infty$ represents an unbounded quantity and is not a number. 8. The fraction $\frac{2\pi}{4}$ simplifies to: $$\frac{2\pi}{4} = \frac{\pi}{2} \approx 1.5708$$ 9. The square root of 5 is approximately: $$\sqrt{5} \approx 2.2361$$ Now, ordering the numerical values from least to greatest (excluding infinity): $$0.6842, 1.5708, 2.2361, 2.85, 20.0855, 20, 45.5, 120$$ Note: The summation is 20, which is less than $e^3 \approx 20.0855$. Final ordered list: $$\frac{13}{19} < \frac{2\pi}{4} < \sqrt{5} < \log_3(23) < \sum_{i=2}^6 i < e^3 < \int_3^{10} x \, dx < 5!$$