1. **State the problem:** Approximate the integral $$\int_0^3 8e^{\sqrt{x}} \, dx$$ using the midpoint rule with $n=6$ subintervals. 2. **Formula for the midpoint rule:** $$M_n = \Delta x \sum_{i=1}^n f\left(x_i^*\right)$$ where $\Delta x = \frac{b-a}{n}$ and $x_i^*$ is the midpoint of the $i$th subinterval. 3. **Calculate $\Delta x$:** $$\Delta x = \frac{3-0}{6} = 0.5$$ 4. **Determine midpoints $x_i^*$:** The subintervals are $[0,0.5], [0.5,1], [1,1.5], [1.5,2], [2,2.5], [2.5,3]$. Midpoints: $$x_1^* = 0.25, x_2^* = 0.75, x_3^* = 1.25, x_4^* = 1.75, x_5^* = 2.25, x_6^* = 2.75$$ 5. **Evaluate the function at each midpoint:** $$f(x) = 8e^{\sqrt{x}}$$ Calculate each: $$f(0.25) = 8e^{\sqrt{0.25}} = 8e^{0.5} \approx 8 \times 1.6487 = 13.1896$$ $$f(0.75) = 8e^{\sqrt{0.75}} = 8e^{0.8660} \approx 8 \times 2.3774 = 19.0192$$ $$f(1.25) = 8e^{\sqrt{1.25}} = 8e^{1.1180} \approx 8 \times 3.0590 = 24.4720$$ $$f(1.75) = 8e^{\sqrt{1.75}} = 8e^{1.3229} \approx 8 \times 3.7536 = 30.0288$$ $$f(2.25) = 8e^{\sqrt{2.25}} = 8e^{1.5} \approx 8 \times 4.4817 = 35.8536$$ $$f(2.75) = 8e^{\sqrt{2.75}} = 8e^{1.6583} \approx 8 \times 5.2503 = 42.0024$$ 6. **Sum the function values:** $$\sum_{i=1}^6 f(x_i^*) = 13.1896 + 19.0192 + 24.4720 + 30.0288 + 35.8536 + 42.0024 = 164.5656$$ 7. **Calculate the midpoint approximation:** $$M_6 = \Delta x \times \sum_{i=1}^6 f(x_i^*) = 0.5 \times 164.5656 = 82.2828$$ **Final answer:** $$\boxed{82.2828}$$ This is the approximate value of the integral using the midpoint rule with $n=6$ subintervals, rounded to four decimal places.