To evaluate the Riemann sum of the function $f(x) = 2x^2 - 7$ over the interval $[0, 3]$ with $n = 6$, taking the sample points to be midpoints, we'll follow these steps:

1. Determine the width of each subinterval:

The interval $[0, 3]$ is divided into $n = 6$ subintervals. The width $\Delta x$ of each subinterval is: $\Delta x = \frac{3 - 0}{6} = \frac{3}{6} = 0.5$

2. Identify the midpoints of each subinterval:

The subintervals are $[0, 0.5]$, $[0.5, 1]$, $[1, 1.5]$, $[1.5, 2]$, $[2, 2.5]$, and $[2.5, 3]$.

The midpoints $x_i^*$ of these subintervals are: $x_1^* = 0.25, \quad x_2^* = 0.75, \quad x_3^* = 1.25, \quad x_4^* = 1.75, \quad x_5^* = 2.25, \quad x_6^* = 2.75$

3. Evaluate the function at each midpoint:

Now we evaluate $f(x)$ at each midpoint $x_i^*$: $f(0.25) = 2(0.25)^2 - 7 = 2(0.0625) - 7 = 0.125 - 7 = -6.875$ $f(0.75) = 2(0.75)^2 - 7 = 2(0.5625) - 7 = 1.125 - 7 = -5.875$ $f(1.25) = 2(1.25)^2 - 7 = 2(1.5625) - 7 = 3.125 - 7 = -3.875$ $f(1.75) = 2(1.75)^2 - 7 = 2(3.0625) - 7 = 6.125 - 7 = -0.875$ $f(2.25) = 2(2.25)^2 - 7 = 2(5.0625) - 7 = 10.125 - 7 = 3.125$ $f(2.75) = 2(2.75)^2 - 7 = 2(7.5625) - 7 = 15.125 - 7 = 8.125$

4. Calculate the Riemann sum:

The Riemann sum is given by: $\text{Riemann Sum} = \sum_{i=1}^{6} f(x_i^*) \Delta x$ Substituting the values we have: $\text{Riemann Sum} = (-6.875)(0.5) + (-5.875)(0.5) + (-3.875)(0.5) + (-0.875)(0.5) + (3.125)(0.5) + (8.125)(0.5)$ $\text{Riemann Sum} = -3.4375 - 2.9375 - 1.9375 - 0.4375 + 1.5625 + 4.0625$ $\text{Riemann Sum} = -12.25 + 5.625 = -6.625$

Final Answer:

The Riemann sum is $\boxed{-6.625}$.

Would you like further details or have any questions?

Here are 5 related questions you might find interesting:

1. How does the accuracy of the Riemann sum change as $n$ increases?
2. What is the difference between a Riemann sum and a definite integral?
3. How would the result change if you used left endpoints instead of midpoints?
4. What does the negative result of the Riemann sum represent in this context?
5. Can you calculate the exact value of the definite integral and compare it with the Riemann sum?

Tip: When using Riemann sums, midpoints often give a better approximation than using left or right endpoints, especially for functions that are not linear.