Let's solve this problem step by step.

Problem Setup:

The function is $f(x) = 27 - 2x^2$, and the interval is $[0, 12]$. We partition it into 4 subintervals of equal length. For each subinterval $[x_{k-1}, x_k]$, the midpoint $c_k$ is given as: $c_k = \frac{2x_{k-1} + x_k}{3}.$

We calculate the Riemann sum $S_4$, which is the sum of the function values at $c_k$ multiplied by the subinterval width.

Step 1: Determine the width of each subinterval

The interval $[0, 12]$ is divided into 4 subintervals, so the width $\Delta x$ is: $\Delta x = \frac{12 - 0}{4} = 3.$

The subintervals are: [ [0, 3], [3, 6], [6, 9], [9, 12]. ]

Step 2: Calculate $c_k$ for each subinterval

Using the formula $c_k = \frac{2x_{k-1} + x_k}{3}$, compute $c_k$ for each subinterval:

1. For $[0, 3]$: $c_1 = \frac{2(0) + 3}{3} = 1$,
2. For $[3, 6]$: $c_2 = \frac{2(3) + 6}{3} = 4$,
3. For $[6, 9]$: $c_3 = \frac{2(6) + 9}{3} = 7$,
4. For $[9, 12]$: $c_4 = \frac{2(9) + 12}{3} = 10$.

Step 3: Evaluate $f(c_k)$ for each $c_k$

The function is $f(x) = 27 - 2x^2$. Compute $f(c_k)$ for each midpoint:

1. $f(c_1) = 27 - 2(1)^2 = 27 - 2 = 25$,
2. $f(c_2) = 27 - 2(4)^2 = 27 - 32 = -5$,
3. $f(c_3) = 27 - 2(7)^2 = 27 - 98 = -71$,
4. $f(c_4) = 27 - 2(10)^2 = 27 - 200 = -173$.

Step 4: Calculate $S_4$

The Riemann sum $S_4$ is: $S_4 = \Delta x \cdot \sum_{k=1}^4 f(c_k).$

Substitute the values: $S_4 = 3 \cdot (f(c_1) + f(c_2) + f(c_3) + f(c_4)).$

Substitute $f(c_1), f(c_2), f(c_3), f(c_4)$: $S_4 = 3 \cdot (25 + (-5) + (-71) + (-173)).$

Simplify: $S_4 = 3 \cdot (-224) = -672.$

Final Answer:

$S_4 = -672.$

Would you like further explanation or a visualization of this calculation? Here are 5 related questions:

1. How do Riemann sums relate to the integral of a function?
2. Can we compute $S_4$ using left or right endpoints instead of midpoints?
3. How does the choice of $c_k$ affect the accuracy of the Riemann sum?
4. What is the geometric interpretation of a negative Riemann sum?
5. How can this calculation be extended to more subintervals or exact integration?

Tip: Always double-check the midpoints $c_k$ and function evaluations to avoid errors in Riemann sum problems.