Use a Riemann sum with 4 rectangles of equal width to approximate the area between y equals x cubed plus 8 and the x-axis on the interval left square bracket short dash 2 comma 1 right square bracket. Use the right-hand endpoint of each subinterval.

23.203125 space u n i t s squared

25.4953125 space u n i t s squared

16.453125 space u n i t s squared

29.42578125 space u n i t s squared

To approximate the area under the curve \( y = x^3 + 8 \) on the interval \([-2, 1]\) using a Riemann sum with 4 rectangles of equal width, and using the right-hand endpoint of each subinterval, we follow these steps:

### Step 1: Determine the width of each rectangle
The interval \([-2, 1]\) has a length of:
\[
\text{Interval Length} = 1 - (-2) = 3
\]
Since we're using 4 rectangles, the width \(\Delta x\) of each rectangle is:
\[
\Delta x = \frac{3}{4} = 0.75
\]

### Step 2: Identify the right-hand endpoints
The right-hand endpoints of the subintervals can be found by starting at the left end of the interval and adding the width of the rectangles incrementally:

- Subinterval 1: \(x_1 = -2 + 0.75 = -1.25\)
- Subinterval 2: \(x_2 = -1.25 + 0.75 = -0.5\)
- Subinterval 3: \(x_3 = -0.5 + 0.75 = 0.25\)
- Subinterval 4: \(x_4 = 0.25 + 0.75 = 1\)

### Step 3: Calculate the function values at these right-hand endpoints
We need to evaluate the function \( y = x^3 + 8 \) at each right-hand endpoint:

- \( f(-1.25) = (-1.25)^3 + 8 = -1.953125 + 8 = 6.046875 \)
- \( f(-0.5) = (-0.5)^3 + 8 = -0.125 + 8 = 7.875 \)
- \( f(0.25) = (0.25)^3 + 8 = 0.015625 + 8 = 8.015625 \)
- \( f(1) = (1)^3 + 8 = 1 + 8 = 9 \)

### Step 4: Compute the Riemann sum
Multiply each function value by the width \(\Delta x = 0.75\) and sum them up to get the Riemann sum:
\[
\text{Riemann Sum} = 0.75 \times (6.046875 + 7.875 + 8.015625 + 9)
\]
\[
= 0.75 \times 30.9375 = 23.203125 \text{ units}^2
\]

### Conclusion:
The correct approximation using the Riemann sum is:

**23.203125 units squared**

Would you like more details on any part of the process or have any other questions?

Here are 5 related questions you might find helpful:
1. How do you compute a left-hand Riemann sum?
2. What is the difference between a Riemann sum and an integral?
3. How does increasing the number of rectangles in a Riemann sum affect the accuracy?
4. Can you explain the trapezoidal rule as an alternative to Riemann sums?
5. How do you set up a Riemann sum for a different function and interval?

**Tip:** When using Riemann sums, increasing the number of rectangles or using more sophisticated methods like the trapezoidal rule can lead to more accurate approximations of the area under a curve.

Learn how to approximate the area under the curve y = x^3 + 8 using a Riemann sum with 4 rectangles and right-hand endpoints on the interval [-2, 1]. Detailed step-by-step solution and calculation included.

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