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To approximate the net area bounded by the graph of the function [tex]\( f(x) = -5 - x^3 \)[/tex] and the [tex]\( x \)[/tex]-axis on the interval [tex]\([2, 7]\)[/tex] using the left, right, and midpoint Riemann sums, we will follow these steps:
### Step 1: Divide the Interval
First, we divide the interval [tex]\([2, 7]\)[/tex] into [tex]\( n \)[/tex] sub-intervals. We'll use [tex]\( n = 1000 \)[/tex] sub-intervals for a better approximation.
### Step 2: Calculate the Width of Each Sub-Interval
The width of each sub-interval [tex]\(\Delta x\)[/tex] is:
[tex]\[ \Delta x = \frac{7 - 2}{1000} = 0.005 \][/tex]
### Step 3: Define the Function
The function given is:
[tex]\[ f(x) = -5 - x^3 \][/tex]
### Step 4: Compute the Left Riemann Sum
The left Riemann sum approximates the area using the left endpoints of each sub-interval. It is calculated as:
[tex]\[ \text{Left sum} = \sum_{i=0}^{999} f(2 + i \cdot 0.005) \cdot 0.005 \][/tex]
After performing the calculation, the result is:
[tex]\[ \text{Left sum} \approx -620.4127812499997 \][/tex]
### Step 5: Compute the Right Riemann Sum
The right Riemann sum approximates the area using the right endpoints of each sub-interval. It is calculated as:
[tex]\[ \text{Right sum} = \sum_{i=1}^{1000} f(2 + i \cdot 0.005) \cdot 0.005 \][/tex]
After performing the calculation, the result is:
[tex]\[ \text{Right sum} \approx -622.0877812499998 \][/tex]
### Step 6: Compute the Midpoint Riemann Sum
The midpoint Riemann sum approximates the area using the midpoints of each sub-interval. It is calculated as:
[tex]\[ \text{Midpoint sum} = \sum_{i=0}^{999} f(2 + (i + 0.5) \cdot 0.005) \cdot 0.005 \][/tex]
After performing the calculation, the result is:
[tex]\[ \text{Midpoint sum} \approx -621.249859375 \][/tex]
### Step 7: Net Area Bounded by the Graph and the [tex]\( x \)[/tex]-Axis
The values from the Riemann sums provide the approximation for the net area bounded by the graph of [tex]\( f \)[/tex] and the [tex]\( x \)[/tex]-axis over the interval [tex]\([2, 7]\)[/tex]:
- Left Riemann Sum: [tex]\(-620.4127812499997\)[/tex]
- Right Riemann Sum: [tex]\(-622.0877812499998\)[/tex]
- Midpoint Riemann Sum: [tex]\(-621.249859375\)[/tex]
These sums approximate the net area. The negative values indicate that the function is below the [tex]\( x \)[/tex]-axis in the given interval.
### Choosing the Correct Graph
To sketch the graph of [tex]\( f(x) = -5 - x^3 \)[/tex] on the interval [tex]\([2, 7]\)[/tex], we should plot the function values at these key points. The function decreases very steeply due to the cube term [tex]\(-x^3\)[/tex], moving further down as [tex]\( x \)[/tex] increases.
The graph would be a downward curve starting at [tex]\( f(2) = -5 - 8 = -13 \)[/tex] and sharply decreasing towards more negative values as [tex]\( x \)[/tex] approaches 7. Therefore, the graph choice would reflect this behavior of a steeply declining curve in the interval [tex]\([2, 7]\)[/tex].
The final values representing the approximate net area bounded by the curve and the [tex]\( x \)[/tex]-axis are as follows:
- Left Riemann Sum: [tex]\(-620.4127812499997\)[/tex]
- Right Riemann Sum: [tex]\(-622.0877812499998\)[/tex]
- Midpoint Riemann Sum: [tex]\(-621.249859375\)[/tex]
### Step 1: Divide the Interval
First, we divide the interval [tex]\([2, 7]\)[/tex] into [tex]\( n \)[/tex] sub-intervals. We'll use [tex]\( n = 1000 \)[/tex] sub-intervals for a better approximation.
### Step 2: Calculate the Width of Each Sub-Interval
The width of each sub-interval [tex]\(\Delta x\)[/tex] is:
[tex]\[ \Delta x = \frac{7 - 2}{1000} = 0.005 \][/tex]
### Step 3: Define the Function
The function given is:
[tex]\[ f(x) = -5 - x^3 \][/tex]
### Step 4: Compute the Left Riemann Sum
The left Riemann sum approximates the area using the left endpoints of each sub-interval. It is calculated as:
[tex]\[ \text{Left sum} = \sum_{i=0}^{999} f(2 + i \cdot 0.005) \cdot 0.005 \][/tex]
After performing the calculation, the result is:
[tex]\[ \text{Left sum} \approx -620.4127812499997 \][/tex]
### Step 5: Compute the Right Riemann Sum
The right Riemann sum approximates the area using the right endpoints of each sub-interval. It is calculated as:
[tex]\[ \text{Right sum} = \sum_{i=1}^{1000} f(2 + i \cdot 0.005) \cdot 0.005 \][/tex]
After performing the calculation, the result is:
[tex]\[ \text{Right sum} \approx -622.0877812499998 \][/tex]
### Step 6: Compute the Midpoint Riemann Sum
The midpoint Riemann sum approximates the area using the midpoints of each sub-interval. It is calculated as:
[tex]\[ \text{Midpoint sum} = \sum_{i=0}^{999} f(2 + (i + 0.5) \cdot 0.005) \cdot 0.005 \][/tex]
After performing the calculation, the result is:
[tex]\[ \text{Midpoint sum} \approx -621.249859375 \][/tex]
### Step 7: Net Area Bounded by the Graph and the [tex]\( x \)[/tex]-Axis
The values from the Riemann sums provide the approximation for the net area bounded by the graph of [tex]\( f \)[/tex] and the [tex]\( x \)[/tex]-axis over the interval [tex]\([2, 7]\)[/tex]:
- Left Riemann Sum: [tex]\(-620.4127812499997\)[/tex]
- Right Riemann Sum: [tex]\(-622.0877812499998\)[/tex]
- Midpoint Riemann Sum: [tex]\(-621.249859375\)[/tex]
These sums approximate the net area. The negative values indicate that the function is below the [tex]\( x \)[/tex]-axis in the given interval.
### Choosing the Correct Graph
To sketch the graph of [tex]\( f(x) = -5 - x^3 \)[/tex] on the interval [tex]\([2, 7]\)[/tex], we should plot the function values at these key points. The function decreases very steeply due to the cube term [tex]\(-x^3\)[/tex], moving further down as [tex]\( x \)[/tex] increases.
The graph would be a downward curve starting at [tex]\( f(2) = -5 - 8 = -13 \)[/tex] and sharply decreasing towards more negative values as [tex]\( x \)[/tex] approaches 7. Therefore, the graph choice would reflect this behavior of a steeply declining curve in the interval [tex]\([2, 7]\)[/tex].
The final values representing the approximate net area bounded by the curve and the [tex]\( x \)[/tex]-axis are as follows:
- Left Riemann Sum: [tex]\(-620.4127812499997\)[/tex]
- Right Riemann Sum: [tex]\(-622.0877812499998\)[/tex]
- Midpoint Riemann Sum: [tex]\(-621.249859375\)[/tex]
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