At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
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]
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
