Use the given graph to estimate the left Riemann sum for the given interval with the stated number of subdivisions. (Round your answer to the nearest integer.) 1)[1, 3.5], n = 5 (The answer is not 27, 54, 36 or -20 ) 2) [2, 14], n = 3 (The answer is not -60, -36, or -32 ) see picture!!!

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Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Use the given graph to estimate the left Riemann sum for the given interval with the stated number of subdivisions. (Round your answer to the nearest integer.)

1)[1, 3.5], n = 5 (The answer is not 27, 54, 36 or -20 )

2) [2, 14], n = 3 (The answer is not -60, -36, or -32 )

see picture!!!

### Using Graphs to Estimate Left Riemann Sums

#### Example 1:
**Interval:** \[1, 3.5\]

**Number of Subdivisions (n):** 5

A graph of a function \( f \) is shown with the x-axis ranging from 0 to 4 and the y-axis ranging from -10 to 10. The curve exhibits various oscillations, including peaks and valleys. The function starts at \( (1, 0) \), rises and falls, and then rises again. 

To estimate the left Riemann sum:
1. Divide the interval [1, 3.5] into 5 equal subintervals.
2. Use the left endpoint of each subinterval to evaluate the function's value.
3. Multiply each function value by the width of the subinterval.
4. Sum these products to approximate the total area under the curve.

#### Example 2:
**Interval:** \[2, 14\]

**Number of Subdivisions (n):** 3

This graph shows a function \( f \) with the x-axis ranging from 0 to 16 and the y-axis ranging from -10 to 10. The curve has significant peaks and troughs, starting from a high point at \( (2, 10) \), dipping into negative values, then rising again to about midway up the graph before finally ascending steeply.

To estimate the left Riemann sum:
1. Divide the interval [2, 14] into 3 equal subintervals.
2. Use the left endpoint of each subinterval to evaluate the function's value.
3. Multiply each function value by the width of the subinterval.
4. Sum these products to approximate the total area under the curve.

### Summary
Left Riemann sum calculations using various step-by-step visual graphs can be highly beneficial in understanding integral approximations in calculus. By partitioning the given interval and summing up these areas, we can get increasingly better estimates of the integral as the number of subdivisions increases.
Transcribed Image Text:### Using Graphs to Estimate Left Riemann Sums #### Example 1: **Interval:** \[1, 3.5\] **Number of Subdivisions (n):** 5 A graph of a function \( f \) is shown with the x-axis ranging from 0 to 4 and the y-axis ranging from -10 to 10. The curve exhibits various oscillations, including peaks and valleys. The function starts at \( (1, 0) \), rises and falls, and then rises again. To estimate the left Riemann sum: 1. Divide the interval [1, 3.5] into 5 equal subintervals. 2. Use the left endpoint of each subinterval to evaluate the function's value. 3. Multiply each function value by the width of the subinterval. 4. Sum these products to approximate the total area under the curve. #### Example 2: **Interval:** \[2, 14\] **Number of Subdivisions (n):** 3 This graph shows a function \( f \) with the x-axis ranging from 0 to 16 and the y-axis ranging from -10 to 10. The curve has significant peaks and troughs, starting from a high point at \( (2, 10) \), dipping into negative values, then rising again to about midway up the graph before finally ascending steeply. To estimate the left Riemann sum: 1. Divide the interval [2, 14] into 3 equal subintervals. 2. Use the left endpoint of each subinterval to evaluate the function's value. 3. Multiply each function value by the width of the subinterval. 4. Sum these products to approximate the total area under the curve. ### Summary Left Riemann sum calculations using various step-by-step visual graphs can be highly beneficial in understanding integral approximations in calculus. By partitioning the given interval and summing up these areas, we can get increasingly better estimates of the integral as the number of subdivisions increases.
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