Difference between revisions of "009B Sample Midterm 2, Problem 1"
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!Foundations: | !Foundations: | ||
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| − | | | + | |Recall: |
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| + | |'''1.''' The height of each rectangle in the left-hand Riemann sum is given by choosing the left endpoint of the interval. | ||
| + | |- | ||
| + | |'''2.''' The height of each rectangle in the right-hand Riemann sum is given by choosing the right endpoint of the interval. | ||
| + | |- | ||
| + | |'''3.''' See the Riemann sums (insert link) for more information. | ||
|} | |} | ||
Revision as of 10:09, 28 March 2016
Consider the region bounded by and the -axis.
- a) Use four rectangles and a Riemann sum to approximate the area of the region . Sketch the region and the rectangles and indicate whether your rectangles overestimate or underestimate the area of .
- b) Find an expression for the area of the region as a limit. Do not evaluate the limit.
| Foundations: |
|---|
| Recall: |
| 1. The height of each rectangle in the left-hand Riemann sum is given by choosing the left endpoint of the interval. |
| 2. The height of each rectangle in the right-hand Riemann sum is given by choosing the right endpoint of the interval. |
| 3. See the Riemann sums (insert link) for more information. |
Solution:
(a)
| Step 1: |
|---|
| Let . Since our interval is and we are using 4 rectangles, each rectangle has width 1. Since the problem doesn't specify, we can choose either right- or left-endpoints. Choosing left-endpoints, the Riemann sum is |
| . |
| Step 2: |
|---|
| Thus, the left-endpoint Riemann sum is |
| . |
| The left-endpoint Riemann sum overestimates the area of . |
(b)
| Step 1: |
|---|
| Let be the number of rectangles used in the left-endpoint Riemann sum for . |
| The width of each rectangle is . |
| Step 2: |
|---|
| So, the left-endpoint Riemann sum is |
| . |
| Now, we let go to infinity to get a limit. |
| So, the area of is equal to . |
| Final Answer: |
|---|
| (a) The left-endpoint Riemann sum is , which overestimates the area of . |
| (b) Using left-endpoint Riemann sums: |