Difference between revisions of "009B Sample Midterm 1, Problem 5"
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!Foundations: | !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. | ||
|} | |} | ||
Revision as of 10:05, 28 March 2016
Let .
- a) Compute the left-hand Riemann sum approximation of with boxes.
- b) Compute the right-hand Riemann sum approximation of with boxes.
- c) Express as a limit of right-hand Riemann sums (as in the definition of the definite integral). 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: |
|---|
| Since our interval is and we are using 3 rectangles, each rectangle has width 1. So, the left-hand Riemann sum is |
| . |
| Step 2: |
|---|
| Thus, the left-hand Riemann sum is |
| . |
(b)
| Step 1: |
|---|
| Since our interval is and we are using 3 rectangles, each rectangle has width 1. So, the right-hand Riemann sum is |
| . |
| Step 2: |
|---|
| Thus, the right-hand Riemann sum is |
| . |
(c)
| Step 1: |
|---|
| Let be the number of rectangles used in the right-hand Riemann sum for . |
| The width of each rectangle is . |
| Step 2: |
|---|
| So, the right-hand Riemann sum is |
| . |
| Finally, we let go to infinity to get a limit. |
| Thus, is equal to . |
| Final Answer: |
|---|
| (a) |
| (b) |
| (c) |