Difference between revisions of "009B Sample Midterm 1, Problem 5"
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|Thus, the left-hand Riemann sum is | |Thus, the left-hand Riemann sum is | ||
|- | |- | ||
| − | | <math | + | | |
| + | <math>\begin{array}{rcl} | ||
| + | \displaystyle{1(f(0)+f(1)+f(2))} & = & \displaystyle{1+0+-3}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{-2.} | ||
| + | \end{array}</math> | ||
|} | |} | ||
| Line 53: | Line 58: | ||
|Thus, the right-hand Riemann sum is | |Thus, the right-hand Riemann sum is | ||
|- | |- | ||
| − | | <math | + | | |
| + | <math>\begin{array}{rcl} | ||
| + | \displaystyle{1(f(1)+f(2)+f(3))} & = & \displaystyle{0+-3+-8}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{-11.} | ||
| + | \end{array}</math> | ||
|} | |} | ||
Revision as of 17:16, 7 February 2017
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: |
|---|
| 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) |