Difference between revisions of "009B Sample Midterm 1, Problem 5"
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|'''3.''' See the Riemann sums (insert link) for more information. | |'''3.''' See the Riemann sums (insert link) for more information. | ||
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|Thus, <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> is equal to <math style="vertical-align: -21px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg).</math> | |Thus, <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> is equal to <math style="vertical-align: -21px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg).</math> | ||
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" |
Revision as of 09:13, 6 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: |
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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: |
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Since our interval is and we are using 3 rectangles, each rectangle has width 1. So, the left-hand Riemann sum is |
Step 2: |
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Thus, the left-hand Riemann sum is |
(b)
Step 1: |
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Since our interval is and we are using 3 rectangles, each rectangle has width 1. So, the right-hand Riemann sum is |
Step 2: |
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Thus, the right-hand Riemann sum is |
(c)
Step 1: |
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Let be the number of rectangles used in the right-hand Riemann sum for |
The width of each rectangle is |
Step 2: |
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So, the right-hand Riemann sum is |
Finally, we let go to infinity to get a limit. |
Thus, is equal to |
Final Answer: |
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(a) |
(b) |
(c) |