Difference between revisions of "009B Sample Midterm 1, Problem 5"
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|Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the left-hand Riemann sum is | |Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the left-hand Riemann sum is | ||
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− | | <math style="vertical-align: 0px">1(f(0)+f(1)+f(2))</math> | + | | <math style="vertical-align: 0px">1(f(0)+f(1)+f(2)).</math> |
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| | | | ||
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|Thus, the left-hand Riemann sum is | |Thus, the left-hand Riemann sum is | ||
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− | | <math style="vertical-align: -4px">1(f(0)+f(1)+f(2))=1+0+-3=-2</math> | + | | <math style="vertical-align: -4px">1(f(0)+f(1)+f(2))=1+0+-3=-2.</math> |
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|Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the right-hand Riemann sum is | |Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the right-hand Riemann sum is | ||
|- | |- | ||
− | | <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))</math> | + | | <math style="vertical-align: -5px">1(f(1)+f(2)+f(3)).</math> |
|- | |- | ||
| | | | ||
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|Thus, the right-hand Riemann sum is | |Thus, the right-hand Riemann sum is | ||
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− | | <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))=0+-3+-8=-11</math> | + | | <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))=0+-3+-8=-11.</math> |
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!Step 1: | !Step 1: | ||
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− | |Let <math style="vertical-align: 0px">n</math> be the number of rectangles used in the right-hand Riemann sum for <math style="vertical-align: -5px">f(x)=1-x^2</math> | + | |Let <math style="vertical-align: 0px">n</math> be the number of rectangles used in the right-hand Riemann sum for <math style="vertical-align: -5px">f(x)=1-x^2.</math> |
|- | |- | ||
− | |The width of each rectangle is <math style="vertical-align: -13px">\Delta x=\frac{3-0}{n}=\frac{3}{n}</math> | + | |The width of each rectangle is <math style="vertical-align: -13px">\Delta x=\frac{3-0}{n}=\frac{3}{n}.</math> |
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|So, the right-hand Riemann sum is | |So, the right-hand Riemann sum is | ||
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− | | <math style="vertical-align: -14px">\Delta x \bigg(f\bigg(1\cdot \frac{3}{n}\bigg)+f\bigg(2\cdot \frac{3}{n}\bigg)+f\bigg(3\cdot \frac{3}{n}\bigg)+\ldots +f(3)\bigg)</math> | + | | <math style="vertical-align: -14px">\Delta x \bigg(f\bigg(1\cdot \frac{3}{n}\bigg)+f\bigg(2\cdot \frac{3}{n}\bigg)+f\bigg(3\cdot \frac{3}{n}\bigg)+\ldots +f(3)\bigg).</math> |
|- | |- | ||
|Finally, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit. | |Finally, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit. | ||
|- | |- | ||
− | |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> |
|} | |} | ||
Revision as of 09:00, 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: |
---|
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: |
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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) |