Difference between revisions of "009B Sample Midterm 2, Problem 1"
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|Thus, the left-endpoint Riemann sum is | |Thus, the left-endpoint Riemann sum is | ||
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− | |<math>1\cdot (f(1)+f(2)+f(3)+f(4))=\bigg(1+\frac{1}{4}+\frac{1}{9}+{1}{16}\bigg)=\frac{205}{144}</math>. | + | | <math>1\cdot (f(1)+f(2)+f(3)+f(4))=\bigg(1+\frac{1}{4}+\frac{1}{9}+{1}{16}\bigg)=\frac{205}{144}</math>. |
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− | |The left-endpoint Riemann sum overestimates the area of <math>S</math>. | + | |The left-endpoint Riemann sum overestimates the area of <math style="vertical-align: 0px">S</math>. |
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!Step 1: | !Step 1: | ||
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− | |Let <math>n</math> be the number of rectangles used in the left-endpoint Riemann sum for <math>f(x)=\frac{1}{x^2}</math>. | + | |Let <math style="vertical-align: 0px">n</math> be the number of rectangles used in the left-endpoint Riemann sum for <math style="vertical-align: -13px">f(x)=\frac{1}{x^2}</math>. |
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− | |The width of each rectangle is <math>\Delta x=\frac{5-1}{n}=\frac{4}{n}</math>. | + | |The width of each rectangle is <math style="vertical-align: -13px">\Delta x=\frac{5-1}{n}=\frac{4}{n}</math>. |
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|So, the left-endpoint Riemann sum is | |So, the left-endpoint Riemann sum is | ||
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− | |<math>\Delta x \bigg(f(1)+f\bigg(1+\frac{4}{n}\bigg)+f\bigg(1+2\frac{4}{n}\bigg)+\ldots +f\bigg(1+(n-1)\frac{4}{n}\bigg)\bigg)</math>. | + | | <math>\Delta x \bigg(f(1)+f\bigg(1+\frac{4}{n}\bigg)+f\bigg(1+2\frac{4}{n}\bigg)+\ldots +f\bigg(1+(n-1)\frac{4}{n}\bigg)\bigg)</math>. |
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− | |Now, we let <math>n</math> go to infinity to get a limit. | + | |Now, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit. |
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− | |So, the area of <math>S</math> is equal to <math>\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}f\bigg(1+i\frac{4}{n}\bigg)</math>. | + | |So, the area of <math style="vertical-align: 0px">S</math> is equal to <math style="vertical-align: -20px">\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}f\bigg(1+i\frac{4}{n}\bigg)</math>. |
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!Final Answer: | !Final Answer: | ||
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− | |'''(a)''' | + | |'''(a)''' The left-endpoint Riemann sum is <math style="vertical-align: -20px">\frac{205}{144}</math>, which overestimates the area of <math style="vertical-align: 0px">S</math>. |
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− | |'''(b)''' Using left-endpoint Riemann sums: <math>\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}f\bigg(1+i\frac{4}{n}\bigg)</math> | + | |'''(b)''' Using left-endpoint Riemann sums: <math style="vertical-align: -20px">\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}f\bigg(1+i\frac{4}{n}\bigg)</math> |
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[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
Revision as of 22:30, 2 February 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: |
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Link to Riemann sums page |
Solution:
(a)
Step 1: |
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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: |
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Thus, the left-endpoint Riemann sum is |
. |
The left-endpoint Riemann sum overestimates the area of . |
(b)
Step 1: |
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Let be the number of rectangles used in the left-endpoint Riemann sum for . |
The width of each rectangle is . |
Step 2: |
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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: |
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(a) The left-endpoint Riemann sum is , which overestimates the area of . |
(b) Using left-endpoint Riemann sums: |