Difference between revisions of "009B Sample Midterm 2, Problem 1"
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 1: | !Step 1: | ||
+ | |- | ||
+ | |Let <math>f(x)=\frac{1}{x^2}</math> | ||
|- | |- | ||
|Since our interval is <math>[1,5]</math> and we are using 4 rectangles, each rectangle has width 1. So, the left-endpoint Riemann sum is | |Since our interval is <math>[1,5]</math> and we are using 4 rectangles, each rectangle has width 1. So, the left-endpoint Riemann sum is | ||
|- | |- | ||
|<math>1(f(1)+f(2)+f(3)+f(4))</math>. | |<math>1(f(1)+f(2)+f(3)+f(4))</math>. | ||
+ | |- | ||
+ | | | ||
|} | |} | ||
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|<math>1(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(f(1)+f(2)+f(3)+f(4))=\bigg(1+\frac{1}{4}+\frac{1}{9}+{1}{16}\bigg)=\frac{205}{144}</math>. | ||
|- | |- | ||
− | | | + | |The left-endpoint Riemann sum overestimates the area of <math>S</math>. |
− | |||
− | |||
|} | |} | ||
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!Step 1: | !Step 1: | ||
|- | |- | ||
− | | | + | |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>. |
|- | |- | ||
− | | | + | |The width of each rectangle is <math>\Delta x=\frac{5-1}{n}=\frac{4}{n}</math>. |
|- | |- | ||
| | | | ||
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!Step 2: | !Step 2: | ||
|- | |- | ||
− | | | + | |So, the left-endpoint Riemann sum is |
|- | |- | ||
− | | | + | |<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>. |
|- | |- | ||
− | | | + | |Now, we let <math>n</math> go to infinity to get a limit. |
|- | |- | ||
− | | | + | |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>. |
|} | |} | ||
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!Final Answer: | !Final Answer: | ||
|- | |- | ||
− | |'''(a)''' <math>\frac{205}{144}</math> | + | |'''(a)''' Left-endpoint Riemann sum: <math>\frac{205}{144}</math>, The left-endpoint Riemann sum overestimates the area of <math>S</math>. |
|- | |- | ||
− | |'''(b)''' | + | |'''(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> |
|} | |} | ||
[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
Revision as of 14:34, 31 January 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 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: |
---|
Let |
Since our interval is and we are using 4 rectangles, each rectangle has width 1. So, the left-endpoint Riemann sum is |
. |
Step 2: |
---|
Thus, the left-endpoint Riemann sum is |
. |
The left-endpoint Riemann sum overestimates the area of . |
(b)
Step 1: |
---|
Let be the number of rectangles used in the left-endpoint Riemann sum for . |
The width of each rectangle is . |
Step 2: |
---|
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) Left-endpoint Riemann sum: , The left-endpoint Riemann sum overestimates the area of . |
(b) Using left-endpoint Riemann sums: |