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:    
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|-
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|Let <math>f(x)=\frac{1}{x^2}</math>
 
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|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  
 
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|<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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|
 
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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>.   
 
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|-
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|The left-endpoint Riemann sum overestimates the area of <math>S</math>.
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|
 
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|
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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>.
 
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|The width of each rectangle is <math>\Delta x=\frac{5-1}{n}=\frac{4}{n}</math>.
 
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|-
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|
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|So, the left-endpoint Riemann sum is
 
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|-
|
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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>.
 
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|-
|  
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|Now, we let <math>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>.
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|'''(a)''' <math>\frac{205}{144}</math>  
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|'''(a)''' Left-endpoint Riemann sum: <math>\frac{205}{144}</math>, The left-endpoint Riemann sum overestimates the area of <math>S</math>.
 
|-
 
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|'''(b)'''  
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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>
 
|}
 
|}
 
[[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:  
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:  
(a) Left-endpoint Riemann sum: , The left-endpoint Riemann sum overestimates the area of .
(b) Using left-endpoint Riemann sums:

Return to Sample Exam