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  
 
|-
 
|-
|<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>.   
+
| &nbsp;&nbsp; <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>.   
 
|-
 
|-
|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: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|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>.
 
|-
 
|-
|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  
 
|-
 
|-
|<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>.
+
| &nbsp;&nbsp; <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.   
+
|Now, we let <math style="vertical-align: 0px">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>.
+
|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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|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|'''(a)''' Left-endpoint Riemann sum: <math>\frac{205}{144}</math>, The left-endpoint Riemann sum overestimates the area of <math>S</math>.  
+
|'''(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>.  
 
|-
 
|-
|'''(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>
 
|}
 
|}
 
[[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:  
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. Since the problem doesn't specify, we can choose either right- or left-endpoints. Choosing left-endpoints, the 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) The left-endpoint Riemann sum is , which overestimates the area of .
(b) Using left-endpoint Riemann sums:

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