Difference between revisions of "009B Sample Midterm 1, Problem 5"

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!Foundations:    
 
!Foundations:    
 
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|Link to Riemann sums content page
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|Link to Riemann sums page
 
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'''Solution:'''
 
'''Solution:'''
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'''(a)'''
 
{| 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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|Since our interval is <math>[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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|-
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|<math>1(f(0)+f(1)+f(2))</math>. 
 
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|-
 
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 2: &nbsp;
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|-
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|Thus, the left-hand Riemann sum is
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|<math>1(f(0)+f(1)+f(2))=1+0+-3=-2</math>. 
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'''(b)'''
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 1: &nbsp;
 
|-
 
|-
|  
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|Since our interval is <math>[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the right-hand Riemann sum is
 
|-
 
|-
|
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|<math>1(f(1)+f(2)+f(3))</math>. 
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 +
|-
 +
|Thus, the right-hand Riemann sum is
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|-
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|<math>1(f(1)+f(2)+f(3))=0+-3+-8=-11</math>. 
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|}
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'''(c)'''
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 1: &nbsp;
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|-
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|Let <math>n</math> be the number of rectangles used in the right-hand Riemann sum for <math>f(x)=1-x^2</math>.
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|-
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|The width of each rectangle is <math>\Delta x=\frac{3-0}{n}=\frac{3}{n}</math>.
 
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|
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 2: &nbsp;
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|-
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|So, the right-hand Riemann sum is
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|-
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|<math>\Delta x \bigg(f\bigg(\frac{3}{n}\bigg)+f\bigg(2\frac{3}{n}\bigg)+f\bigg(3\frac{3}{n}\bigg)+\ldots +f(3)\bigg)</math>.
 
|-
 
|-
|
+
|Now, we let <math>n</math> go to infinity to get a limit. 
 
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|-
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|So, the area of <math>S</math> is equal to <math>\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)</math>.
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|'''(a)''' <math>-2</math>
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|'''(b)''' <math>-11</math>
 
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|-
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|'''(c)''' <math>\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)</math>
 
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[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 14:47, 31 January 2016

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:  
Link to Riemann sums page

Solution:

(a)

Step 1:  
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:  
So, the right-hand Riemann sum is
.
Now, we let go to infinity to get a limit.
So, the area of is equal to .
Final Answer:  
(a)
(b)
(c)

Return to Sample Exam