Difference between revisions of "009B Sample Midterm 2, Problem 1"

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<span class="exam">Consider the region <math>S</math> bounded by <math>x=1,x=5,y=\frac{1}{x^2}</math> and the <math>x</math>-axis.
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<span class="exam"> Consider the region <math style="vertical-align: 0px">S</math> bounded by <math style="vertical-align: -13px">x=1,x=5,y=\frac{1}{x^2}</math> and the <math>x</math>-axis.
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::<span class="exam">a) Use four rectangles and a Riemann sum to approximate the area of the region <math style="vertical-align: 0px">S</math>. Sketch the region <math style="vertical-align: 0px">S</math> and the rectangles and indicate whether your rectangles overestimate or underestimate the area of <math style="vertical-align: 0px">S</math>.
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::<span class="exam">b) Find an expression for the area of the region <math style="vertical-align: 0px">S</math> as a limit. Do not evaluate the limit.
  
::<span class="exam">a) Use four rectangles and a Riemann sum to approximate the area of the region <math>S</math>. Sketch the region <math>S</math> and the rectangles and indicate whether your rectangles overestimate or underestimate the area of <math>S</math>.
 
::<span class="exam">b) Find an expression for the area of the region <math>S</math> as a limit. Do not evaluate the limit.
 
  
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|Let <math>f(x)=\frac{1}{x^2}</math>
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|Let <math style="vertical-align: -13px">f(x)=\frac{1}{x^2}</math>Since our interval is <math style="vertical-align: -5px">[1,5]</math> 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  
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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  
 
 
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|<math>1(f(1)+f(2)+f(3)+f(4))</math>.   
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| &nbsp;&nbsp; <math>1\cdot (f(1)+f(2)+f(3)+f(4))</math>.   
 
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|Thus, the left-endpoint Riemann sum is  
 
|Thus, the left-endpoint Riemann sum is  
 
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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>.   
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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>.   
 
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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>S</math>.

Revision as of 21:40, 1 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) Left-endpoint Riemann sum: , The left-endpoint Riemann sum overestimates the area of .
(b) Using left-endpoint Riemann sums:

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