Difference between revisions of "009B Sample Final 3, Problem 1"

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!Step 1:    
 
!Step 1:    
 
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|Since our interval is &nbsp;<math style="vertical-align: -5px">[-1,1]</math>&nbsp; and we are using &nbsp;<math style="vertical-align: -1px">4</math>&nbsp; rectangles, each rectangle has width &nbsp;<math style="vertical-align: -13px">\frac{1}{2}.</math>
 
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|Let &nbsp;<math style="vertical-align: -6px">f(x)=1-x^2.</math>
 
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|So, the left-endpoint Riemann sum is
 
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| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: 0px">S=\frac{1}{2}\bigg(f(-1)+f\bigg(-\frac{1}{2}\bigg)+f(0)+f\bigg(\frac{1}{2}\bigg)\bigg).</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|Thus, the left-endpoint Riemann sum is
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{S} & = & \displaystyle{\frac{1}{2}\bigg(0+\frac{3}{4}+1+\frac{3}{4}\bigg)}\\
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&&\\
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& = & \displaystyle{\frac{5}{4}.}
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\end{array}</math>
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{5}{4}</math>
 
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[[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 17:29, 28 February 2017

Divide the interval    into four subintervals of equal length    and compute the left-endpoint Riemann sum of  

Foundations:  
The height of each rectangle in the left-endpoint Riemann sum is given by choosing the left endpoint of the interval.


Solution:

Step 1:  
Since our interval is    and we are using    rectangles, each rectangle has width  
Let  
So, the left-endpoint Riemann sum is
      
Step 2:  
Thus, the left-endpoint Riemann sum is

       


Final Answer:  
       

Return to Sample Exam