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

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|Thus, the left-hand Riemann sum is  
 
|Thus, the left-hand Riemann sum is  
 
|-
 
|-
| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: -4px">1(f(0)+f(1)+f(2))=1+0+-3=-2.</math>
+
|  
 +
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{1(f(0)+f(1)+f(2))} & = & \displaystyle{1+0+-3}\\
 +
&&\\
 +
& = & \displaystyle{-2.}
 +
\end{array}</math>
 
|}
 
|}
  
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|Thus, the right-hand Riemann sum is  
 
|Thus, the right-hand Riemann sum is  
 
|-
 
|-
| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))=0+-3+-8=-11.</math>
+
|  
 +
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{1(f(1)+f(2)+f(3))} & = & \displaystyle{0+-3+-8}\\
 +
&&\\
 +
& = & \displaystyle{-11.}
 +
\end{array}</math>
 
|}
 
|}
  

Revision as of 17:16, 7 February 2017

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:  
1. The height of each rectangle in the left-hand Riemann sum is given by choosing the left endpoint of the interval.
2. The height of each rectangle in the right-hand Riemann sum is given by choosing the right endpoint of the interval.
3. See the Riemann sums (insert link) for more information.


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
      
Finally, we let go to infinity to get a limit.
Thus, is equal to


Final Answer:  
    (a)    
    (b)    
    (c)    

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