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

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|'''3.''' See the Riemann sums (insert link) for more information.
 
|'''3.''' See the Riemann sums (insert link) for more information.
 
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'''Solution:'''
 
'''Solution:'''
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|Thus, <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> is equal to <math style="vertical-align: -21px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg).</math>
 
|Thus, <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> is equal to <math style="vertical-align: -21px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg).</math>
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

Revision as of 09:13, 6 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:  
Recall:
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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