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

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|Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the left-hand Riemann sum is  
 
|Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the left-hand Riemann sum is  
 
|-
 
|-
| &nbsp;&nbsp; <math style="vertical-align: 0px">1(f(0)+f(1)+f(2))</math>
+
| &nbsp;&nbsp; <math style="vertical-align: 0px">1(f(0)+f(1)+f(2)).</math>
 
|-
 
|-
 
|
 
|
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|Thus, the left-hand Riemann sum is  
 
|Thus, the left-hand Riemann sum is  
 
|-
 
|-
| &nbsp;&nbsp; <math style="vertical-align: -4px">1(f(0)+f(1)+f(2))=1+0+-3=-2</math>.  
+
| &nbsp;&nbsp; <math style="vertical-align: -4px">1(f(0)+f(1)+f(2))=1+0+-3=-2.</math>   
 
|}
 
|}
  
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|Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the right-hand Riemann sum is
 
|Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the right-hand Riemann sum is
 
|-
 
|-
| &nbsp;&nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))</math>
+
| &nbsp;&nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3)).</math>  
 
|-
 
|-
 
|
 
|
Line 54: Line 54:
 
|Thus, the right-hand Riemann sum is  
 
|Thus, the right-hand Riemann sum is  
 
|-
 
|-
| &nbsp;&nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))=0+-3+-8=-11</math>.  
+
| &nbsp;&nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))=0+-3+-8=-11.</math>   
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|Let <math style="vertical-align: 0px">n</math> be the number of rectangles used in the right-hand Riemann sum for <math style="vertical-align: -5px">f(x)=1-x^2</math>.
+
|Let <math style="vertical-align: 0px">n</math> be the number of rectangles used in the right-hand Riemann sum for <math style="vertical-align: -5px">f(x)=1-x^2.</math>
 
|-
 
|-
|The width of each rectangle is <math style="vertical-align: -13px">\Delta x=\frac{3-0}{n}=\frac{3}{n}</math>.
+
|The width of each rectangle is <math style="vertical-align: -13px">\Delta x=\frac{3-0}{n}=\frac{3}{n}.</math>
 
|-
 
|-
 
|
 
|
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|So, the right-hand Riemann sum is  
 
|So, the right-hand Riemann sum is  
 
|-
 
|-
| &nbsp;&nbsp; <math style="vertical-align: -14px">\Delta x \bigg(f\bigg(1\cdot \frac{3}{n}\bigg)+f\bigg(2\cdot \frac{3}{n}\bigg)+f\bigg(3\cdot \frac{3}{n}\bigg)+\ldots +f(3)\bigg)</math>.
+
| &nbsp;&nbsp; <math style="vertical-align: -14px">\Delta x \bigg(f\bigg(1\cdot \frac{3}{n}\bigg)+f\bigg(2\cdot \frac{3}{n}\bigg)+f\bigg(3\cdot \frac{3}{n}\bigg)+\ldots +f(3)\bigg).</math>
 
|-
 
|-
 
|Finally, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit.   
 
|Finally, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit.   
 
|-
 
|-
|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>
 
|}
 
|}
  

Revision as of 09:00, 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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