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

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|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, the area of <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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|}
  

Revision as of 08:33, 1 February 2016

Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=1-x^2} .

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:  
Link to Riemann sums page

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
   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 )}} .
Finally, we let go to infinity to get a limit.
Thus, is equal to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)} .
Final Answer:  
(a)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -2}
(b)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -11}
(c)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)}

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