Difference between revisions of "009B Sample Midterm 1, Problem 2"

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!Foundations:    
 
!Foundations:    
 
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|-
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|The average value of a function <math style="vertical-align: -5px">f(x)</math> on an interval <math style="vertical-align: -5px">[a,b]</math> is given by <math style="vertical-align: -18px">f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx</math>.  
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|The average value of a function <math style="vertical-align: -5px">f(x)</math> on an interval <math style="vertical-align: -5px">[a,b]</math> is given by  
 +
|-
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|&nbsp; &nbsp; <math style="vertical-align: -18px">f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx</math>.  
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 +
|-
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|This problem wants us to find the average value of <math>s(t)</math> over the interval <math>[0,5].</math>
 
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|-
 
|Using the formula given in Foundations, we have:
 
|Using the formula given in Foundations, we have:
 
|-
 
|-
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| &nbsp; &nbsp;<math style="vertical-align: 0px">f_{\text{avg}}=</math>  
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| &nbsp; &nbsp;<math style="vertical-align: 0px">s_{\text{avg}}=\frac{1}{5-0} \int_0^5 t(25-5t)+18~dt.</math>  
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|-
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|
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|First, we distribute to get
 
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|-
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|
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|&nbsp; &nbsp; <math>s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-t^2+18~dt.</math>
 
|-
 
|-
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|
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|Then, we integrate to get
 +
|-
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|&nbsp; &nbsp; <math>s_{\text{avg}}=\left. \frac{1}{5}\bigg[\frac{25t^2}{2}-\frac{5t^3}{3}+18t\bigg]\right|_0^5.</math>
 
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!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
|-
 
|-
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|
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|We now evaluate to get
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|-
 
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|-
 
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|-
 
|-
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|&nbsp; &nbsp;<math>s_{\text{avg}}=\frac{1}{5}\bigg[\frac{25(5)^2}{2}-\frac{5(5)^3}{3}+18(5)\bigg]-0=\frac{233}{6}.</math>
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
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| &nbsp; &nbsp; <math></math>
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| &nbsp; &nbsp; <math>\frac{233}{6}</math>
 
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|-
 
|  
 
|  
 
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[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 09:48, 6 February 2017

Otis Taylor plots the price per share of a stock that he owns as a function of time

and finds that it can be approximated by the function

where is the time (in years) since the stock was purchased.

Find the average price of the stock over the first five years.


Foundations:  
The average value of a function on an interval is given by
    .


Solution:

Step 1:  
This problem wants us to find the average value of over the interval 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 [0,5].}
Using the formula given in Foundations, we have:
   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 s_{\text{avg}}=\frac{1}{5-0} \int_0^5 t(25-5t)+18~dt.}
Step 2:  
First, we distribute to get
    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 s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-t^2+18~dt.}
Then, we integrate to get
    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 s_{\text{avg}}=\left. \frac{1}{5}\bigg[\frac{25t^2}{2}-\frac{5t^3}{3}+18t\bigg]\right|_0^5.}
Step 3:  
We now evaluate to get
   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 s_{\text{avg}}=\frac{1}{5}\bigg[\frac{25(5)^2}{2}-\frac{5(5)^3}{3}+18(5)\bigg]-0=\frac{233}{6}.}


Final Answer:  
    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 \frac{233}{6}}

Return to Sample Exam

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