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

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
|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>.  
+
|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  
 +
|-
 +
|&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;  
 +
|-
 +
|This problem wants us to find the average value of <math>s(t)</math> over the interval <math>[0,5].</math>
 
|-
 
|-
 
|Using the formula given in Foundations, we have:
 
|Using the formula given in Foundations, we have:
 
|-
 
|-
| &nbsp; &nbsp;<math style="vertical-align: 0px">f_{\text{avg}}=</math>  
+
| &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;
 
|-
 
|-
|
+
|First, we distribute to get
 
|-
 
|-
|
+
|&nbsp; &nbsp; <math>s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-t^2+18~dt.</math>
 
|-
 
|-
|
+
|Then, we integrate to get
 +
|-
 +
|&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;
 
|-
 
|-
|
+
|We now evaluate to get
|-
 
|
 
|-
 
|
 
 
|-
 
|-
|
+
|&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;  
 
|-
 
|-
| &nbsp; &nbsp; <math></math>
+
| &nbsp; &nbsp; <math>\frac{233}{6}</math>
 
|-
 
|-
 
|  
 
|  
 
|}
 
|}
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 10:48, 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
Using the formula given in Foundations, we have:
   
Step 2:  
First, we distribute to get
   
Then, we integrate to get
   
Step 3:  
We now evaluate to get
   


Final Answer:  
   

Return to Sample Exam