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

From Grad Wiki
Jump to navigation Jump to search
Line 50: Line 50:
 
\displaystyle{s_{\text{avg}}} & = & \displaystyle{\frac{1}{5}\bigg[\frac{25(5)^2}{2}-\frac{5(5)^3}{3}+18(5)\bigg]-0}\\
 
\displaystyle{s_{\text{avg}}} & = & \displaystyle{\frac{1}{5}\bigg[\frac{25(5)^2}{2}-\frac{5(5)^3}{3}+18(5)\bigg]-0}\\
 
&&\\
 
&&\\
& = & \displaystyle{\frac{233}{6}.}
+
& = & \displaystyle{\frac{233}{6}}\\
 +
&&\\
 +
& \approx & \displaystyle{$38.83.}
  
 
\end{array}</math>
 
\end{array}</math>
Line 59: Line 61:
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
| &nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{233}{6}</math>
+
| &nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{233}{6}\approx $38.83</math>
 
|-
 
|-
 
|  
 
|  
 
|}
 
|}
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 10:54, 18 March 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 average value formula, 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