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

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<span class="exam">Find the average value of the function on the given interval.
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<span class="exam"> Otis Taylor plots the price per share of a stock that he owns as a function of time
  
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::<math>f(x)=2x^3(1+x^2)^4,~~~[0,2]</math>
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<span class="exam">and finds that it can be approximated by the function
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::<math>s(t)=t(25-5t)+18</math>
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<span class="exam">where &nbsp;<math style="vertical-align: 0px">t</math>&nbsp; is the time (in years) since the stock was purchased.
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<span class="exam">Find the average price of the stock over the first five years.
  
  
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!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
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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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|The average value of a function &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; on an interval &nbsp;<math style="vertical-align: -5px">[a,b]</math>&nbsp; is given by  
 
|-
 
|-
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|
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|&nbsp; &nbsp; &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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::<math>f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.</math>
 
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|Using the formula given in Foundations, we have:
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|This problem wants us to find the average value of &nbsp;<math style="vertical-align: -5px">s(t)</math>&nbsp; over the interval &nbsp;<math style="vertical-align: -5px">[0,5].</math>
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|-
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|Using the average value formula, we have
 
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|-
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|
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| &nbsp; &nbsp; &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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::<math>\begin{array}{rcl}
 
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\displaystyle{f_{\text{avg}}} & = & \displaystyle{\frac{1}{2-0}\int_0^2 2x^3(1+x^2)^4~dx}\\
 
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&&\\
 
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& = & \displaystyle{\int_0^2 x^3(1+x^2)^4~dx.}\\
 
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\end{array}</math>
 
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|Now, we use <math>u</math>-substitution. Let <math style="vertical-align: -2px">u=1+x^2.</math> Then, <math style="vertical-align: 0px">du=2x dx</math> and <math style="vertical-align: -13px">\frac{du}{2}=xdx.</math> Also, <math style="vertical-align: 0px">x^2=u-1.</math>
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|First, we distribute to get
 
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|-
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|We need to change the bounds on the integral. We have <math style="vertical-align: -4px">u_1=1+0^2=1</math> and <math style="vertical-align: -3px">u_2=1+2^2=5.</math>
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-5t^2+18~dt.</math>
 
|-
 
|-
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|So, the integral becomes
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|Then, we integrate to get
 
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|-
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|
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|&nbsp; &nbsp; &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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::<math>\begin{array}{rcl}
 
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\displaystyle{f_{\text{avg}}} & = & \displaystyle{\int_0^2 x\cdot x^2 (1+x^2)^4~dx}\\
 
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&&\\
 
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& = & \displaystyle{\frac{1}{2}\int_1^5(u-1)u^4~du}\\
 
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&&\\
 
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& = & \displaystyle{\frac{1}{2}\int_1^5(u^5-u^4)~du.}\\
 
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\end{array}</math>
 
 
|}
 
|}
  
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!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
|-
 
|-
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|We integrate to get
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|We now evaluate to get  
 
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|-
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|  
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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::<math>\begin{array}{rcl}
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\displaystyle{s_{\text{avg}}} & = & \displaystyle{\frac{1}{5}\bigg[\frac{25(5)^2}{2}-\frac{5(5)^3}{3}+18(5)\bigg]-0}\\
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\displaystyle{f_{\text{avg}}} & = & \displaystyle{\left.\frac{u^6}{12}-\frac{u^5}{10}\right|_{1}^5}\\
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&&\\
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& = & \displaystyle{\frac{233}{6}}\\
 
&&\\
 
&&\\
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& = & \displaystyle{\left.u^5\bigg(\frac{u}{12}-\frac{1}{10}\bigg)\right|_{1}^5.}\\
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& \approx & \displaystyle{$38.83.}
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\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 4: &nbsp;
 
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|-
 
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|We evaluate to get
 
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|-
 
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|
 
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::<math>\begin{array}{rcl}
 
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\displaystyle{f_{\text{avg}}} & = & \displaystyle{5^5\bigg(\frac{5}{12}-\frac{1}{10}\bigg)-1^5\bigg(\frac{1}{12}-\frac{1}{10}\bigg)}\\
 
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&&\\
 
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& = & \displaystyle{3125\bigg(\frac{19}{60}\bigg)-\frac{-1}{60}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{59376}{60}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{4948}{5}.}\\
 
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\end{array}</math>
 
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|}
 
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
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|&nbsp;&nbsp; <math>\frac{4948}{5}</math>
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| &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 10:59, 9 April 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

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