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>
 +
|-
 +
|Using the average value formula, we have
 
|-
 
|-
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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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|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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|&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  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 t}   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  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)}   on an 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 [a,b]}   is given by
        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_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.}


Solution:

Step 1:  
This problem wants us to find the average value of  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(t)}   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 average value formula, 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-5t^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 \begin{array}{rcl} \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}}\\ &&\\ & \approx & \displaystyle{$38.83.} \end{array}}


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}\approx $38.83}

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