Difference between revisions of "009B Sample Midterm 1, Problem 2"
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!Foundations: | !Foundations: | ||
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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>. | + | |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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 1: | !Step 1: | ||
| + | |- | ||
| + | |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: | ||
|- | |- | ||
| − | | <math style="vertical-align: 0px"> | + | | <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: | !Step 2: | ||
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| − | | | + | |First, we distribute to get |
|- | |- | ||
| − | | | + | | <math>s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-t^2+18~dt.</math> |
|- | |- | ||
| − | | | + | |Then, we integrate to get |
| + | |- | ||
| + | | <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: | !Step 3: | ||
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| − | | | + | |We now evaluate to get |
| − | |||
| − | |||
| − | |||
| − | |||
|- | |- | ||
| − | | | + | | <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: | !Final Answer: | ||
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| − | | <math></math> | + | | <math>\frac{233}{6}</math> |
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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 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
- 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)=t(25-5t)+18}
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 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}} |