Difference between revisions of "009B Sample Midterm 1, Problem 2"
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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 | |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> | + | | <math style="vertical-align: -18px">f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.</math> |
|} | |} | ||
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|This problem wants us to find the average value of <math style="vertical-align: -5px">s(t)</math> over the interval <math style="vertical-align: -5px">[0,5].</math> | |This problem wants us to find the average value of <math style="vertical-align: -5px">s(t)</math> over the interval <math style="vertical-align: -5px">[0,5].</math> | ||
|- | |- | ||
| − | |Using the formula | + | |Using the average value formula, we have |
|- | |- | ||
| − | | <math style="vertical-align: 0px">s_{\text{avg}}=\frac{1}{5-0} \int_0^5 t(25-5t)+18~dt.</math> | + | | <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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|First, we distribute to get | |First, we distribute to get | ||
|- | |- | ||
| − | | <math>s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-t^2+18~dt.</math> | + | | <math>s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-t^2+18~dt.</math> |
|- | |- | ||
|Then, we integrate to get | |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> | + | | <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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|We now evaluate to get | |We now evaluate to get | ||
|- | |- | ||
| − | | <math>\begin{array}{rcl} | + | | <math>\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{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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!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | <math>\frac{233}{6}</math> | + | | <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 16:36, 7 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 |
| 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.} |
![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
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-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 \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}.} \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}} |