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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 | + | <span class="exam">Evaluate the indefinite and definite integrals. |
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| − | <span class="exam">and finds that it can be approximated by the function | + | <span class="exam">(a) <math>\int x^2\sqrt{1+x^3}~dx</math> |
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| − | ::<math>s(t)=t(25-5t)+18</math>
| + | <span class="exam">(b) <math>\int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~dx</math> |
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| | + | (insert picture of handwritten solution) |
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| − | <span class="exam">where <math style="vertical-align: 0px">t</math> is the time (in years) since the stock was purchased. | + | [[009B Sample Midterm 1, Problem 2 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']] |
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| − | <span class="exam">Find the average price of the stock over the first five years.
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !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
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| − | | <math style="vertical-align: -18px">f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.</math>
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| − | '''Solution:'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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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>
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| − | |Using the average value formula, we have
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| − | | <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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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |First, we distribute to get
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| − | | <math>s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-5t^2+18~dt.</math>
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| − | |Then, we integrate to get
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| − | | <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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |We now evaluate to get
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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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| − | &&\\
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| − | & = & \displaystyle{\frac{233}{6}}\\
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| − | &&\\
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| − | & \approx & \displaystyle{$38.83.}
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | <math>\frac{233}{6}\approx $38.83</math>
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| − | |}
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| | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
Evaluate the indefinite and definite integrals.
(a) 
(b) 
(insert picture of handwritten solution)
Detailed Solution for this Problem
Return to Sample Exam