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| − | <span class="exam">Evaluate the indefinite and definite integrals. | + | <span class="exam"> A population grows at a rate |
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| − | ::<span class="exam">a) <math>\int x^2 e^x~dx</math> | + | ::<math>P'(t)=500e^{-t}</math> |
| − | ::<span class="exam">b) <math>\int_{1}^{e} x^3\ln x~dx</math>
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| | + | <span class="exam">where <math style="vertical-align: -5px">P(t)</math> is the population after <math style="vertical-align: 0px">t</math> months. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <span class="exam">(a) Find a formula for the population size after <math style="vertical-align: 0px">t</math> months, given that the population is <math style="vertical-align: 0px">2000</math> at <math style="vertical-align: 0px">t=0.</math> |
| − | !Foundations:
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| − | |Review integration by parts.
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| − | |}
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| − | '''Solution:''' | + | <span class="exam">(b) Use your answer to part (a) to find the size of the population after one month. |
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| | + | [[009B Sample Midterm 1, Problem 3 Solution|'''<u>Solution</u>''']] |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |We proceed using integration by parts. Let <math style="vertical-align: 0px">u=x^2</math> and <math style="vertical-align: 0px">dv=e^xdx</math>. Then, <math style="vertical-align: 0px">du=2xdx</math> and <math style="vertical-align: 0px">v=e^x</math>.
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| − | |Therefore, we have
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| − | | <math style="vertical-align: -12px">\int x^2 e^x~dx=x^2e^x-\int 2xe^x~dx</math>.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009B Sample Midterm 1, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Step 2:
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| − | |Now, we need to use integration by parts again. Let <math style="vertical-align: 0px">u=2x</math> and <math style="vertical-align: 0px">dv=e^xdx</math>. Then, <math style="vertical-align: 0px">du=2dx</math> and <math style="vertical-align: 0px">v=e^x</math>. | |
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| − | |Building on the previous step, we have
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| − | | <math style="vertical-align: -15px">\int x^2 e^x~dx=x^2e^x-\bigg(2xe^x-\int 2e^x~dx\bigg)=x^2e^x-2xe^x+2e^x+C</math>.
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| − | |}
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |We proceed using integration by parts. Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x^3dx</math>. Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -14px">v=\frac{x^4}{4}</math>.
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| − | |Therefore, we have
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| − | | <math style="vertical-align: -20px">\int_{1}^{e} x^3\ln x~dx=\left.\ln x \bigg(\frac{x^4}{4}\bigg)\right|_{1}^{e}-\int_1^e \frac{x^3}{4}~dx=\left.\ln x \bigg(\frac{x^4}{4}\bigg)-\frac{x^4}{16}\right|_{1}^{e}</math>.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Now, we evaluate to get
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| − | | <math style="vertical-align: -17px">\int_{1}^{e} x^3\ln x~dx=\bigg((\ln e) \frac{e^4}{4}-\frac{e^4}{16}\bigg)-\bigg((\ln 1) \frac{1^4}{4}-\frac{1^4}{16}\bigg)=\frac{e^4}{4}-\frac{e^4}{16}+\frac{1}{16}=\frac{3e^4+1}{16}</math>.
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |'''(a)''' <math>x^2e^x-2xe^x+2e^x+C</math>
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| − | |'''(b)''' <math>\frac{3e^4+1}{16}</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>''']] |
A population grows at a rate
where 
is the population after 
months.
(a) Find a formula for the population size after months, given that the population is 
at 
(b) Use your answer to part (a) to find the size of the population after one month.
Solution
Detailed Solution
Return to Sample Exam