Difference between revisions of "009B Sample Midterm 1, Problem 3"

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<span class="exam">Evaluate the indefinite and definite integrals.
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<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>
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::<math>P'(t)=500e^{-t}</math>
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::<span class="exam">b) <math>\int_{1}^{e} x^3\ln x~dx</math>
 
  
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<span class="exam">where &nbsp;<math style="vertical-align: -5px">P(t)</math>&nbsp; is the population after &nbsp;<math style="vertical-align: 0px">t</math>&nbsp; months.
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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<span class="exam">(a) &nbsp; Find a formula for the population size after &nbsp;<math style="vertical-align: 0px">t</math>&nbsp; months, given that the population is &nbsp;<math style="vertical-align: 0px">2000</math>&nbsp; at &nbsp;<math style="vertical-align: 0px">t=0.</math>
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!Foundations: &nbsp;  
 
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|'''1.''' Integration by parts tells us that <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math>
 
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|'''2.''' How would you integrate <math style="vertical-align: -12px">\int x\ln x~dx?</math>
 
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::You could use integration by parts.
 
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::Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x~dx.</math> Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -12px">v=\frac{x^2}{2}.</math> Thus,
 
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::<math>\begin{array}{rcl}
 
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\displaystyle{\int x\ln x~dx} & = & \displaystyle{\frac{x^2\ln x}{2}-\int \frac{x}{2}~dx}\\
 
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&&\\
 
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& = & \displaystyle{\frac{x^2\ln x}{2}-\frac{x^2}{4}+C.}\\
 
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\end{array}</math>
 
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|}
 
  
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'''Solution:'''
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<span class="exam">(b) &nbsp; Use your answer to part (a) to find the size of the population after one month.
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<hr>
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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: &nbsp;
 
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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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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[[009B Sample Midterm 1, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
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!Step 2: &nbsp;
 
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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>\begin{array}{rcl}
 
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\displaystyle{\int x^2 e^x~dx} & = & \displaystyle{x^2e^x-\bigg(2xe^x-\int 2e^x~dx\bigg)}\\
 
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&&\\
 
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& = & \displaystyle{x^2e^x-2xe^x+2e^x+C.}\\
 
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\end{array}</math>
 
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'''(b)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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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>\begin{array}{rcl}
 
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\displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\left.\ln x \bigg(\frac{x^4}{4}\bigg)\right|_{1}^{e}-\int_1^e \frac{x^3}{4}~dx }\\
 
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&&\\
 
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& = & \displaystyle{\left.\ln x \bigg(\frac{x^4}{4}\bigg)-\frac{x^4}{16}\right|_{1}^{e}.}\\
 
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\end{array}</math>
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|Now, we evaluate to get
 
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::<math>\begin{array}{rcl}
 
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\displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\bigg((\ln e) \frac{e^4}{4}-\frac{e^4}{16}\bigg)-\bigg((\ln 1) \frac{1^4}{4}-\frac{1^4}{16}\bigg)}\\
 
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&&\\
 
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& = & \displaystyle{\frac{e^4}{4}-\frac{e^4}{16}+\frac{1}{16}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{3e^4+1}{16}.}\\
 
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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: &nbsp;
 
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|&nbsp;&nbsp; '''(a)''' &nbsp; <math>x^2e^x-2xe^x+2e^x+C</math>
 
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|&nbsp;&nbsp; '''(b)''' &nbsp; <math>\frac{3e^4+1}{16}</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>''']]

Latest revision as of 10:04, 20 November 2017

A population grows at a rate

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 P(t)}   is the population after  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}   months.

(a)   Find a formula for the population size after  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}   months, given that the population is  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 2000}   at  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=0.}

(b)   Use your answer to part (a) to find the size of the population after one month.

Solution


Detailed Solution


Return to Sample Exam

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