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

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(Created page with "<span class="exam">Evaluate the indefinite and definite integrals. ::<span class="exam">a) <math>\int x^2 e^x~dx</math> ::<span class="exam">b) <math>\int_{1}^{e} x^3\ln x~d...")
 
 
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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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|Review integration by parts.
 
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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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| &nbsp;&nbsp; <math style="vertical-align: -12px">\int x^2 e^x~dx=x^2e^x-\int 2xe^x~dx</math>.
 
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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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| &nbsp;&nbsp; <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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'''(b)'''
 
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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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| &nbsp;&nbsp; <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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!Step 2: &nbsp;
 
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|Now, we evaluate to get
 
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| &nbsp;&nbsp; <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: &nbsp;
 
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|'''(a)''' &nbsp; <math>x^2e^x-2xe^x+2e^x+C</math>
 
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|'''(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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