Difference between revisions of "009B Sample Midterm 2, Problem 3"

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<span class="exam"> Evaluate
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<span class="exam"> A particle moves along a straight line with velocity given by:
  
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::<span class="exam">a) <math style="vertical-align: -14px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt</math>  
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::::::<math>v(t)=-32t+200</math>
  
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::<span class="exam">b) <math style="vertical-align: -14px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx</math>
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<span class="exam">feet per second. Determine the total distance traveled by the particle
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<span class="exam">from time <math>t=0</math> to time <math>t=10.</math>
  
  
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!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
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|How would you integrate <math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx?</math>
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::You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=x^2+x.</math> Then, <math style="vertical-align: -4px">du=(2x+1)~dx.</math>
 
 
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::Thus, <math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx=\int \sqrt{u}=\frac{2}{3}u^{3/2}+C=\frac{2}{3}(x^2+x)^{3/2}+C.</math>
 
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|We multiply the product inside the integral to get
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| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\int_1^2 \bigg(8t^3-10+12-\frac{15}{t^3}\bigg)~dt=\int_1^2 (8t^3+2-15t^{-3})~dt</math>. 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|We integrate to get
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| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\left. 2t^4+2t+\frac{15}{2}t^{-2}\right|_1^2</math>.
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|We now evaluate to get
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| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=2(2)^4+2(2)+\frac{15}{2(2)^2}-\bigg(2+2+\frac{15}{2}\bigg)=36+\frac{15}{8}-4-\frac{15}{2}=\frac{211}{8}</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 use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=x^4+2x^2+4</math>. Then, <math style="vertical-align: -5px">du=(4x^3+4x)dx</math> and <math style="vertical-align: -14px">\frac{du}{4}=(x^3+x)dx</math>. Also, we need to change the bounds of integration.
 
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|Plugging in our values into the equation <math style="vertical-align: -2px">u=x^4+2x^2+4</math>, we get <math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math> and <math style="vertical-align: -4px">u_2=2^4+2(2)^2+4=28</math>.
 
 
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|Therefore, the integral becomes&nbsp; <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du</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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|We now have:
 
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| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{1}{4}\int_4^{28}\sqrt{u}~du=\left.\frac{1}{6}u^{\frac{3}{2}}\right|_4^{28}=\frac{1}{6}(28^{\frac{3}{2}}-4^{\frac{3}{2}})=\frac{1}{6}((\sqrt{28})^3-(\sqrt{4})^3)=\frac{1}{6}((2\sqrt{7})^3-2^3)</math>.
 
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|So, we have
 
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| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}</math>.
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|'''(a)''' &nbsp; <math>\frac{211}{8}</math>
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|'''(b)''' &nbsp; <math>\frac{28\sqrt{7}-4}{3}</math>
 
 
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[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 10:21, 5 February 2017

A particle moves along a straight line with velocity given by:

feet per second. Determine the total distance traveled by the particle

from time to time 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=10.}


Foundations:  

Solution:

(a)

Step 1:  
Step 2:  


Final Answer:  

Return to Sample Exam

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