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

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<span class="exam"> Compute the following integrals:
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<span class="exam"> Find a curve &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; with the following properties:  
  
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::<span class="exam">a) <math>\int x^2\sin (x^3) dx</math>
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<span class="exam">(i) &nbsp; <math style="vertical-align: -5px">f''(x)=6x</math>  
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::<span class="exam">b) <math>\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)dx</math>
 
  
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<span class="exam">(ii) &nbsp; Its graph passes through the point &nbsp;<math style="vertical-align: -5px">(0,1)</math>&nbsp; and has a horizontal tangent there.
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[[009B Sample Midterm 3, Problem 3 Solution|'''<u>Solution</u>''']]
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Foundations: &nbsp;
 
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| u substitution
 
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'''Solution:'''
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[[009B Sample Midterm 3, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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'''(a)'''
 
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!Step 1: &nbsp;
 
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|We proceed using u substitution. Let <math>u=x^3</math>. Then, <math>du=3x^2dx</math>.
 
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|Therefore, we have
 
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|<math>\int x^2\sin (x^3) dx=\int \frac{\sin(u)}{3}du</math>
 
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!Step 2: &nbsp;
 
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|We integrate to get
 
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|<math>\int x^2\sin (x^3) dx=\frac{-1}{3}\cos(u)+C=\frac{-1}{3}\cos(x^3)+C</math>
 
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'''(b)'''
 
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!Step 1: &nbsp;
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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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)''' <math>\frac{-1}{3}\cos(x^3)+C</math>
 
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|'''(b)'''
 
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[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 17:28, 23 November 2017

Find a curve    with the following properties:

(i)  

(ii)   Its graph passes through the point    and has a horizontal tangent there.

Solution


Detailed Solution


Return to Sample Exam

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