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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<hr>
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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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|-
 
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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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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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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|}
 
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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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|-
 
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|Again, we proceed using u substitution. Let <math>u=\cos(x)</math>. Then, <math>du=-\sin(x)dx</math>.
 
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|Since this is a definite integral, we need to change the bounds of integration.
 
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|We have <math>u_1=\cos\bigg(-\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math> and <math>u_2=\cos\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math>.
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|-
 
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|So, we get
 
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|<math>\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx=\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} -u^2~du=\left.\frac{-u^3}{3}\right|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}}=0</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)''' <math>\frac{-1}{3}\cos(x^3)+C</math>
 
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|'''(b)''' <math>0</math>
 
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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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