Difference between revisions of "009B Sample Final 1, Problem 2"

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<span class="exam"> We would like to evaluate
 
<span class="exam"> We would like to evaluate
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:::::<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg).</math>
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::<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg).</math>
  
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<span class="exam">a) Compute <math style="vertical-align: -15px">f(x)=\int_{-1}^{x} \sin(t^2)2t\,dt</math>.
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<span class="exam">(a) Compute &nbsp;<math style="vertical-align: -15px">f(x)=\int_{-1}^{x} \sin(t^2)2t\,dt.</math>
  
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<span class="exam">b) Find <math style="vertical-align: -5px">f'(x)</math>.
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<span class="exam">(b) Find &nbsp;<math style="vertical-align: -5px">f'(x).</math>
  
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<span class="exam">c) State the Fundamental Theorem of Calculus.
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<span class="exam">(c) State the Fundamental Theorem of Calculus.
  
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<span class="exam">d) Use the Fundamental Theorem of Calculus to compute&thinsp; <math style="vertical-align: -15px">\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg)</math> &thinsp;without first computing the integral.
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<span class="exam">(d) Use the Fundamental Theorem of Calculus to compute &nbsp;<math style="vertical-align: -15px">\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg)</math>&nbsp; without first computing the integral.
  
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<span class="exam">d) Use the Fundamental Theorem of Calculus to compute&thinsp; <math style="vertical-align: -15px">\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2tdt\bigg)</math> &thinsp;without first computing the integral.
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<hr>
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[[009B Sample Final 1, Problem 2 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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|-
 
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|How would you integrate <math>\int e^{x^2}2x~dx</math>?
 
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|-
 
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|
 
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::You could use <math style="vertical-align: -1px">u</math>-substitution. Let <math style="vertical-align: 0px">u=x^2</math>. Then, <math style="vertical-align: 0px">du=2xdx</math>.
 
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::So, we get <math style="vertical-align: -14px">\int e^u~du=e^u+C=e^{x^2}+C</math>.
 
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|}
 
  
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'''Solution:'''
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[[009B Sample Final 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
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== Tempa ==
 
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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 <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: 0px">u=t^2</math>. Then, <math style="vertical-align: 0px">du=2t\,dt</math>.
 
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|-
 
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|Since this is a definite integral, we need to change the bounds of integration.
 
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|-
 
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|Plugging our values into the equation <math style="vertical-align: 0px">u=t^2</math>, we get <math style="vertical-align: -5px">u_1=(-1)^2=1</math> and <math style="vertical-align: -3px">u_2=x^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 have
 
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|
 
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::<math>\begin{array}{rcl}
 
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f(x) & = & \displaystyle{\int_{-1}^{x} \sin(t^2)2t~dt}\\
 
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&&\\
 
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& = & \displaystyle{\int_{1}^{x^2} \sin(u)~du}\\
 
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&&\\
 
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& = & \displaystyle{-\cos(u)\bigg|_{1}^{x^2}}\\
 
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&&\\
 
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& = & \displaystyle{-\cos(x^2)+\cos(1)}.\\
 
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\end{array}</math>
 
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|}
 
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== Tempb ==
 
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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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|From part (a), we have <math style="vertical-align: -5px">f(x)=-\cos(x^2)+\cos(1)</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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|If we take the derivative, we get <math style="vertical-align: -5px">f'(x)=\sin(x^2)2x</math>, since <math style="vertical-align: -5px">\cos(1)</math> is just a constant.
 
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|}
 
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== Tempc ==
 
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'''(c)'''
 
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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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|The Fundamental Theorem of Calculus has two parts.
 
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|-
 
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|'''The Fundamental Theorem of Calculus, Part 1'''
 
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|-
 
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|Let <math>f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -14px">F(x)=\int_a^x f(t)~dt</math>.
 
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|-
 
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|Then, <math style="vertical-align: 0px">F</math> is a differentiable function on <math style="vertical-align: -5px">(a,b)</math> and <math style="vertical-align: -5px">F'(x)=f(x)</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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|'''The Fundamental Theorem of Calculus, Part 2'''
 
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|-
 
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|Let <math>f</math> be continuous on <math>[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math>f</math>.
 
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|Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
 
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|}
 
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== Tempd ==
 
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'''(d)'''
 
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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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|By the '''Fundamental Theorem of Calculus, Part 1''',
 
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|-
 
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|
 
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::<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t~dt\bigg)=\sin(x^2)2x</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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|-
 
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|'''(a)''' <math>f(x)=-\cos(x^2)+\cos(1)</math>
 
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|-
 
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|'''(b)''' <math>f'(x)=\sin(x^2)2x</math>
 
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|-
 
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|'''(c)''' '''The Fundamental Theorem of Calculus, Part 1'''
 
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|-
 
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|Let <math>f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -14px">F(x)=\int_a^x f(t)~dt</math>.
 
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|-
 
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|Then, <math style="vertical-align: 0px">F</math> is a differentiable function on <math style="vertical-align: -5px">(a,b)</math> and <math style="vertical-align: -5px">F'(x)=f(x)</math>. 
 
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|-
 
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|'''The Fundamental Theorem of Calculus, Part 2'''
 
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|-
 
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|Let <math>f</math> be continuous on <math>[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math>f</math>.
 
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|-
 
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|Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
 
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|-
 
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|'''(d)''' <math style="vertical-align: -5px">\sin(x^2)2x</math>
 
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|}
 
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 17:15, 2 December 2017

We would like to evaluate

(a) Compute  

(b) Find  

(c) State the Fundamental Theorem of Calculus.

(d) Use the Fundamental Theorem of Calculus to compute    without first computing the integral.

Solution


Detailed Solution


Return to Sample Exam

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