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
:::::<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg).</math>
+
::<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg).</math>
  
::<span class="exam">a) Compute <math style="vertical-align: -15px">f(x)=\int_{-1}^{x} \sin(t^2)2t\,dt.</math>
+
<span class="exam">(a) Compute &nbsp;<math style="vertical-align: -15px">f(x)=\int_{-1}^{x} \sin(t^2)2t\,dt.</math>
  
::<span class="exam">b) Find <math style="vertical-align: -5px">f'(x).</math>
+
<span class="exam">(b) Find &nbsp;<math style="vertical-align: -5px">f'(x).</math>
  
::<span class="exam">c) State the Fundamental Theorem of Calculus.
+
<span class="exam">(c) State the Fundamental Theorem of Calculus.
  
::<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.
+
<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.
 
 
::<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.
 
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
|How would you integrate <math>\int e^{x^2}2x~dx?</math>
+
|How would you integrate &nbsp;<math>\int e^{x^2}2x~dx?</math>
 
|-
 
|-
 
|
 
|
::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>
+
&nbsp; &nbsp; &nbsp; &nbsp;You could use &nbsp;<math style="vertical-align: -1px">u</math>-substitution.  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;Let &nbsp;<math style="vertical-align: 0px">u=x^2.</math>&nbsp; Then, &nbsp;<math style="vertical-align: 0px">du=2xdx.</math>
 
|-
 
|-
 
|
 
|
::So, we get <math style="vertical-align: -14px">\int e^u~du=e^u+C=e^{x^2}+C.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;So, we get &nbsp;<math style="vertical-align: -14px">\int e^u~du=e^u+C=e^{x^2}+C.</math>
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|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>
+
|We proceed using &nbsp;<math style="vertical-align: 0px">u</math>-substitution.  
 +
|-
 +
|Let &nbsp;<math style="vertical-align: 0px">u=t^2.</math>&nbsp; Then, &nbsp;<math style="vertical-align: 0px">du=2t\,dt.</math>  
 
|-
 
|-
 
|Since this is a definite integral, we need to change the bounds of integration.  
 
|Since this is a definite integral, we need to change the bounds of integration.  
 
|-
 
|-
|Plugging our values into the equation <math style="vertical-align: -4px">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>
+
|Plugging our values into the equation &nbsp;<math style="vertical-align: -4px">u=t^2,</math>&nbsp; we get  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">u_1=(-1)^2=1</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">u_2=x^2.</math>
 
|}
 
|}
  
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|-
 
|-
 
|
 
|
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
f(x) & = & \displaystyle{\int_{-1}^{x} \sin(t^2)2t~dt}\\
 
f(x) & = & \displaystyle{\int_{-1}^{x} \sin(t^2)2t~dt}\\
 
&&\\
 
&&\\
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|From part (a), we have <math style="vertical-align: -5px">f(x)=-\cos(x^2)+\cos(1).</math>
+
|From part (a), we have &nbsp;<math style="vertical-align: -5px">f(x)=-\cos(x^2)+\cos(1).</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|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.
+
|If we take the derivative, we get &nbsp;<math style="vertical-align: -5px">f'(x)=\sin(x^2)2x,</math>&nbsp; since &nbsp;<math style="vertical-align: -5px">\cos(1)</math>&nbsp; is a constant.
 
|}
 
|}
  
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|The Fundamental Theorem of Calculus has two parts.  
 
|The Fundamental Theorem of Calculus has two parts.  
 
|-
 
|-
|'''<u>The Fundamental Theorem of Calculus, Part 1</u>'''
+
|'''The Fundamental Theorem of Calculus, Part 1'''
 
|-
 
|-
|
+
|&nbsp; &nbsp; &nbsp; &nbsp;Let &nbsp;<math>f</math>&nbsp; be continuous on &nbsp;<math style="vertical-align: -5px">[a,b]</math>&nbsp; and let &nbsp;<math style="vertical-align: -14px">F(x)=\int_a^x f(t)~dt.</math>
: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>
 
 
|-
 
|-
|
+
|&nbsp; &nbsp; &nbsp; &nbsp;Then, &nbsp;<math style="vertical-align: 0px">F</math>&nbsp; is a differentiable function on &nbsp;<math style="vertical-align: -5px">(a,b)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">F'(x)=f(x).</math>
: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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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|'''<u>The Fundamental Theorem of Calculus, Part 2</u>'''
+
|'''The Fundamental Theorem of Calculus, Part 2'''
 
|-
 
|-
|
+
|&nbsp; &nbsp; &nbsp; &nbsp;Let &nbsp;<math>f</math>&nbsp; be continuous on &nbsp;<math>[a,b]</math>&nbsp; and let &nbsp;<math style="vertical-align: 0px">F</math>&nbsp; be any antiderivative of &nbsp;<math>f.</math>
: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>
 
 
|-
 
|-
|
+
|&nbsp; &nbsp; &nbsp; &nbsp;Then, &nbsp;<math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
:Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 
 
|}
 
|}
  
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|-
 
|-
 
|
 
|
::<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t~dt\bigg)\,=\,\sin(x^2)2x.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t~dt\bigg)\,=\,\sin(x^2)2x.</math>
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|&nbsp;&nbsp; '''(a)''' &nbsp;<math>f(x)=-\cos(x^2)+\cos(1)</math>
+
|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math>f(x)=-\cos(x^2)+\cos(1)</math>
|-
 
|&nbsp;&nbsp; '''(b)''' &nbsp;<math>f'(x)=\sin(x^2)2x</math>
 
|-
 
|&nbsp;&nbsp; '''(c)''' &nbsp;'''<u>The Fundamental Theorem of Calculus, Part 1</u>'''
 
|-
 
|&nbsp;&nbsp;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>.
 
|-
 
|&nbsp;&nbsp;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>. 
 
|-
 
|&nbsp;&nbsp; '''<u>The Fundamental Theorem of Calculus, Part 2</u>'''
 
 
|-
 
|-
|&nbsp;&nbsp;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>.
+
|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;<math>f'(x)=\sin(x^2)2x</math>
 
|-
 
|-
|&nbsp;&nbsp;Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
+
|&nbsp; &nbsp;'''(c)''' &nbsp; &nbsp;See above
 
|-
 
|-
|&nbsp;&nbsp; '''(d)''' &nbsp;<math style="vertical-align: -5px">\sin(x^2)2x</math>
+
|&nbsp; &nbsp;'''(d)''' &nbsp; &nbsp;<math style="vertical-align: -5px">\sin(x^2)2x</math>
 
|}
 
|}
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 08:41, 10 April 2017

We would like to evaluate

(a) Compute  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=\int _{-1}^{x}\sin(t^{2})2t\,dt.}

(b) Find  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f'(x).}

(c) State the Fundamental Theorem of Calculus.

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

Foundations:  
How would you integrate  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int e^{x^{2}}2x~dx?}

       You could use  -substitution.

       Let    Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=2xdx.}

       So, we get  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int e^{u}~du=e^{u}+C=e^{x^{2}}+C.}


Solution:

(a)

Step 1:  
We proceed using  -substitution.
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=t^{2}.}   Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=2t\,dt.}
Since this is a definite integral, we need to change the bounds of integration.
Plugging our values into the equation  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=t^{2},}   we get
         and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u_{2}=x^{2}.}
Step 2:  
So, we have

       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}f(x)&=&\displaystyle {\int _{-1}^{x}\sin(t^{2})2t~dt}\\&&\\&=&\displaystyle {\int _{1}^{x^{2}}\sin(u)~du}\\&&\\&=&\displaystyle {-\cos(u){\bigg |}_{1}^{x^{2}}}\\&&\\&=&\displaystyle {-\cos(x^{2})+\cos(1)}.\\\end{array}}}


(b)

Step 1:  
From part (a), we have  
Step 2:  
If we take the derivative, we get  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f'(x)=\sin(x^{2})2x,}   since  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \cos(1)}   is a constant.

(c)

Step 1:  
The Fundamental Theorem of Calculus has two parts.
The Fundamental Theorem of Calculus, Part 1
       Let    be continuous on    and let  
       Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle F}   is a differentiable function on    and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle F'(x)=f(x).}
Step 2:  
The Fundamental Theorem of Calculus, Part 2
       Let    be continuous on    and let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle F}   be any antiderivative of  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f.}
       Then,  
(d)  
By the Fundamental Theorem of Calculus, Part 1,

       


Final Answer:  
   (a)    
   (b)    Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f'(x)=\sin(x^{2})2x}
   (c)    See above
   (d)    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 \sin(x^2)2x}

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