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

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::<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.
  
 
{| 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. 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>
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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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>.
+
|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: 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>.
+
|Plugging our values into the equation &nbsp;<math style="vertical-align: -4px">u=t^2,</math>&nbsp; we get  
 +
|-
 +
|<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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|-
 
|-
 
|
 
|
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::<math>\begin{array}{rcl}
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&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;  
 
|-
 
|-
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|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;
 
|-
 
|-
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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.
+
|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 just 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'''
 
|-
 
|-
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|&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; &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>
 
|-
 
|-
−
|&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; &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>
 
|}
 
|}
  
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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;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;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>
 
|-
 
|-
−
|&nbsp;&nbsp;Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
+
|&nbsp; &nbsp; &nbsp; &nbsp;Then, &nbsp;<math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 
|}
 
|}
  
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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>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t~dt\bigg)\,=\,\sin(x^2)2x.</math>
 
|}
 
|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
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|'''(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>
−
|-
 
−
|'''(b)''' &nbsp;<math>f'(x)=\sin(x^2)2x</math>
 
−
|-
 
−
|'''(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>. 
 
−
|-
 
−
|'''<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
 
|-
 
|-
−
|'''(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>''']]

Revision as of 10:50, 27 February 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.

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 (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 \int e^u~du=e^u+C=e^{x^2}+C.}


Solution:

(a)

Step 1:  
We proceed using  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 u} -substitution.
Let  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 u=t^2.}   Then,  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 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 (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 u=t^2,}   we get
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 u_1=(-1)^2=1}   and  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 u_2=x^2.}
Step 2:  
So, we have

       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 \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  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 f(x)=-\cos(x^2)+\cos(1).}
Step 2:  
If we take the derivative, we get  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 f'(x)=\sin(x^2)2x,}   since  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 \cos(1)}   is just a constant.

(c)

Step 1:  
The Fundamental Theorem of Calculus has two parts.
The Fundamental Theorem of Calculus, Part 1
       Let  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 f}   be continuous on  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 [a,b]}   and let  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 F(x)=\int_a^x f(t)~dt.}
       Then,  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 F}   is a differentiable function on  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 (a,b)}   and  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 F'(x)=f(x).}
Step 2:  
The Fundamental Theorem of Calculus, Part 2
       Let  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 f}   be continuous on  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 [a,b]}   and let  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 F}   be any antiderivative of  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 f.}
       Then,  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 \int_a^b f(x)~dx=F(b)-F(a).}
(d)  
By the Fundamental Theorem of Calculus, Part 1,

       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 \frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t~dt\bigg)\,=\,\sin(x^2)2x.}


Final Answer:  
   (a)    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 f(x)=-\cos(x^2)+\cos(1)}
   (b)    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 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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