009B Sample Final 1, Problem 2 - Revision history

MathAdmin at 16:41, 10 April 2017

2017-04-10T16:41:31Z

MathAdmin at 22:10, 18 April 2016

2016-04-18T22:10:20Z

MathAdmin at 19:51, 18 April 2016

2016-04-18T19:51:23Z

MathAdmin: Created page with " We would like to evaluate ::::: $\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg).$ a) Compute
2016-02-27T05:38:40Z

Created page with " We would like to evaluate :::::<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg).</math> a) Compute <math style="vertical-ali..."

New page
 We would like to evaluate
:::::<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg).</math>

a) Compute <math style="vertical-align: -15px">f(x)=\int_{-1}^{x} \sin(t^2)2t\,dt</math>.

b) Find <math style="vertical-align: -5px">f'(x)</math>.

c) State the Fundamental Theorem of Calculus.

d) Use the Fundamental Theorem of Calculus to compute  <math style="vertical-align: -15px">\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg)</math>  without first computing the integral.

d) Use the Fundamental Theorem of Calculus to compute  <math style="vertical-align: -15px">\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2tdt\bigg)</math>  without first computing the integral.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|How would you integrate <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>.
|-
|
::So, we get <math style="vertical-align: -14px">\int e^u~du=e^u+C=e^{x^2}+C</math>.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|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>.
|-
|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>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
||So, we have
|-
|
::<math>\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}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|From part (a), we have <math style="vertical-align: -5px">f(x)=-\cos(x^2)+\cos(1)</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|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.
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|The Fundamental Theorem of Calculus has two parts.
|-
|'''The Fundamental Theorem of Calculus, Part 1'''
|-
|  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>.
|-
|  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>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|'''The Fundamental Theorem of Calculus, Part 2'''
|-
|  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>.
|-
|  Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(d)  
|-
|By the '''Fundamental Theorem of Calculus, Part 1''',
|-
|
::<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;"
!Final Answer:  
|-
|'''(a)'''  <math>f(x)=-\cos(x^2)+\cos(1)</math>
|-
|'''(b)'''  <math>f'(x)=\sin(x^2)2x</math>
|-
|'''(c)'''  '''The Fundamental Theorem of Calculus, Part 1'''
|-
|  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>.
|-
|  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>.
|-
|'''The Fundamental Theorem of Calculus, Part 2'''
|-
|  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>.
|-
|  Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
|-
|'''(d)'''  <math style="vertical-align: -5px">\sin(x^2)2x</math>
|}
[[009B_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 1: / Line 1: @@
 <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.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !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.
 |-
+|
-::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:'''
@@ Line 31: / Line 32: @@
 !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.
 |-
 |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
 |}
@@ Line 44: / Line 49: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 f(x) & = & \displaystyle{\int_{-1}^{x} \sin(t^2)2t~dt}\\
 &&\\
@@ Line 61: / Line 66: @@
 !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>
 |}
@@ Line 67: / Line 72: @@
 !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.
 |}
@@ Line 77: / Line 82: @@
 |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>
 |-
+|&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>
 |}
@@ Line 89: / Line 92: @@
 !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>
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;Then, &nbsp;<math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 |}
@@ Line 104: / Line 105: @@
 |-
+|
-::<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;"
 !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;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>''']]

@@ Line 2: / Line 2: @@
 :::::<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 <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 <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&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&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.
+::<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;"
 !Foundations: &nbsp;
 |-
-|How would you integrate <math>\int e^{x^2}2x~dx</math>?
+|How would you integrate <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>.
+::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>
 |-
+|
-::So, we get <math style="vertical-align: -14px">\int e^u~du=e^u+C=e^{x^2}+C</math>.
+::So, we get <math style="vertical-align: -14px">\int e^u~du=e^u+C=e^{x^2}+C.</math>
 |}
@@ Line 31: / Line 31: @@
 !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 <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>
 |-
 |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 <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>
 |}
@@ Line 61: / Line 61: @@
 !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 <math style="vertical-align: -5px">f(x)=-\cos(x^2)+\cos(1).</math>
 |}
@@ Line 67: / Line 67: @@
 !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 <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.
 |}
@@ Line 79: / Line 79: @@
 |'''<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>.
 |}
@@ Line 89: / Line 91: @@
 |'''<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;Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
 |}

@@ Line 110: / Line 110: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp;<math>f(x)=-\cos(x^2)+\cos(1)</math>
+|&nbsp;&nbsp; '''(a)''' &nbsp;<math>f(x)=-\cos(x^2)+\cos(1)</math>
 |-
-|'''(b)''' &nbsp;<math>f'(x)=\sin(x^2)2x</math>
+|&nbsp;&nbsp; '''(b)''' &nbsp;<math>f'(x)=\sin(x^2)2x</math>
 |-
-|'''(c)''' &nbsp;'''<u>The Fundamental Theorem of Calculus, Part 1</u>'''
+|&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>.
@@ Line 120: / Line 120: @@
 |&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; '''<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>.
@@ Line 126: / Line 126: @@
 |&nbsp;&nbsp;Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
 |-
-|'''(d)''' &nbsp;<math style="vertical-align: -5px">\sin(x^2)2x</math>
+|&nbsp;&nbsp; '''(d)''' &nbsp;<math style="vertical-align: -5px">\sin(x^2)2x</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]