009B Sample Final 1, Problem 2 - Revision history

Kayla Murray at 00:15, 3 December 2017

2017-12-03T00:15:18Z

Kayla Murray at 19:39, 18 March 2017

2017-03-18T19:39:51Z

Kayla Murray at 17:50, 27 February 2017

2017-02-27T17:50:00Z

Kayla Murray at 02:10, 19 February 2017

2017-02-19T02:10:27Z

Kayla Murray at 02:09, 19 February 2017

2017-02-19T02:09:57Z

Grad at 19:49, 25 February 2016

2016-02-25T19:49:27Z

Grad: /* Tempd */

2016-02-25T19:48:25Z

Tempd

Grad at 19:43, 25 February 2016

2016-02-25T19:43:33Z

Grad: /* Tempb */

2016-02-25T19:42:30Z

Tempb

Grad: /* Tempa */

2016-02-25T19:41:00Z

Tempa

@@ Line 10: / Line 10: @@
 <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;"
+<hr>
-!Foundations: &nbsp;
+[[009B Sample Final 1, Problem 2 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009B Sample Final 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 16: / Line 16: @@
 |-
+|
-&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>
+&nbsp; &nbsp; &nbsp; &nbsp;You could use &nbsp;<math style="vertical-align: -1px">u</math>-substitution.
 |-
+|
@@ Line 38: / Line 40: @@
 |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>
+|&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>
 |}
@@ Line 70: / Line 72: @@
 !Step 2: &nbsp;
 |-
-|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.
+|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 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 &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">(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. 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>
 |}
@@ Line 30: / Line 30: @@
 !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: 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
 |}
@@ Line 43: / Line 47: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 f(x) & = & \displaystyle{\int_{-1}^{x} \sin(t^2)2t~dt}\\
 &&\\
@@ Line 60: / Line 64: @@
 !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 66: / Line 70: @@
 !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 just a constant.
 |}
@@ Line 76: / Line 80: @@
 |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;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>
 |}
@@ Line 86: / Line 90: @@
 !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>
 |}
@@ Line 99: / Line 103: @@
 |-
+|
-::<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>
 |}
@@ Line 106: / Line 110: @@
 !Final Answer: &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
 |-
-|'''(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>''']]

← Older revision		Revision as of 02:10, 19 February 2017
Line 9:		Line 9:

	<span class="exam">(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.		<span class="exam">(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.
−
−	<span class="exam">(e) 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;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ 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 <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">(e) 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;"
@@ Line 23: / Line 23: @@
 ::So, we get <math style="vertical-align: -14px">\int e^u~du=e^u+C=e^{x^2}+C</math>.
 |}
 '''Solution:'''
@@ Line 102: / Line 103: @@
 ::<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;"

@@ Line 94: / Line 94: @@
 |}
 == Tempd ==
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
+!(d) &nbsp;
 |-
 |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>
+::<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t~dt\bigg)\,=\,\sin(x^2)2x.</math>
 |}
@@ Line 108: / Line 106: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' <math>f(x)=-\cos(x^2)+\cos(1)</math>
+|'''(a)''' &nbsp;<math>f(x)=-\cos(x^2)+\cos(1)</math>
 |-
-|'''(b)''' <math>f'(x)=\sin(x^2)2x</math>
+|'''(b)''' &nbsp;<math>f'(x)=\sin(x^2)2x</math>
 |-
-|'''(c)''' '''The Fundamental Theorem of Calculus, Part 1'''
+|'''(c)''' &nbsp;'''<u>The Fundamental Theorem of Calculus, Part 1</u>'''
 |-
-|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;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>.
+|&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>.
 |-
-|'''The Fundamental Theorem of Calculus, Part 2'''
+|'''<u>The Fundamental Theorem of Calculus, Part 2</u>'''
 |-
-|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;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>.
+|&nbsp;&nbsp;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>
+|'''(d)''' &nbsp;<math style="vertical-align: -5px">\sin(x^2)2x</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 25: / Line 25: @@
 '''Solution:'''
-== Tempa ==
 '''(a)'''
@@ Line 55: / Line 55: @@
 |}
-== Tempb ==
 '''(b)'''
@@ Line 70: / Line 70: @@
 |}
 '''(c)'''

@@ 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=2tdt</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 in 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: 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>.
 |}
@@ Line 51: / Line 51: @@
 & = & \displaystyle{-\cos(u)\bigg|_{1}^{x^2}}\\
 &&\\
-& = & \displaystyle{-\cos(x^2)+\cos(1)}\\
+& = & \displaystyle{-\cos(x^2)+\cos(1)}.\\
 \end{array}</math>
 |}
 == Tempb ==
 '''(b)'''