009B Sample Midterm 2, Problem 2 - Revision history

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MathAdmin: Created page with " This problem has three parts: ::a) State the Fundamental Theorem of Calculus. ::b) Compute
2016-02-03T06:57:27Z

Created page with " This problem has three parts: ::a) State the Fundamental Theorem of Calculus. ::b) Compute <math style="ve..."

New page
 This problem has three parts:

::a) State the Fundamental Theorem of Calculus.

::b) Compute   <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt</math>.

::c) Evaluate <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Review the Fundamental Theorem of Calculus.
|}

'''Solution:'''

'''(a)'''

{| 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 style="vertical-align: -4px">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 style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">f</math>.
|-
|Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
|}
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let <math style="vertical-align: -15px">F(x)=\int_0^{\cos (x)}\sin (t)~dt</math>. The problem is asking us to find <math style="vertical-align: -5px">F'(x)</math>.
|-
|Let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -14px">G(x)=\int_0^x \sin(t)~dt</math>.
|-
|Then, <math style="vertical-align: -5px">F(x)=G(g(x))</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|If we take the derivative of both sides of the last equation, we get <math style="vertical-align: -5px">F'(x)=G'(g(x))g'(x)</math> by the Chain Rule.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, <math style="vertical-align: -5px">g'(x)=-\sin(x)</math> and <math style="vertical-align: -5px">G'(x)=\sin(x)</math> by the '''Fundamental Theorem of Calculus, Part 1'''.
|-
|Since <math style="vertical-align: -6px">G'(g(x))=\sin(g(x))=\sin(\cos(x))</math>, we have <math style="vertical-align: -5px">F'(x)=G'(g(x))\cdot g'(x)=\sin(\cos(x))\cdot(-\sin(x))</math>.
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
| Using the '''Fundamental Theorem of Calculus, Part 2''', we have
|-
|    <math>\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan(x)\biggr|_{0}^{\pi/4}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, we get
|-
|    <math style="vertical-align: -16px">\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan \bigg(\frac{\pi}{4}\bigg)-\tan (0)=1</math>.
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''
|-
|'''The Fundamental Theorem of Calculus, Part 1'''
|-
|Let <math style="vertical-align: -4px">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 style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">f</math>.
|-
|Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
|-
|'''(b)'''   <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt\,=\,\sin(\cos(x))\cdot(-\sin(x))</math>.
|-
|'''(c)''' <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx\,=\,1</math>.
|}
[[009B_Sample_Midterm_2|'''Return to Sample Exam''']]

← Older revision		Revision as of 01:09, 13 November 2017
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	<span class="exam">(b)   <math style="vertical-align: -14px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx</math>		<span class="exam">(b)   <math style="vertical-align: -14px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx</math>

		+	<hr>
		+	[[009B Sample Midterm 2, Problem 2 Solution\|'''<u>Solution</u>''']]

−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Foundations:  ~~
−	\|-
−	~~\|How would you integrate  <math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx?</math>~~
−	\|-
−	\|
−	~~        You can use  <math style="vertical-align: 0px">u</math>-substitution.~~
−	\|-
−	~~\|        Let  <math style="vertical-align: -2px">u=x^2+x.</math>~~
−	\|-
−	~~\|        Then,  <math style="vertical-align: -4px">du=(2x+1)~dx.</math>~~
−	\|-
−	\|
−	~~        Thus,~~
−	\|-
−	\|
−	~~        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\int (2x+1)\sqrt{x^2+x}~dx} & = & \displaystyle{\int \sqrt{u}~du}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{2}{3}u^{3/2}+C}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{2}{3}(x^2+x)^{3/2}+C.}~~
−	~~\end{array}</math>~~
−	\|}

		+	[[009B Sample Midterm 2, Problem 2 Detailed Solution\|'''<u>Detailed Solution</u>''']]

−	~~'''Solution:'''~~

−	~~'''(a)'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|We multiply the product inside the integral to get~~
−	\|-
−	\|
−	~~        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt} & = & \displaystyle{\int_1^2 \bigg(8t^3-10+12-\frac{15}{t^3}\bigg)~dt}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\int_1^2 (8t^3+2-15t^{-3})~dt.}~~
−	~~\end{array}</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|We integrate to get~~
−	\|-
−	~~\|        <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\left. 2t^4+2t+\frac{15}{2}t^{-2}\right\|_1^2.</math>~~
−	\|-
−	~~\|We now evaluate to get~~
−	\|-
−	\|
−	~~        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt} & = & \displaystyle{2(2)^4+2(2)+\frac{15}{2(2)^2}-\bigg(2+2+\frac{15}{2}\bigg)}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{36+\frac{15}{8}-4-\frac{15}{2}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{211}{8}.}~~
−	~~\end{array}</math>~~
−	\|}
−
−	~~'''(b)'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|We use  <math style="vertical-align: 0px">u</math>-substitution.~~
−	\|-
−	~~\|Let  <math style="vertical-align: -2px">u=x^4+2x^2+4.</math>~~
−	\|-
−	~~\|Then,  <math style="vertical-align: -5px">du=(4x^3+4x)dx</math>  and  <math style="vertical-align: -14px">\frac{du}{4}=(x^3+x)dx.</math>~~
−	\|-
−	~~\|Also, we need to change the bounds of integration.~~
−	\|-
−	~~\|Plugging in our values into the equation  <math style="vertical-align: -4px">u=x^4+2x^2+4,</math>  we get~~
−	\|-
−	~~\|       <math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math>  and  <math style="vertical-align: -5px">u_2=2^4+2(2)^2+4=28.</math>~~
−	\|-
−	~~\|Therefore, the integral becomes~~
−	\|-
−	~~\|        <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du.</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|We now have~~
−	\|-
−	\|
−	~~        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx} & = & \displaystyle{\frac{1}{4}\int_4^{28}\sqrt{u}~du}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\left.\frac{1}{6}u^{\frac{3}{2}}\right\|_4^{28}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{1}{6}(28^{\frac{3}{2}}-4^{\frac{3}{2}})}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{1}{6}((\sqrt{28})^3-(\sqrt{4})^3)}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{1}{6}((2\sqrt{7})^3-2^3).}~~
−	~~\end{array}</math>~~
−	\|-
−	~~\|Therefore,~~
−	\|-
−	~~\|         <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}.</math>~~
−	\|}
−
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Final Answer:  ~~
−	\|-
−	~~\|    '''(a)'''     <math>\frac{211}{8}</math>~~
−	\|-
−	~~\|    '''(b)'''     <math>\frac{28\sqrt{7}-4}{3}</math>~~
−	\|}
	[[009B_Sample_Midterm_2\|'''<u>Return to Sample Exam</u>''']]		[[009B_Sample_Midterm_2\|'''<u>Return to Sample Exam</u>''']]

@@ Line 103: / Line 103: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp;&nbsp; '''(a)'''
 |-
-|'''The Fundamental Theorem of Calculus, Part 1'''
+|&nbsp;&nbsp; '''The Fundamental Theorem of Calculus, Part 1'''
 |-
-|Let <math style="vertical-align: -4px">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 style="vertical-align: -4px">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'''
+|&nbsp;&nbsp; '''The Fundamental Theorem of Calculus, Part 2'''
 |-
-|Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">f.</math>
+|&nbsp;&nbsp; Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">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>
 |-
-|'''(b)''' &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt\,=\,\sin(\cos(x))\cdot(-\sin(x))</math>
+|&nbsp;&nbsp; '''(b)''' &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt\,=\,\sin(\cos(x))\cdot(-\sin(x))</math>
 |-
-|'''(c)''' <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx\,=\,1</math>
+|&nbsp;&nbsp; '''(c)''' <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx\,=\,1</math>
 |}
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 11: / Line 11: @@
 !Foundations: &nbsp;
 |-
-|Review the Fundamental Theorem of Calculus.
+|'''1.''' What does Part 1 of the Fundamental Theorem of Calculus say about&nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt?</math>
 |}

@@ Line 33: / Line 33: @@
 |'''The Fundamental Theorem of Calculus, Part 1'''
 |-
-|Let <math style="vertical-align: -4px">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>.
 |}
@@ Line 43: / Line 45: @@
 |'''The Fundamental Theorem of Calculus, Part 2'''
 |-
-|Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">f</math>.
 |-
-|Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
 |}
 '''(b)'''
@@ Line 52: / Line 56: @@
 !Step 1: &nbsp;
 |-
-|Let <math style="vertical-align: -15px">F(x)=\int_0^{\cos (x)}\sin (t)~dt</math>. The problem is asking us to find <math style="vertical-align: -5px">F'(x)</math>.
+|Let <math style="vertical-align: -15px">F(x)=\int_0^{\cos (x)}\sin (t)~dt.</math> The problem is asking us to find <math style="vertical-align: -5px">F'(x).</math>
 |-
-|Let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -14px">G(x)=\int_0^x \sin(t)~dt</math>.
+|Let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -14px">G(x)=\int_0^x \sin(t)~dt.</math>
 |-
-|Then, <math style="vertical-align: -5px">F(x)=G(g(x))</math>.
+|Then, <math style="vertical-align: -5px">F(x)=G(g(x)).</math>
 |}
@@ Line 70: / Line 74: @@
 |Now, <math style="vertical-align: -5px">g'(x)=-\sin(x)</math> and <math style="vertical-align: -5px">G'(x)=\sin(x)</math> by the '''Fundamental Theorem of Calculus, Part 1'''.
 |-
-|Since <math style="vertical-align: -6px">G'(g(x))=\sin(g(x))=\sin(\cos(x))</math>, we have <math style="vertical-align: -5px">F'(x)=G'(g(x))\cdot g'(x)=\sin(\cos(x))\cdot(-\sin(x))</math>.
+|Since <math style="vertical-align: -6px">G'(g(x))=\sin(g(x))=\sin(\cos(x)),</math> we have
 |}
@@ Line 80: / Line 87: @@
 | Using the '''Fundamental Theorem of Calculus, Part 2''', we have
 |-
-| &nbsp;&nbsp; <math>\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan(x)\biggr|_{0}^{\pi/4}</math>
 |}
@@ Line 87: / Line 95: @@
 |-
 |So, we get
 |-
+|
 |}
@@ Line 100: / Line 107: @@
 |'''The Fundamental Theorem of Calculus, Part 1'''
 |-
-|Let <math style="vertical-align: -4px">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>.
+|Let <math style="vertical-align: -4px">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>.
+|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 style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">f</math>.
+|Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">f.</math>
 |-
-|Then, <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>
 |-
-|'''(b)''' &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt\,=\,\sin(\cos(x))\cdot(-\sin(x))</math>.
+|'''(b)''' &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt\,=\,\sin(\cos(x))\cdot(-\sin(x))</math>
 |-
-|'''(c)''' <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx\,=\,1</math>.
+|'''(c)''' <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx\,=\,1</math>
 |}
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]