009B Sample Midterm 2, Problem 1 - Revision history

Kayla Murray at 00:58, 13 November 2017

2017-11-13T00:58:58Z

Kayla Murray at 01:29, 5 November 2017

2017-11-05T01:29:37Z

Kayla Murray at 18:04, 18 March 2017

2017-03-18T18:04:10Z

Kayla Murray at 01:06, 27 February 2017

2017-02-27T01:06:31Z

Kayla Murray at 00:11, 19 February 2017

2017-02-19T00:11:34Z

Kayla Murray at 02:03, 8 February 2017

2017-02-08T02:03:39Z

Kayla Murray at 02:03, 8 February 2017

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Kayla Murray at 02:02, 8 February 2017

2017-02-08T02:02:22Z

Kayla Murray at 16:34, 7 February 2017

2017-02-07T16:34:23Z

Kayla Murray at 16:33, 7 February 2017

2017-02-07T16:33:42Z

@@ Line 9: / Line 9: @@
 <hr>
-(insert picture of handwritten solution)
+[[009B Sample Midterm 2, Problem 1 Solution|'''<u>Solution</u>''']]
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam"> This problem has three parts:
-<span class="exam">(a) State the Fundamental Theorem of Calculus.
+<span class="exam">(a) State the both parts of the fundamental theorem of calculus.
-<span class="exam">(b) Compute &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt.</math>
+<span class="exam">(b) Compute &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt</math>.
-<span class="exam">(c) Evaluate &nbsp;<math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx.</math>
+<span class="exam">(c) Evaluate &nbsp;<math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx</math>.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+(insert picture of handwritten solution)
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 121: / Line 121: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' See solution above.
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; See solution above.
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &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; &nbsp; <math style="vertical-align: -5px">\sin(\cos(x))\cdot(-\sin(x))</math>
 |-
-|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx\,=\,1.</math>
+|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math style="vertical-align: -3px">1</math>
 |}
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 3: / Line 3: @@
 <span class="exam">(a) State the Fundamental Theorem of Calculus.
-<span class="exam">(b) Compute &thinsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt.</math>
+<span class="exam">(b) Compute &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt.</math>
-<span class="exam">(c) Evaluate <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx.</math>
+<span class="exam">(c) Evaluate &nbsp;<math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx.</math>
@@ Line 11: / Line 11: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' What does Part 1 of the Fundamental Theorem of Calculus say about <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt?</math>
+|'''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 18: / Line 18: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt=\sin(x).</math>
 |-
-|'''2.''' What does Part 2 of the Fundamental Theorem of Calculus say about <math style="vertical-align: -15px">\int_a^b\sec^2x~dx</math> where <math style="vertical-align: -5px">a,b</math> are constants?
+|'''2.''' What does Part 2 of the Fundamental Theorem of Calculus say about &nbsp;<math style="vertical-align: -15px">\int_a^b\sec^2x~dx</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">a,b</math>&nbsp; are constants?
 |-
+|
 &nbsp; &nbsp; &nbsp; &nbsp; Part 2 of the Fundamental Theorem of Calculus says that
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\int_a^b\sec^2x~dx=F(b)-F(a)</math> where <math style="vertical-align: 0px">F</math> is any antiderivative of <math style="vertical-align: 0px">\sec^2x.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\int_a^b\sec^2x~dx=F(b)-F(a)</math>&nbsp; where &nbsp;<math style="vertical-align: 0px">F</math>&nbsp; is any antiderivative of &nbsp;<math style="vertical-align: 0px">\sec^2x.</math>
 |}
@@ Line 38: / Line 38: @@
 |'''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 &nbsp;<math style="vertical-align: -4px">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>
 |-
-|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, &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 48: / Line 48: @@
 |'''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 &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; be continuous on &nbsp;<math style="vertical-align: -5px">[a,b]</math>&nbsp; and let &nbsp;<math style="vertical-align: 0px">F</math>&nbsp; be any antiderivative of &nbsp;<math style="vertical-align: -4px">f.</math>
 |-
 |Then,
@@ Line 60: / Line 60: @@
 !Step 1: &nbsp;
 |-
-|Let <math style="vertical-align: -15px">F(x)=\int_0^{\cos (x)}\sin (t)~dt.</math>
+|Let &nbsp;<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>
+|The problem is asking us to find &nbsp;<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 &nbsp;<math style="vertical-align: -5px">g(x)=\cos(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">G(x)=\int_0^x \sin(t)~dt.</math>
 |-
 |Then,
@@ Line 86: / Line 86: @@
 !Step 3: &nbsp;
 |-
-|Now, <math style="vertical-align: -5px">g'(x)=-\sin(x)</math> and <math style="vertical-align: -5px">G'(x)=\sin(x)</math>
+|Now, &nbsp;<math style="vertical-align: -5px">g'(x)=-\sin(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">G'(x)=\sin(x)</math>
 |-
 |by the '''Fundamental Theorem of Calculus, Part 1'''.
@@ Line 121: / Line 121: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;
+|&nbsp; &nbsp; '''(a)''' See solution above.
 |-
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt\,=\,\sin(\cos(x))\cdot(-\sin(x)).</math>

@@ Line 1: / Line 1: @@
 <span class="exam"> This problem has three parts:
-::<span class="exam">a) State the Fundamental Theorem of Calculus.
+<span class="exam">(a) State the Fundamental Theorem of Calculus.
-::<span class="exam">b) Compute &thinsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt.</math>
+<span class="exam">(b) Compute &thinsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt.</math>
-::<span class="exam">c) Evaluate <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx.</math>
+<span class="exam">(c) Evaluate <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx.</math>

@@ Line 78: / Line 78: @@
 |we get
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">F'(x)=G'(g(x))g'(x)</math> by the Chain Rule.
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">F'(x)=G'(g(x))g'(x)</math>
 |}

@@ Line 76: / Line 76: @@
 |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.
+|we get
 |}

@@ Line 14: / Line 14: @@
 |-
+|
-&nbsp; &nbsp; Part 1 of the Fundamental Theorem of Calculus says that
+&nbsp; &nbsp; &nbsp; &nbsp; Part 1 of the Fundamental Theorem of Calculus says that
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt=\sin(x).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt=\sin(x).</math>
 |-
 |'''2.''' What does Part 2 of the Fundamental Theorem of Calculus say about <math style="vertical-align: -15px">\int_a^b\sec^2x~dx</math> where <math style="vertical-align: -5px">a,b</math> are constants?
 |-
+|
-&nbsp; &nbsp; Part 2 of the Fundamental Theorem of Calculus says that
+&nbsp; &nbsp; &nbsp; &nbsp; Part 2 of the Fundamental Theorem of Calculus says that
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -15px">\int_a^b\sec^2x~dx=F(b)-F(a)</math> where <math style="vertical-align: 0px">F</math> is any antiderivative of <math style="vertical-align: 0px">\sec^2x.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\int_a^b\sec^2x~dx=F(b)-F(a)</math> where <math style="vertical-align: 0px">F</math> is any antiderivative of <math style="vertical-align: 0px">\sec^2x.</math>
 |}
@@ Line 52: / Line 52: @@
 |Then,
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 |}
@@ Line 68: / Line 68: @@
 |Then,
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -5px">F(x)=G(g(x)).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">F(x)=G(g(x)).</math>
 |}
@@ Line 88: / Line 88: @@
 |Since
 |-
-|&nbsp; &nbsp;<math style="vertical-align: -4px">G'(g(x))=\sin(g(x))=\sin(\cos(x)),</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -4px">G'(g(x))=\sin(g(x))=\sin(\cos(x)),</math>
 |-
 |we have
 |-
-|&nbsp; &nbsp;<math style="vertical-align: -5px">F'(x)=G'(g(x))\cdot g'(x)=\sin(\cos(x))\cdot(-\sin(x)).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">F'(x)=G'(g(x))\cdot g'(x)=\sin(\cos(x))\cdot(-\sin(x)).</math>
 |}
@@ Line 102: / Line 102: @@
 | 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>
+| &nbsp; &nbsp; &nbsp; &nbsp; <math>\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan(x)\biggr|_{0}^{\pi/4}.</math>
 |}
@@ Line 110: / Line 110: @@
 |So, we get
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan \bigg(\frac{\pi}{4}\bigg)-\tan (0)=1.</math>
+| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -16px">\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan \bigg(\frac{\pi}{4}\bigg)-\tan (0)=1.</math>
 |}
@@ Line 117: / Line 117: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;
 |-
-|'''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; 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; &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt\,=\,\sin(\cos(x))\cdot(-\sin(x)).</math>
 |-
-|'''(c)''' &nbsp; <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx\,=\,1.</math>
+|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <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>''']]