009B Sample Midterm 3, Problem 2 - Revision history

Kayla Murray at 01:29, 13 November 2017

2017-11-13T01:29:18Z

Kayla Murray at 01:23, 5 November 2017

2017-11-05T01:23:46Z

Kayla Murray at 01:36, 27 February 2017

2017-02-27T01:36:31Z

Kayla Murray at 01:34, 8 February 2017

2017-02-08T01:34:02Z

Kayla Murray at 17:14, 7 February 2017

2017-02-07T17:14:39Z

Kayla Murray at 16:48, 6 February 2017

2017-02-06T16:48:53Z

Kayla Murray at 01:01, 19 April 2016

2016-04-19T01:01:27Z

Kayla Murray at 01:18, 30 March 2016

2016-03-30T01:18:41Z

Kayla Murray at 01:09, 30 March 2016

2016-03-30T01:09:15Z

Kayla Murray at 23:38, 29 March 2016

2016-03-29T23:38:10Z

@@ Line 4: / Line 4: @@
 <hr>
-(insert picture of handwritten solution)
+[[009B Sample Midterm 3, Problem 2 Solution|'''<u>Solution</u>''']]
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 3: / Line 3: @@
 ::<math>F(x)=\int_{\cos (x)}^5 \frac{1}{1+u^{10}}~du.</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009B Sample Midterm 3, Problem 2 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 9: / Line 9: @@
 |What does Part 1 of the Fundamental Theorem of Calculus
 |-
-|say is the derivative of <math style="vertical-align: -16px">G(x)=\int_x^5 \frac{1}{1+u^{10}}~du?</math>
+|say is the derivative of &nbsp;<math style="vertical-align: -16px">G(x)=\int_x^5 \frac{1}{1+u^{10}}~du?</math>
 |-
+|
@@ Line 15: / Line 15: @@
 |-
+|
-&nbsp;&nbsp;&nbsp;&nbsp; So, we have <math style="vertical-align: -16px">G(x)=-\int_5^x \frac{1}{1+u^{10}}~du.</math>
+&nbsp;&nbsp;&nbsp;&nbsp; So, we have &nbsp;<math style="vertical-align: -16px">G(x)=-\int_5^x \frac{1}{1+u^{10}}~du.</math>
 |-
+|
-&nbsp;&nbsp;&nbsp;&nbsp; By Part 1 of the Fundamental Theorem of Calculus, <math style="vertical-align: -16px">G'(x)=-\frac{1}{1+x^{10}}.</math>
+&nbsp;&nbsp;&nbsp;&nbsp; By Part 1 of the Fundamental Theorem of Calculus, &nbsp;<math style="vertical-align: -16px">G'(x)=-\frac{1}{1+x^{10}}.</math>
 |}
@@ Line 29: / Line 29: @@
 |-
+|
-Let <math style="vertical-align: -5px">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: -5px">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: -1px">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: -1px">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>
 |-
 |'''The Fundamental Theorem of Calculus, Part 2'''
 |-
+|
-Let <math style="vertical-align: -5px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -1px">F</math> be any antiderivative of <math style="vertical-align: -5px">f.</math> Then,
+Let &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; be continuous on &nbsp;<math style="vertical-align: -5px">[a,b]</math>&nbsp; and let &nbsp;<math style="vertical-align: -1px">F</math>&nbsp; be any antiderivative of &nbsp;<math style="vertical-align: -5px">f.</math> Then,
 |-
+|
@@ Line 50: / Line 50: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">F(x)=-\int_5^{\cos (x)} \frac{1}{1+u^{10}}~du.</math>
 |-
-|Now, let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -15px">G(x)=\int_5^x \frac{1}{1+u^{10}}~du.</math>
+|Now, let &nbsp;<math style="vertical-align: -5px">g(x)=\cos(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -15px">G(x)=\int_5^x \frac{1}{1+u^{10}}~du.</math>
 |-
 |Therefore,
@@ Line 90: / Line 90: @@
 !Final Answer: &nbsp;
 |-
-|'''The Fundamental Theorem of Calculus, Part 1'''
+|&nbsp; &nbsp; &nbsp; &nbsp; See Step 1 above
 |-
 |&nbsp;&nbsp; &nbsp; &nbsp; <math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
 |}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|What does Part 1 of the Fundamental Theorem of Calculus say is the derivative of <math style="vertical-align: -16px">G(x)=\int_x^5 \frac{1}{1+u^{10}}~du?</math>
+|What does Part 1 of the Fundamental Theorem of Calculus
 |-
+|
-&nbsp; First, we need to switch the bounds of integration.
+&nbsp;&nbsp;&nbsp;&nbsp; First, we need to switch the bounds of integration.
 |-
+|
-&nbsp; So, we have <math style="vertical-align: -16px">G(x)=-\int_5^x \frac{1}{1+u^{10}}~du.</math>
+&nbsp;&nbsp;&nbsp;&nbsp; So, we have <math style="vertical-align: -16px">G(x)=-\int_5^x \frac{1}{1+u^{10}}~du.</math>
 |-
+|
-&nbsp; By Part 1 of the Fundamental Theorem of Calculus, <math style="vertical-align: -16px">G'(x)=-\frac{1}{1+x^{10}}.</math>
+&nbsp;&nbsp;&nbsp;&nbsp; By Part 1 of the Fundamental Theorem of Calculus, <math style="vertical-align: -16px">G'(x)=-\frac{1}{1+x^{10}}.</math>
 |}
@@ Line 38: / Line 40: @@
 |-
+|
-&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 46: / Line 48: @@
 |First,
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -15px">F(x)=-\int_5^{\cos (x)} \frac{1}{1+u^{10}}~du.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">F(x)=-\int_5^{\cos (x)} \frac{1}{1+u^{10}}~du.</math>
 |-
 |Now, let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -15px">G(x)=\int_5^x \frac{1}{1+u^{10}}~du.</math>
@@ Line 53: / Line 55: @@
 |-
+|
-&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>
 |-
 |Hence,
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -5px">F'(x)=-G'(g(x))g'(x)</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">F'(x)=-G'(g(x))g'(x)</math>
 |-
 |by the Chain Rule.
@@ Line 65: / Line 67: @@
 !Step 3: &nbsp;
 |-
-|Now, <math style="vertical-align: -5px">g'(x)=-\sin(x).</math>
+|Now,
 |-
 |By the Fundamental Theorem of Calculus,
 |-
+|
-&nbsp; &nbsp; <math style="vertical-align: -15px">G'(x)=\frac{1}{1+x^{10}}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">G'(x)=\frac{1}{1+x^{10}}.</math>
 |-
 |Hence,
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{F'(x)} & = & \displaystyle{-\frac{1}{1+\cos^{10}x}(-\sin(x))}\\
 &&\\
@@ Line 100: / Line 104: @@
 |-
+|
-&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>
 |-
-|&nbsp;&nbsp; <math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
+|&nbsp;&nbsp; &nbsp; &nbsp; <math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
 |}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 10: / Line 10: @@
 |-
+|
-::First, we need to switch the bounds of integration.
+&nbsp; First, we need to switch the bounds of integration.
 |-
+|
-::So, we have <math style="vertical-align: -16px">G(x)=-\int_5^x \frac{1}{1+u^{10}}~du.</math>
+&nbsp; So, we have <math style="vertical-align: -16px">G(x)=-\int_5^x \frac{1}{1+u^{10}}~du.</math>
 |-
+|
-::By Part 1 of the Fundamental Theorem of Calculus, <math style="vertical-align: -16px">G'(x)=-\frac{1}{1+x^{10}}.</math>
+&nbsp; By Part 1 of the Fundamental Theorem of Calculus, <math style="vertical-align: -16px">G'(x)=-\frac{1}{1+x^{10}}.</math>
 |}
@@ Line 27: / Line 27: @@
 |-
+|
-:Let <math style="vertical-align: -5px">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: -5px">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: -1px">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: -1px">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: -5px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -1px">F</math> be any antiderivative of <math style="vertical-align: -5px">f.</math>
+Let <math style="vertical-align: -5px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -1px">F</math> be any antiderivative of <math style="vertical-align: -5px">f.</math> Then,
 |-
+|
-:Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
+&nbsp; &nbsp; <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 |}
@@ Line 44: / Line 44: @@
 !Step 2: &nbsp;
 |-
-|First, we have
+|First,
 |-
+|&nbsp; &nbsp; <math style="vertical-align: -15px">F(x)=-\int_5^{\cos (x)} \frac{1}{1+u^{10}}~du.</math>
 |-
 |Now, let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -15px">G(x)=\int_5^x \frac{1}{1+u^{10}}~du.</math>
 |-
-|So,
+|Therefore,
 |-
+|
-::<math style="vertical-align: -5px">F(x)=-G(g(x)).</math>
+&nbsp; &nbsp; <math style="vertical-align: -5px">F(x)=-G(g(x)).</math>
 |-
-|Hence, <math style="vertical-align: -5px">F'(x)=-G'(g(x))g'(x)</math> by the Chain Rule.
+|by the Chain Rule.
 |}
@@ Line 67: / Line 70: @@
 |-
+|
-::<math style="vertical-align: -15px">G'(x)=\frac{1}{1+x^{10}}.</math>
+&nbsp; &nbsp; <math style="vertical-align: -15px">G'(x)=\frac{1}{1+x^{10}}.</math>
 |-
 |Hence,
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{F'(x)} & = & \displaystyle{-\frac{1}{1+\cos^{10}x}(-\sin(x))}\\
 &&\\
@@ Line 94: / Line 97: @@
 |-
+|
-Let <math style="vertical-align: -5px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -1px">F</math> be any antiderivative of <math style="vertical-align: -5px">f.</math>
+Let <math style="vertical-align: -5px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -1px">F</math> be any antiderivative of <math style="vertical-align: -5px">f.</math> Then,
 |-
+|
-Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
+&nbsp; &nbsp; <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 |-
 |&nbsp;&nbsp; <math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
 |}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 25: / Line 25: @@
 |'''The Fundamental Theorem of Calculus, Part 1'''
 |-
 |Let <math style="vertical-align: -5px">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: -1px">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: -5px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -1px">F</math> be any antiderivative of <math style="vertical-align: -5px">f.</math>
 |-
 |Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 |}
@@ Line 39: / Line 43: @@
 !Step 2: &nbsp;
 |-
-|First, we have <math style="vertical-align: -15px">F(x)=-\int_5^{\cos (x)} \frac{1}{1+u^{10}}~du.</math>
+|First, we have
 |-
 |Now, let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -15px">G(x)=\int_5^x \frac{1}{1+u^{10}}~du.</math>
 |-
-|So, <math style="vertical-align: -5px">F(x)=-G(g(x)).</math>
+|So,
 |-
 |Hence, <math style="vertical-align: -5px">F'(x)=-G'(g(x))g'(x)</math> by the Chain Rule.
@@ Line 53: / Line 63: @@
 |Now, <math style="vertical-align: -5px">g'(x)=-\sin(x).</math>
 |-
-| By the Fundamental Theorem of Calculus, <math style="vertical-align: -15px">G'(x)=\frac{1}{1+x^{10}}.</math>
+|By the Fundamental Theorem of Calculus,
 |-
-|Hence, <math style="vertical-align: -15px">F'(x)=-\frac{1}{1+\cos^{10}x}(-\sin(x))=\frac{\sin(x)}{1+\cos^{10}x}.</math>
 |-
+|
 |}
@@ Line 63: / Line 81: @@
 !Final Answer: &nbsp;
 |-
-|'''The Fundamental Theorem of Calculus, Part 1'''
+|&nbsp;&nbsp; '''The Fundamental Theorem of Calculus, Part 1'''
 |-
 |Let <math style="vertical-align: -5px">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: -1px">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: -5px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -1px">F</math> be any antiderivative of <math style="vertical-align: -5px">f.</math>
 |-
 |Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 |-
-|<math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
+|&nbsp;&nbsp; <math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
 |}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 27: / Line 27: @@
 |'''The Fundamental Theorem of Calculus, Part 1'''
 |-
-|Let <math>f</math> be continuous on <math>[a,b]</math> and let <math>F(x)=\int_a^x f(t)~dt</math>.
+|Let <math>f</math> be continuous on <math>[a,b]</math> and let <math>F(x)=\int_a^x f(t)~dt.</math>
 |-
-|Then, <math>F</math> is a differentiable function on <math>(a,b)</math> and <math>F'(x)=f(x)</math>.
+|Then, <math>F</math> is a differentiable function on <math>(a,b)</math> and <math>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>F</math> be any antiderivative of <math>f</math>.
+|Let <math>f</math> be continuous on <math>[a,b]</math> and let <math>F</math> be any antiderivative of <math>f.</math>
 |-
-|Then, <math>\int_a^b f(x)~dx=F(b)-F(a)</math>
+|Then, <math>\int_a^b f(x)~dx=F(b)-F(a).</math>
 |}
@@ Line 41: / Line 41: @@
 !Step 2: &nbsp;
 |-
-|First, we have <math>F(x)=-\int_5^{\cos (x)} \frac{1}{1+u^{10}}~du</math>.
+|First, we have <math>F(x)=-\int_5^{\cos (x)} \frac{1}{1+u^{10}}~du.</math>
 |-
-|Now, let <math>g(x)=\cos(x)</math> and <math>G(x)=\int_5^x \frac{1}{1+u^{10}}~du</math>
+|Now, let <math>g(x)=\cos(x)</math> and <math>G(x)=\int_5^x \frac{1}{1+u^{10}}~du.</math>
 |-
-|So, <math>F(x)=-G(g(x))</math>.
+|So, <math>F(x)=-G(g(x)).</math>
 |-
 |Hence, <math>F'(x)=-G'(g(x))g'(x)</math> by the Chain Rule.
@@ Line 53: / Line 53: @@
 !Step 3: &nbsp;
 |-
-|Now, <math>g'(x)=-\sin(x)</math>.
+|Now, <math>g'(x)=-\sin(x).</math>
 |-
-| By the Fundamental Theorem of Calculus, <math>G'(x)=\frac{1}{1+x^{10}}</math>.
+| By the Fundamental Theorem of Calculus, <math>G'(x)=\frac{1}{1+x^{10}}.</math>
 |-
-|Hence, <math>F'(x)=-\frac{1}{1+\cos^{10}x}(-\sin(x))=\frac{\sin(x)}{1+\cos^{10}x}</math>
+|Hence, <math>F'(x)=-\frac{1}{1+\cos^{10}x}(-\sin(x))=\frac{\sin(x)}{1+\cos^{10}x}.</math>
 |-
+|
@@ Line 67: / Line 67: @@
 |'''The Fundamental Theorem of Calculus, Part 1'''
 |-
-|Let <math>f</math> be continuous on <math>[a,b]</math> and let <math>F(x)=\int_a^x f(t)~dt</math>.
+|Let <math>f</math> be continuous on <math>[a,b]</math> and let <math>F(x)=\int_a^x f(t)~dt.</math>
 |-
-|Then, <math>F</math> is a differentiable function on <math>(a,b)</math> and <math>F'(x)=f(x)</math>.
+|Then, <math>F</math> is a differentiable function on <math>(a,b)</math> and <math>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>F</math> be any antiderivative of <math>f</math>.
+|Let <math>f</math> be continuous on <math>[a,b]</math> and let <math>F</math> be any antiderivative of <math>f.</math>
 |-
-|Then, <math>\int_a^b f(x)~dx=F(b)-F(a)</math>
+|Then, <math>\int_a^b f(x)~dx=F(b)-F(a).</math>
 |-
-| <math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
+|<math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
 |}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]