009B Sample Final 3, Problem 2 - Revision history

Kayla Murray at 00:41, 3 December 2017

2017-12-03T00:41:39Z

Kayla Murray at 20:00, 18 March 2017

2017-03-18T20:00:55Z

Kayla Murray at 21:42, 12 March 2017

2017-03-12T21:42:44Z

Kayla Murray at 23:41, 1 March 2017

2017-03-01T23:41:39Z

Kayla Murray at 23:34, 1 March 2017

2017-03-01T23:34:28Z

Kayla Murray at 23:33, 1 March 2017

2017-03-01T23:33:55Z

Kayla Murray at 23:26, 1 March 2017

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Kayla Murray at 23:22, 1 March 2017

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Kayla Murray at 23:18, 1 March 2017

2017-03-01T23:18:21Z

Kayla Murray at 23:07, 1 March 2017

2017-03-01T23:07:02Z

@@ Line 7: / Line 7: @@
 <span class="exam">(c) &nbsp;<math>\int_1^e \frac{\cos(\ln(x))}{x}~dx</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009B Sample Final 3, Problem 2 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009B Sample Final 3, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 10: / Line 10: @@
 !Foundations: &nbsp;
 |-
-|'''1.'''
+|'''1.''' Recall
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>\int \frac{1}{1+x^2}~dx=\arctan(x)+C</math>
@@ Line 92: / Line 92: @@
 !Step 1: &nbsp;
 |-
-|We use &nbsp;<math style="vertical-align: 0px">u</math>-substitution. Let &nbsp;<math style="vertical-align: -2px">u=1+x^3.</math>
+|We use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
 |Then, &nbsp;<math style="vertical-align: 0px">du=3x^2dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">\frac{du}{3}=x^2dx.</math>
@@ Line 134: / Line 136: @@
 |Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=\ln(x),</math>
 |-
-|we get &nbsp;<math style="vertical-align: -6px">u_1=\ln(1)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -6px">u_2=\ln(e)=1.</math>
+|we get
 |-
 |Therefore, the integral becomes

@@ Line 57: / Line 57: @@
 |Also, we need to change the bounds of integration.
 |-
-|Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=4x,</math>
+|Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=4x,</math>&nbsp; we get
 |-
-|we get &nbsp;<math style="vertical-align: -6px">u_1=4(0)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -16px">u_2=4\bigg(\frac{\sqrt{3}}{4}\bigg)=\sqrt{3}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -6px">u_1=4(0)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -16px">u_2=4\bigg(\frac{\sqrt{3}}{4}\bigg)=\sqrt{3}.</math>
 |-
 |Therefore, the integral becomes

@@ Line 51: / Line 51: @@
 |Now, we use &nbsp;<math>u</math>-substitution.
 |-
-|Let &nbsp;<math style="vertical-align: -5px">u=4x.</math>
+|Let &nbsp;<math style="vertical-align: -2px">u=4x.</math>
 |-
-|Then, &nbsp;<math style="vertical-align: -5px">du=4dx</math>&nbsp; and &nbsp;<math>\frac{du}{4}=dx.</math>
+|Then, &nbsp;<math style="vertical-align: -1px">du=4dx</math>&nbsp; and &nbsp;<math>\frac{du}{4}=dx.</math>
 |-
 |Also, we need to change the bounds of integration.
@@ Line 59: / Line 59: @@
 |Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=4x,</math>
 |-
-|we get &nbsp;<math style="vertical-align: -15px">u_1=4(0)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -16px">u_2=4\bigg(\frac{\sqrt{3}}{4}\bigg)=\sqrt{3}.</math>
+|we get &nbsp;<math style="vertical-align: -6px">u_1=4(0)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -16px">u_2=4\bigg(\frac{\sqrt{3}}{4}\bigg)=\sqrt{3}.</math>
 |-
 |Therefore, the integral becomes
@@ Line 128: / Line 128: @@
 |Let &nbsp;<math style="vertical-align: -5px">u=\ln(x).</math>
 |-
-|Then, &nbsp;<math style="vertical-align: -5px">du=\frac{1}{x}dx.</math>
+|Then, &nbsp;<math style="vertical-align: -15px">du=\frac{1}{x}dx.</math>
 |-
 |Also, we need to change the bounds of integration.
@@ Line 134: / Line 134: @@
 |Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=\ln(x),</math>
 |-
-|we get &nbsp;<math style="vertical-align: -15px">u_1=\ln(1)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -16px">u_2=\ln(e)=1.</math>
+|we get &nbsp;<math style="vertical-align: -6px">u_1=\ln(1)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -6px">u_2=\ln(e)=1.</math>
 |-
 |Therefore, the integral becomes

@@ Line 138: / Line 138: @@
 |Therefore, the integral becomes
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -19px">\int_0^2 \cos(u)~du.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -19px">\int_0^1 \cos(u)~du.</math>
 |-
+|
@@ Line 150: / Line 150: @@
+|
 &nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int_1^e \frac{\cos(\ln(x))}{x}~dx} & = & \displaystyle{\int_0^2 \cos(u)~du}\\
+\displaystyle{\int_1^e \frac{\cos(\ln(x))}{x}~dx} & = & \displaystyle{\int_0^1 \cos(u)~du}\\
 &&\\
 & = & \displaystyle{\sin(u)\bigg|_0^1}\\

@@ Line 124: / Line 124: @@
 !Step 1: &nbsp;
 |-
+|We use &nbsp;<math>u</math>-substitution.
 |-
+|
@@ Line 131: / Line 145: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |-
+|
@@ Line 145: / Line 170: @@
 |&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;<math>-\frac{1}{3(1+x^3)}+C</math>
 |-
-|&nbsp; &nbsp;'''(c)''' &nbsp; &nbsp;
+|&nbsp; &nbsp;'''(c)''' &nbsp; &nbsp;<math>\sin(1)</math>
 |}
 [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 98: / Line 98: @@
 |Therefore, the integral becomes
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\frac{1}{3}\int \sqrt{u}~du.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\frac{1}{3}\int \frac{1}{u^2}~du.</math>
 |-
+|
@@ Line 109: / Line 109: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int x^2\sqrt{1+x^3}~dx} & = & \displaystyle{\frac{1}{3}\int \sqrt{u}~du}\\
+\displaystyle{\int \frac{x^2}{(1+x^3)^2}~dx} & = & \displaystyle{\frac{1}{3}\int \frac{1}{u^2}~du}\\
 &&\\
-& = & \displaystyle{\frac{2}{9}u^{\frac{3}{2}}+C}\\
+& = & \displaystyle{-\frac{1}{3u}+C}\\
 &&\\
-& = & \displaystyle{\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C.}
+& = & \displaystyle{-\frac{1}{3(1+x^3)}+C.}
 \end{array}</math>
 |-
@@ Line 143: / Line 143: @@
 |&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math>\frac{\pi}{12}</math>
 |-
-|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;
+|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;<math>-\frac{1}{3(1+x^3)}+C</math>
 |-
 |&nbsp; &nbsp;'''(c)''' &nbsp; &nbsp;
 |}
 [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 3: / Line 3: @@
 <span class="exam">(a) &nbsp;<math>\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx</math>
-<span class="exam">(b) &nbsp;<math>\int \frac{x^2}{(1+x^3)^2}</math>
+<span class="exam">(b) &nbsp;<math>\int \frac{x^2}{(1+x^3)^2}~dx</math>
 <span class="exam">(c) &nbsp;<math>\int_1^e \frac{\cos(\ln(x))}{x}~dx</math>
@@ Line 92: / Line 92: @@
 !Step 1: &nbsp;
 |-
+|We use &nbsp;<math style="vertical-align: 0px">u</math>-substitution. Let &nbsp;<math style="vertical-align: -2px">u=1+x^3.</math>
 |-
+|
@@ Line 100: / Line 106: @@
 !Step 2: &nbsp;
 |-
+|We now have
 |-
+|

@@ Line 45: / Line 45: @@
 !Step 1: &nbsp;
 |-
+|First, we notice
 |-
+|Therefore, the integral becomes
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -19px">\frac{1}{4}\int_0^{\sqrt{3}} \frac{1}{1+u^2}~du.</math>
 |-
+|
@@ Line 57: / Line 71: @@
 !Step 2: &nbsp;
 |-
+|We now have
 |-
+|
 |}
@@ Line 106: / Line 127: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math>\frac{\pi}{12}</math>
 |-
-|'''(b)'''
+|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;
 |-
-|'''(c)'''
+|&nbsp; &nbsp;'''(c)''' &nbsp; &nbsp;
 |}
 [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]