009B Sample Midterm 3, Problem 3 - Revision history

Kayla Murray at 00:28, 24 November 2017

2017-11-24T00:28:48Z

Kayla Murray at 18:04, 27 March 2017

2017-03-27T18:04:12Z

Kayla Murray at 18:12, 18 March 2017

2017-03-18T18:12:29Z

Kayla Murray at 01:38, 27 February 2017

2017-02-27T01:38:11Z

Kayla Murray at 00:14, 19 February 2017

2017-02-19T00:14:37Z

Kayla Murray at 01:38, 8 February 2017

2017-02-08T01:38:04Z

Kayla Murray at 17:17, 7 February 2017

2017-02-07T17:17:00Z

Kayla Murray at 16:51, 6 February 2017

2017-02-06T16:51:26Z

Kayla Murray at 16:37, 6 February 2017

2017-02-06T16:37:12Z

Kayla Murray at 01:03, 19 April 2016

2016-04-19T01:03:01Z

@@ Line 1: / Line 1: @@
-<span class="exam"> Compute the following integrals:
+<span class="exam"> Find a curve &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; with the following properties:
-<span class="exam">(a) &nbsp; <math>\int x^2\sin (x^3) ~dx</math>
+<span class="exam">(i) &nbsp; <math style="vertical-align: -5px">f''(x)=6x</math>
-<span class="exam">(b) &nbsp; <math>\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx</math>
+<span class="exam">(ii) &nbsp; Its graph passes through the point &nbsp;<math style="vertical-align: -5px">(0,1)</math>&nbsp; and has a horizontal tangent there.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009B Sample Midterm 3, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 74: / Line 74: @@
 |Since this is a definite integral, we need to change the bounds of integration.
 |-
-|We have &nbsp;<math style="vertical-align: -15px">u_1=\cos\bigg(-\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math>&nbsp; and &nbsp;<math style="vertical-align: -15px">u_2=\cos\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}.</math>
+|We have
 |}

@@ Line 56: / Line 56: @@
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{\int x^2\sin (x^3) ~dx} & = & \displaystyle{\frac{-1}{3}\cos(u)+C}\\
+\displaystyle{\int x^2\sin (x^3) ~dx} & = & \displaystyle{-\frac{1}{3}\cos(u)+C}\\
 &&\\
-& = & \displaystyle{\frac{-1}{3}\cos(x^3)+C.}\\
+& = & \displaystyle{-\frac{1}{3}\cos(x^3)+C.}\\
 \end{array}</math>
 |}
@@ Line 96: / Line 96: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{-1}{3}\cos(x^3)+C</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>-\frac{1}{3}\cos(x^3)+C</math>
 |-
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>0</math>
 |}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 9: / Line 9: @@
 !Foundations: &nbsp;
 |-
-|How would you integrate <math style="vertical-align: -5px">2x(x^2+1)^3~dx?</math>
+|How would you integrate &nbsp;<math style="vertical-align: -5px">2x(x^2+1)^3~dx?</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.
+&nbsp; &nbsp; &nbsp; &nbsp; You could use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -3px">u=x^2+1.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -3px">u=x^2+1.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -1px">du=2x~dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=2x~dx.</math>
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; Thus,
@@ Line 37: / Line 37: @@
 !Step 1: &nbsp;
 |-
-|We proceed using <math style="vertical-align: 0px">u</math>-substitution.
+|We proceed using &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let <math style="vertical-align: -1px">u=x^3.</math>
+|Let &nbsp;<math style="vertical-align: -1px">u=x^3.</math>
 |-
-|Then, <math style="vertical-align: -1px">du=3x^2~dx</math> and <math style="vertical-align: -14px">\frac{du}{3}=x^2~dx.</math>
+|Then, &nbsp;<math style="vertical-align: -1px">du=3x^2~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{3}=x^2~dx.</math>
 |-
 |Therefore, we have
@@ Line 68: / Line 68: @@
 |We proceed using u substitution.
 |-
-|Let <math style="vertical-align: -5px">u=\cos(x).</math>
+|Let &nbsp;<math style="vertical-align: -5px">u=\cos(x).</math>
 |-
-|Then, <math style="vertical-align: -5px">du=-\sin(x)~dx.</math>
+|Then, &nbsp;<math style="vertical-align: -5px">du=-\sin(x)~dx.</math>
 |-
 |Since this is a definite integral, we need to change the bounds of integration.
 |-
-|We have <math style="vertical-align: -15px">u_1=\cos\bigg(-\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math> and <math style="vertical-align: -15px">u_2=\cos\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}.</math>
+|We have &nbsp;<math style="vertical-align: -15px">u_1=\cos\bigg(-\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math>&nbsp; and &nbsp;<math style="vertical-align: -15px">u_2=\cos\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}.</math>
 |}

@@ Line 1: / Line 1: @@
 <span class="exam"> Compute the following integrals:
-::<span class="exam">a) &nbsp; <math>\int x^2\sin (x^3) ~dx</math>
+<span class="exam">(a) &nbsp; <math>\int x^2\sin (x^3) ~dx</math>
-::<span class="exam">b) &nbsp; <math>\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx</math>

@@ Line 11: / Line 11: @@
 |-
+|
-&nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.
+&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.
 |-
-|&nbsp; &nbsp; Let <math style="vertical-align: -3px">u=x^2+1.</math> Then, <math style="vertical-align: -1px">du=2x~dx.</math> Thus,
+|&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -3px">u=x^2+1.</math>
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp;&nbsp;&nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int 2x(x^2+1)^3~dx} & = & \displaystyle{\int u^3~du}\\
 &&\\
@@ Line 34: / Line 38: @@
 |We proceed using <math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let <math style="vertical-align: -1px">u=x^3.</math> Then, <math style="vertical-align: -1px">du=3x^2~dx</math> and <math style="vertical-align: -14px">\frac{du}{3}=x^2~dx.</math>
+|Let <math style="vertical-align: -1px">u=x^3.</math>
 |-
 |Therefore, we have
 |-
+|
-&nbsp; &nbsp;<math>\int x^2\sin (x^3) ~dx=\int \frac{\sin(u)}{3}~du.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\int x^2\sin (x^3) ~dx=\int \frac{\sin(u)}{3}~du.</math>
 |}
@@ Line 48: / Line 54: @@
 |-
+|
-&nbsp;&nbsp;<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\int x^2\sin (x^3) ~dx} & = & \displaystyle{\frac{-1}{3}\cos(u)+C}\\
 &&\\
@@ Line 61: / Line 67: @@
 |We proceed using u substitution.
 |-
-|Let <math style="vertical-align: -5px">u=\cos(x).</math> Then, <math style="vertical-align: -5px">du=-\sin(x)~dx.</math>
+|Let <math style="vertical-align: -5px">u=\cos(x).</math>
 |-
 |Since this is a definite integral, we need to change the bounds of integration.
@@ Line 74: / Line 82: @@
 |-
+|
-&nbsp; &nbsp;<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx} & = & \displaystyle{\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} -u^2~du}\\
 &&\\
@@ Line 87: / Line 95: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp; <math>\frac{-1}{3}\cos(x^3)+C</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{-1}{3}\cos(x^3)+C</math>
 |-
-|'''(b)''' &nbsp; <math>0</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>0</math>
 |}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 79: / Line 79: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' <math>\frac{-1}{3}\cos(x^3)+C</math>
+|&nbsp;&nbsp; '''(a)''' <math>\frac{-1}{3}\cos(x^3)+C</math>
 |-
-|'''(b)''' <math>0</math>
+|&nbsp;&nbsp; '''(b)''' <math>0</math>
 |}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]