009B Sample Midterm 1, Problem 4 - Revision history

Kayla Murray at 23:03, 12 November 2017

2017-11-12T23:03:48Z

Kayla Murray at 01:53, 5 November 2017

2017-11-05T01:53:03Z

Kayla Murray at 20:33, 14 March 2017

2017-03-14T20:33:21Z

Kayla Murray at 00:59, 27 February 2017

2017-02-27T00:59:06Z

Kayla Murray at 23:58, 7 February 2017

2017-02-07T23:58:37Z

Kayla Murray at 23:57, 7 February 2017

2017-02-07T23:57:28Z

Kayla Murray at 23:56, 7 February 2017

2017-02-07T23:56:39Z

Kayla Murray at 23:54, 7 February 2017

2017-02-07T23:54:20Z

Kayla Murray at 16:20, 7 February 2017

2017-02-07T16:20:18Z

Kayla Murray at 03:54, 7 February 2017

2017-02-07T03:54:53Z

@@ Line 7: / Line 7: @@
 <hr>
-(insert picture of handwritten solution)
+[[009B Sample Midterm 1, Problem 4 Solution|'''<u>Solution</u>''']]
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Evaluate the integral:
+<span class="exam"> Evaluate the indefinite and definite integrals.
-::<math>\int \sin^3x \cos^2x~dx</math>
+<span class="exam">(a) &nbsp; <math>\int x^2 e^x~dx</math>
-'''Solution:'''
+[[009B Sample Midterm 1, Problem 4 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 14: / Line 14: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; You could use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
+&nbsp; &nbsp; &nbsp; &nbsp; You can use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -2px">u=\sin x.</math>
@@ Line 43: / Line 43: @@
 |Using the identity &nbsp;<math style="vertical-align: -4px">\sin^2x+\cos^2x=1,</math>
 |-
-|we get &nbsp;<math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>
+|we get
 |-
 |If we use this identity, we have

@@ Line 11: / Line 11: @@
 | &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>
 |-
-|'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sin^2x\cos x~dx?</math>
+|'''2.''' How would you integrate &nbsp;<math style="vertical-align: -14px">\int \sin^2x\cos x~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: -2px">u=\sin x.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -2px">u=\sin x.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -1px">du=\cos x~dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=\cos x~dx.</math>
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; Thus,
@@ Line 41: / Line 41: @@
 | &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx.</math>
 |-
-|Using the identity <math style="vertical-align: -4px">\sin^2x+\cos^2x=1,</math>
+|Using the identity &nbsp;<math style="vertical-align: -4px">\sin^2x+\cos^2x=1,</math>
 |-
-|we get <math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>
+|we get &nbsp;<math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>
 |-
 |If we use this identity, we have
@@ Line 58: / Line 58: @@
 !Step 2: &nbsp;
 |-
-|Now, we use <math style="vertical-align: 0px">u</math>-substitution.
+|Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-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>
 |-
 |Therefore,

@@ Line 18: / Line 18: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -2px">u=\sin x.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -1px">du=\cos x~dx.</math> Thus,
+|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -1px">du=\cos x~dx.</math>
 |-
+|

@@ Line 14: / Line 14: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=\sin x.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -1px">du=\cos x~dx.</math> Thus,
@@ Line 76: / Line 78: @@
 !Final Answer: &nbsp;
 |-
-| &nbsp;&nbsp; &nbsp; &nbsp;<math>\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>
+| &nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>
 |}
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' Recall the trig identity:
+|'''1.''' Recall the trig identity
 |-
-| &nbsp; &nbsp; <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>
+| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>
 |-
 |'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sin^2x\cos x~dx?</math>
 |-
+|
-&nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=\sin x.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=\sin x.</math>
 |-
-|&nbsp; &nbsp; Then, <math style="vertical-align: -1px">du=\cos x~dx.</math> Thus,
+|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -1px">du=\cos x~dx.</math> Thus,
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int \sin^2x\cos x~dx} & = & \displaystyle{\int u^2~du}\\
 &&\\
@@ Line 35: / Line 35: @@
 |First, we write
 |-
-| &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx.</math>
+| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx.</math>
 |-
 |Using the identity <math style="vertical-align: -4px">\sin^2x+\cos^2x=1,</math>
@@ Line 44: / Line 44: @@
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int (\sin x) (1-\cos^2x)\cos^2x~dx}\\
 &&\\
@@ Line 56: / Line 56: @@
 |Now, we use <math style="vertical-align: 0px">u</math>-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>
 |-
 |Therefore,
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int -(u^2-u^4)~du}\\
 &&\\
@@ Line 74: / Line 76: @@
 !Final Answer: &nbsp;
 |-
-| &nbsp;&nbsp; <math>\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>
+| &nbsp;&nbsp; &nbsp; &nbsp;<math>\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</math>
 |}
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]