009B Sample Midterm 2, Problem 4 - Revision history

Kayla Murray at 01:02, 13 November 2017

2017-11-13T01:02:32Z

Kayla Murray: Replaced content with " Evaluate the integral: :: $\int e^{-2x}\sin (2x)~dx$
(insert picture of handwritten solution) 009B Sample Midterm 2, Problem 4 Deta..."

2017-11-05T01:43:21Z

Replaced content with "<span class="exam"> Evaluate the integral: ::<math>\int e^{-2x}\sin (2x)~dx</math> <hr> (insert picture of handwritten solution) 009B Sample Midterm 2, Problem 4 Deta..."

Kayla Murray at 20:41, 14 March 2017

2017-03-14T20:41:26Z

Kayla Murray at 01:28, 27 February 2017

2017-02-27T01:28:49Z

Kayla Murray at 01:54, 8 February 2017

2017-02-08T01:54:06Z

Kayla Murray at 22:48, 7 February 2017

2017-02-07T22:48:23Z

Kayla Murray at 22:45, 7 February 2017

2017-02-07T22:45:39Z

Kayla Murray at 16:52, 7 February 2017

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Kayla Murray at 03:56, 7 February 2017

2017-02-07T03:56:33Z

Kayla Murray at 16:32, 6 February 2017

2017-02-06T16:32:33Z

@@ Line 4: / Line 4: @@
 <hr>
-(insert picture of handwritten solution)
+[[009B Sample Midterm 2, Problem 4 Solution|'''<u>Solution</u>''']]
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 3: / Line 3: @@
 ::<math>\int e^{-2x}\sin (2x)~dx</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009B Sample Midterm 2, Problem 4 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 14: / Line 14: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; You could use integration by parts.
+&nbsp; &nbsp; &nbsp; &nbsp; You can use integration by parts.
 |-
+|

@@ Line 11: / Line 11: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\int u~dv=uv-\int v~du.</math>
 |-
-|'''2.''' How would you integrate <math style="vertical-align: -15px">\int e^x\sin x~dx?</math>
+|'''2.''' How would you integrate &nbsp;<math style="vertical-align: -15px">\int e^x\sin x~dx?</math>
 |-
+|
@@ Line 17: / Line 17: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -5px">u=\sin(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -5px">u=\sin(x)</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^x~dx.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -5px">du=\cos(x)~dx</math> and <math style="vertical-align: 0px">v=e^x.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -5px">du=\cos(x)~dx</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=e^x.</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; Thus, <math style="vertical-align: -15px">\int e^x\sin x~dx=e^x\sin(x)-\int e^x\cos(x)~dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Thus, &nbsp;<math style="vertical-align: -15px">\int e^x\sin x~dx=e^x\sin(x)-\int e^x\cos(x)~dx.</math>
 |-
+|
@@ Line 28: / Line 28: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -5px">u=\cos(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -5px">u=\cos(x)</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^x~dx.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -5px">du=-\sin(x)~dx</math> and <math style="vertical-align: 0px">v=e^x.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -5px">du=-\sin(x)~dx</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=e^x.</math>
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; Therefore,
@@ Line 50: / Line 50: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; Hence, <math style="vertical-align: -15px">\int e^x\sin (x)~dx=\frac{e^x}{2}(\sin(x)-\cos(x))+C.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Hence, &nbsp;<math style="vertical-align: -15px">\int e^x\sin (x)~dx=\frac{e^x}{2}(\sin(x)-\cos(x))+C.</math>
 |}
@@ Line 60: / Line 60: @@
 |We proceed using integration by parts.
 |-
-|Let <math style="vertical-align: -5px">u=\sin(2x)</math> and <math style="vertical-align: 0px">dv=e^{-2x}dx.</math>
+|Let &nbsp;<math style="vertical-align: -5px">u=\sin(2x)</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-2x}dx.</math>
 |-
-|Then, <math style="vertical-align: -5px">du=2\cos(2x)dx</math> and <math style="vertical-align: -13px">v=\frac{e^{-2x}}{-2}.</math>
+|Then, &nbsp;<math style="vertical-align: -5px">du=2\cos(2x)dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">v=\frac{e^{-2x}}{-2}.</math>
 |-
 |Thus, we get
@@ Line 79: / Line 79: @@
 |Now, we need to use integration by parts again.
 |-
-|Let <math style="vertical-align: -5px">u=\cos(2x)</math> and <math style="vertical-align: 0px">dv=e^{-2x}dx.</math>
+|Let &nbsp;<math style="vertical-align: -5px">u=\cos(2x)</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-2x}dx.</math>
 |-
-|Then, <math style="vertical-align: -5px">du=-2\sin(2x)dx</math> and <math style="vertical-align: -13px">v=\frac{e^{-2x}}{-2}.</math>
+|Then, &nbsp;<math style="vertical-align: -5px">du=-2\sin(2x)dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">v=\frac{e^{-2x}}{-2}.</math>
 |-
 |Therefore, we get

@@ Line 9: / Line 9: @@
 |'''1.''' Integration by parts tells us
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -15px">\int u~dv=uv-\int v~du.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\int u~dv=uv-\int v~du.</math>
 |-
 |'''2.''' How would you integrate <math style="vertical-align: -15px">\int e^x\sin x~dx?</math>
 |-
+|
-&nbsp; &nbsp; You could use integration by parts.
+&nbsp; &nbsp; &nbsp; &nbsp; You could use integration by parts.
 |-
+|
-&nbsp; &nbsp; Let <math style="vertical-align: -5px">u=\sin(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -5px">u=\sin(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math>
 |-
-|&nbsp; &nbsp; Then, <math style="vertical-align: -5px">du=\cos(x)~dx</math> and <math style="vertical-align: 0px">v=e^x.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -5px">du=\cos(x)~dx</math> and <math style="vertical-align: 0px">v=e^x.</math>
 |-
+|
-&nbsp; &nbsp; Thus, <math style="vertical-align: -15px">\int e^x\sin x~dx=e^x\sin(x)-\int e^x\cos(x)~dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Thus, <math style="vertical-align: -15px">\int e^x\sin x~dx=e^x\sin(x)-\int e^x\cos(x)~dx.</math>
 |-
+|
-&nbsp; &nbsp; Now, we need to use integration by parts a second time.
+&nbsp; &nbsp; &nbsp; &nbsp; Now, we need to use integration by parts a second time.
 |-
+|
-&nbsp; &nbsp; Let <math style="vertical-align: -5px">u=\cos(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math> Then, <math style="vertical-align: -5px">du=-\sin(x)~dx</math> and <math style="vertical-align: 0px">v=e^x.</math> So,
+&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -5px">u=\cos(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math>
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int e^x\sin x~dx} & = & \displaystyle{e^x\sin(x)-(e^x\cos(x)-\int -e^x\sin(x)~dx}\\
 &&\\
@@ Line 38: / Line 42: @@
 |-
+|
-&nbsp; &nbsp; Notice, we are back where we started. So, adding the last term on the right hand side to the opposite side,
+&nbsp; &nbsp; &nbsp; &nbsp; Notice, we are back where we started.
 |-
+|
-&nbsp; &nbsp; we get <math style="vertical-align: -13px">2\int e^x\sin (x)~dx=e^x(\sin(x)-\cos(x)).</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">2\int e^x\sin (x)~dx=e^x(\sin(x)-\cos(x)).</math>
 |-
+|
-&nbsp; &nbsp; Hence, <math style="vertical-align: -15px">\int e^x\sin (x)~dx=\frac{e^x}{2}(\sin(x)-\cos(x))+C.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Hence, <math style="vertical-align: -15px">\int e^x\sin (x)~dx=\frac{e^x}{2}(\sin(x)-\cos(x))+C.</math>
 |}
@@ Line 58: / Line 64: @@
 |Then, <math style="vertical-align: -5px">du=2\cos(2x)dx</math> and <math style="vertical-align: -13px">v=\frac{e^{-2x}}{-2}.</math>
 |-
-|So, we get
+|Thus, we get
 |-
+|
@@ Line 80: / Line 86: @@
 |-
+|
-&nbsp;&nbsp; <math style="vertical-align: -13px">\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}+\frac{\cos(2x)e^{-2x}}{-2}-\int e^{-2x}\sin(2x)~dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}+\frac{\cos(2x)e^{-2x}}{-2}-\int e^{-2x}\sin(2x)~dx.</math>
 |}
@@ Line 94: / Line 98: @@
 |Thus, if we add the integral on the right to the other side of the equation, we get
 |-
-| &nbsp;&nbsp; <math>2\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}+\frac{\cos(2x)e^{-2x}}{-2}.</math>
+| &nbsp; &nbsp; &nbsp; &nbsp; <math>2\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}+\frac{\cos(2x)e^{-2x}}{-2}.</math>
 |-
 |Now, we divide both sides by 2 to get
 |-
-| &nbsp;&nbsp; <math>\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-4}+\frac{\cos(2x)e^{-2x}}{-4}.</math>
+| &nbsp; &nbsp; &nbsp; &nbsp; <math>\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-4}+\frac{\cos(2x)e^{-2x}}{-4}.</math>
 |-
 |Thus, the final answer is
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -13px">\int e^{-2x}\sin (2x)~dx=\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\int e^{-2x}\sin (2x)~dx=\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C.</math>
 |}

@@ Line 109: / Line 109: @@
 !Final Answer: &nbsp;
 |-
-| &nbsp;&nbsp; <math>\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C</math>
+| &nbsp;&nbsp; &nbsp; &nbsp; <math>\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C</math>
 |}
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' Integration by parts tells us <math style="vertical-align: -15px">\int u~dv=uv-\int v~du.</math>
+|'''1.''' Integration by parts tells us
 |-
 |'''2.''' How would you integrate <math style="vertical-align: -15px">\int e^x\sin x~dx?</math>
 |-
+|
-::You could use integration by parts.
+&nbsp; &nbsp; You could use integration by parts.
 |-
+|
-::Let <math style="vertical-align: -5px">u=\sin(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math> Then, <math style="vertical-align: -5px">du=\cos(x)~dx</math> and <math style="vertical-align: 0px">v=e^x.</math>
+&nbsp; &nbsp; Let <math style="vertical-align: -5px">u=\sin(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math>
 |-
+|
-::Thus, <math style="vertical-align: -15px">\int e^x\sin x~dx=e^x\sin(x)-\int e^x\cos(x)~dx.</math>
+&nbsp; &nbsp; Thus, <math style="vertical-align: -15px">\int e^x\sin x~dx=e^x\sin(x)-\int e^x\cos(x)~dx.</math>
 |-
+|
-::Now, we need to use integration by parts a second time.
+&nbsp; &nbsp; Now, we need to use integration by parts a second time.
 |-
+|
-::Let <math style="vertical-align: -5px">u=\cos(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math> Then, <math style="vertical-align: -5px">du=-\sin(x)~dx</math> and <math style="vertical-align: 0px">v=e^x.</math> So,
+&nbsp; &nbsp; Let <math style="vertical-align: -5px">u=\cos(x)</math> and <math style="vertical-align: 0px">dv=e^x~dx.</math> Then, <math style="vertical-align: -5px">du=-\sin(x)~dx</math> and <math style="vertical-align: 0px">v=e^x.</math> So,
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int e^x\sin x~dx} & = & \displaystyle{e^x\sin(x)-(e^x\cos(x)-\int -e^x\sin(x)~dx}\\
 &&\\
@@ Line 34: / Line 38: @@
 |-
+|
-::Notice, we are back where we started. So, adding the last term on the right hand side to the opposite side,
+&nbsp; &nbsp; Notice, we are back where we started. So, adding the last term on the right hand side to the opposite side,
 |-
+|
-::we get <math style="vertical-align: -13px">2\int e^x\sin (x)~dx=e^x(\sin(x)-\cos(x)).</math>
+&nbsp; &nbsp; we get <math style="vertical-align: -13px">2\int e^x\sin (x)~dx=e^x(\sin(x)-\cos(x)).</math>
 |-
+|
-::Hence, <math style="vertical-align: -15px">\int e^x\sin (x)~dx=\frac{e^x}{2}(\sin(x)-\cos(x))+C.</math>
+&nbsp; &nbsp; Hence, <math style="vertical-align: -15px">\int e^x\sin (x)~dx=\frac{e^x}{2}(\sin(x)-\cos(x))+C.</math>
 |}
@@ Line 48: / Line 52: @@
 !Step 1: &nbsp;
 |-
-|We proceed using integration by parts. Let <math style="vertical-align: -5px">u=\sin(2x)</math> and <math style="vertical-align: 0px">dv=e^{-2x}dx.</math> Then, <math style="vertical-align: -5px">du=2\cos(2x)dx</math> and <math style="vertical-align: -13px">v=\frac{e^{-2x}}{-2}.</math>
+|We proceed using integration by parts.
 |-
 |So, we get
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -14px">\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}-\int \frac{e^{-2x}2\cos(2x)~dx}{-2}=\frac{\sin(2x)e^{-2x}}{-2}+\int e^{-2x}\cos(2x)~dx.</math>
 |}
@@ Line 58: / Line 71: @@
 !Step 2: &nbsp;
 |-
-|Now, we need to use integration by parts again. Let <math style="vertical-align: -5px">u=\cos(2x)</math> and <math style="vertical-align: 0px">dv=e^{-2x}dx.</math> Then, <math style="vertical-align: -5px">du=-2\sin(2x)dx</math> and <math style="vertical-align: -13px">v=\frac{e^{-2x}}{-2}.</math>
+|Now, we need to use integration by parts again.
 |-
-|So, we get
+|Let <math style="vertical-align: -5px">u=\cos(2x)</math> and <math style="vertical-align: 0px">dv=e^{-2x}dx.</math>
 |-
 | &nbsp;&nbsp; <math style="vertical-align: -13px">\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}+\frac{\cos(2x)e^{-2x}}{-2}-\int e^{-2x}\sin(2x)~dx.</math>
 |-
+|
@@ Line 70: / Line 88: @@
 !Step 3: &nbsp;
 |-
-|Notice that the integral on the right of the last equation in Step 2 is the same integral that we had at the beginning of the problem.
+|Notice that the integral on the right of the last equation in Step 2
 |-
-|So, if we add the integral on the right to the other side of the equation, we get
+|is the same integral that we had at the beginning of the problem.
 |-
 | &nbsp;&nbsp; <math>2\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-2}+\frac{\cos(2x)e^{-2x}}{-2}.</math>
@@ Line 80: / Line 100: @@
 | &nbsp;&nbsp; <math>\int e^{-2x}\sin (2x)~dx=\frac{\sin(2x)e^{-2x}}{-4}+\frac{\cos(2x)e^{-2x}}{-4}.</math>
 |-
-|Thus, the final answer is <math style="vertical-align: -13px">\int e^{-2x}\sin (2x)~dx=\frac{e^{-2x}}{-4}((\sin(2x)+\cos(2x))+C.</math>
+|Thus, the final answer is
 |}