009B Sample Midterm 3, Problem 4 - Revision history

Kayla Murray at 01:30, 13 November 2017

2017-11-13T01:30:44Z

Kayla Murray at 01:26, 5 November 2017

2017-11-05T01:26:15Z

Kayla Murray at 01:41, 27 February 2017

2017-02-27T01:41:40Z

Kayla Murray at 01:40, 27 February 2017

2017-02-27T01:40:01Z

Kayla Murray at 00:15, 19 February 2017

2017-02-19T00:15:03Z

Kayla Murray at 01:40, 8 February 2017

2017-02-08T01:40:03Z

Kayla Murray at 17:19, 7 February 2017

2017-02-07T17:19:26Z

Kayla Murray at 15:37, 7 February 2017

2017-02-07T15:37:54Z

Kayla Murray at 15:32, 7 February 2017

2017-02-07T15:32:02Z

Kayla Murray at 15:30, 7 February 2017

2017-02-07T15:30:27Z

@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp; <math>\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx</math>
 <hr>
-(insert picture of handwritten solution)
+[[009B Sample Midterm 3, Problem 4 Solution|'''<u>Solution</u>''']]
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam"> The rate of reaction to a drug is given by:
+<span class="exam"> Compute the following integrals:
-::<math>r'(t)=2t^2e^{-t}</math>
+<span class="exam">(a) &nbsp; <math>\int x^2\sin (x^3) ~dx</math>
-<span class="exam">where &nbsp;<math style="vertical-align: 0px">t</math>&nbsp; is the number of hours since the drug was administered.
+<span class="exam">(b) &nbsp; <math>\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx</math>
-<span class="exam">Find the total reaction to the drug from &nbsp;<math style="vertical-align: -1px">t=1</math>&nbsp; to &nbsp;<math style="vertical-align: 0px">t=6.</math>
+[[009B Sample Midterm 3, Problem 4 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 3: / Line 3: @@
 ::<math>r'(t)=2t^2e^{-t}</math>
-<span class="exam">where &nbsp;<math>t</math>&nbsp; is the number of hours since the drug was administered.
+<span class="exam">where &nbsp;<math style="vertical-align: 0px">t</math>&nbsp; is the number of hours since the drug was administered.
 <span class="exam">Find the total reaction to the drug from &nbsp;<math style="vertical-align: -1px">t=1</math>&nbsp; to &nbsp;<math style="vertical-align: 0px">t=6.</math>

@@ Line 3: / Line 3: @@
 ::<math>r'(t)=2t^2e^{-t}</math>
-<span class="exam">where <math>t</math> is the number of hours since the drug was administered.
+<span class="exam">where &nbsp;<math>t</math>&nbsp; is the number of hours since the drug was administered.
-<span class="exam">Find the total reaction to the drug from <math style="vertical-align: -1px">t=1</math> to <math style="vertical-align: 0px">t=6.</math>
+<span class="exam">Find the total reaction to the drug from &nbsp;<math style="vertical-align: -1px">t=1</math>&nbsp; to &nbsp;<math style="vertical-align: 0px">t=6.</math>
@@ Line 11: / Line 11: @@
 !Foundations: &nbsp;
 |-
-|If we calculate <math style="vertical-align: -14px">\int_a^b r'(t)~dt,</math> what are we calculating?
+|If we calculate &nbsp;<math style="vertical-align: -14px">\int_a^b r'(t)~dt,</math>&nbsp; what are we calculating?
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; We are calculating <math style="vertical-align: -5px">r(b)-r(a).</math> This is the total reaction to the
+&nbsp; &nbsp; &nbsp; &nbsp; We are calculating &nbsp;<math style="vertical-align: -5px">r(b)-r(a).</math>&nbsp; This is the total reaction to the
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; drug from <math style="vertical-align: 0px">t=a</math> to <math style="vertical-align: 0px">t=b.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; drug from &nbsp;<math style="vertical-align: 0px">t=a</math>&nbsp; to &nbsp;<math style="vertical-align: 0px">t=b.</math>
 |}
@@ Line 25: / Line 25: @@
 !Step 1: &nbsp;
 |-
-|To calculate the total reaction to the drug from <math style="vertical-align: -1px">t=1</math> to <math style="vertical-align: -4px">t=6,</math>
+|To calculate the total reaction to the drug from &nbsp;<math style="vertical-align: -1px">t=1</math>&nbsp; to &nbsp;<math style="vertical-align: -4px">t=6,</math>
 |-
 |we need to calculate
@@ Line 38: / Line 38: @@
 |We proceed using integration by parts.
 |-
-|Let <math style="vertical-align: 0px">u=2t^2</math> and <math style="vertical-align: 0px">dv=e^{-t}dt.</math>
+|Let &nbsp;<math style="vertical-align: 0px">u=2t^2</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-t}dt.</math>
 |-
-|Then, <math style="vertical-align: -1px">du=4t~dt</math> and <math style="vertical-align: 0px">v=-e^{-t}.</math>
+|Then, &nbsp;<math style="vertical-align: -1px">du=4t~dt</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=-e^{-t}.</math>
 |-
 |Then, we have
@@ Line 52: / Line 52: @@
 |Now, we need to use integration by parts again.
 |-
-|Let <math style="vertical-align: 0px">u=4t</math> and <math style="vertical-align: 0px">dv=e^{-t}dt.</math>
+|Let &nbsp;<math style="vertical-align: 0px">u=4t</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-t}dt.</math>
 |-
-|Then, <math style="vertical-align: -1px">du=4dt</math> and <math style="vertical-align: 0px">v=-e^{-t}.</math>
+|Then, &nbsp;<math style="vertical-align: -1px">du=4dt</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=-e^{-t}.</math>
 |-
 |Thus, we get

@@ Line 1: / Line 1: @@
 <span class="exam"> The rate of reaction to a drug is given by:
-::::::<math>r'(t)=2t^2e^{-t}</math>
+::<math>r'(t)=2t^2e^{-t}</math>
 <span class="exam">where <math>t</math> is the number of hours since the drug was administered.

@@ Line 14: / Line 14: @@
 |-
+|
-&nbsp; &nbsp; We are calculating <math style="vertical-align: -5px">r(b)-r(a).</math> This is the total reaction to the
+&nbsp; &nbsp; &nbsp; &nbsp; We are calculating <math style="vertical-align: -5px">r(b)-r(a).</math> This is the total reaction to the
 |-
+|
-&nbsp; &nbsp; drug from <math style="vertical-align: 0px">t=a</math> to <math style="vertical-align: 0px">t=b.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; drug from <math style="vertical-align: 0px">t=a</math> to <math style="vertical-align: 0px">t=b.</math>
 |}
@@ Line 25: / Line 25: @@
 !Step 1: &nbsp;
 |-
-|To calculate the total reaction to the drug from <math style="vertical-align: -1px">t=1</math> to <math style="vertical-align: -5px">t=6,</math>
+|To calculate the total reaction to the drug from <math style="vertical-align: -1px">t=1</math> to <math style="vertical-align: -4px">t=6,</math>
 |-
 |we need to calculate
 |-
+|
-&nbsp; &nbsp; <math>\int_1^6 r'(t)~dt=\int_1^6 2t^2e^{-t}~dt.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\int_1^6 r'(t)~dt=\int_1^6 2t^2e^{-t}~dt.</math>
 |}
@@ Line 44: / Line 44: @@
 |Then, we have
 |-
-|&nbsp;&nbsp; <math style="vertical-align: -14px">\int_1^62t^2e^{-t}~dt=\left. -2t^2e^{-t}\right|_1^6+\int_1^6 4te^{-t}~dt.</math>
+|&nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\int_1^62t^2e^{-t}~dt=\left. -2t^2e^{-t}\right|_1^6+\int_1^6 4te^{-t}~dt.</math>
 |}
@@ Line 59: / Line 59: @@
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int_1^62t^2e^{-t}~dt} & = & \displaystyle{\left. -2t^2e^{-t}-4te^{-t}\right|_1^6+\int_1^6 4e^{-t}}\\
 &&\\
@@ Line 74: / Line 74: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; <math>\frac{-100+10e^5}{e^6}</math>
+|&nbsp; &nbsp; &nbsp;&nbsp; <math>\frac{-100+10e^5}{e^6}</math>
 |}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 25: / Line 25: @@
 !Step 1: &nbsp;
 |-
-|To calculate the total reaction to the drug from <math>t=1</math> to <math>t=6,</math> we need to calculate
+|To calculate the total reaction to the drug from <math style="vertical-align: -1px">t=1</math> to <math style="vertical-align: -5px">t=6,</math> we need to calculate
 |-
+|
@@ Line 34: / Line 34: @@
 !Step 2: &nbsp;
 |-
-|We proceed using integration by parts. Let <math style="vertical-align: -2px">u=2t^2</math> and <math style="vertical-align: 0px">dv=e^{-t}dt.</math> Then, <math style="vertical-align: -5px">du=4t~dt</math> and <math style="vertical-align: -2px">v=-e^{-t}.</math>
+|We proceed using integration by parts. Let <math style="vertical-align: 0px">u=2t^2</math> and <math style="vertical-align: 0px">dv=e^{-t}dt.</math> Then, <math style="vertical-align: -1px">du=4t~dt</math> and <math style="vertical-align: 0px">v=-e^{-t}.</math>
 |-
 |Then, we have
 |-
 |&nbsp;&nbsp; <math style="vertical-align: -14px">\int_1^62t^2e^{-t}~dt=\left. -2t^2e^{-t}\right|_1^6+\int_1^6 4te^{-t}~dt.</math>
@@ Line 46: / Line 44: @@
 !Step 3: &nbsp;
 |-
-|Now, we need to use integration by parts again. Let <math style="vertical-align: -2px">u=4t</math> and <math style="vertical-align: 0px">dv=e^{-t}dt.</math> Then, <math style="vertical-align: -5px">du=4dt</math> and <math style="vertical-align: -2px">v=-e^{-t}.</math>
+|Now, we need to use integration by parts again. Let <math style="vertical-align: 0px">u=4t</math> and <math style="vertical-align: 0px">dv=e^{-t}dt.</math> Then, <math style="vertical-align: -1px">du=4dt</math> and <math style="vertical-align: 0px">v=-e^{-t}.</math>
 |-
 |Thus, we get

← Older revision		Revision as of 15:32, 7 February 2017
Line 5:		Line 5:
	<span class="exam">where <math>t</math> is the number of hours since the drug was administered.		<span class="exam">where <math>t</math> is the number of hours since the drug was administered.

−	<span class="exam">Find the total reaction to the drug from <math>t=1</math> to <math>t=6.</math>	+	<span class="exam">Find the total reaction to the drug from <math style="vertical-align: -1px">t=1</math> to <math style="vertical-align: 0px">t=6.</math>