022 Exam 2 Sample B, Problem 7 - Revision history

MathAdmin at 00:28, 18 May 2015

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MathAdmin at 00:27, 18 May 2015

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MathAdmin at 18:04, 17 May 2015

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MathAdmin at 18:04, 17 May 2015

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MathAdmin at 01:49, 16 May 2015

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MathAdmin at 01:33, 16 May 2015

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MathAdmin at 01:28, 16 May 2015

2015-05-16T01:28:28Z

MathAdmin: Created page with "Find the antiderivatives: ::(a) $\int xe^{3x^2+1}\,dx.$
::(b) $\int_2^54x - 5\,dx.$ {| c..."

2015-05-16T01:10:37Z

Created page with "Find the antiderivatives: ::(a) <math> \int xe^{3x^2+1}\,dx.</math> ::(b) <math>\int_2^54x - 5\,dx.</math> {| c..."

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Find the antiderivatives:

::(a) <math> \int xe^{3x^2+1}\,dx.</math>
 
::(b) <math>\int_2^54x - 5\,dx.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|This problem requires two rules of integration. In particular, you need
|-
|'''Integration by substitution (''u'' - sub):''' If <math style="vertical-align: -20%">u = g(x)</math>  is a differentiable functions whose range is in the domain of <math style="vertical-align: -20%">f</math>, then
|-
|
::<math>\int g'(x)f(g(x)) dx \,=\, \int f(u) du.</math>
|-
|We also need the derivative of the natural log since we will recover natural log from integration:
|-
|
::<math>\left(ln(x)\right)' \,=\, \frac{1}{x}.</math>
|}

 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Use a ''u''-substitution with <math style="vertical-align: -8%">u = 3x + 2.</math> This means <math style="vertical-align: 0%">du = 3\,dx</math>, or <math style="vertical-align: -20%">dx=du/3</math>. After substitution we have
::<math>\int \frac{1}{3x + 2} \,=\, \int \frac{1}{u}\,\frac{du}{3}\,=\,\frac{1}{3}\int\frac{1}{u}\,du.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We can now take the integral remembering the special rule:
|-
|
::<math>\frac{1}{3}\int\frac{1}{u}\,du\,=\, \frac{\log(u)}{3}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
| Now we need to substitute back into our original variables using our original substitution <math style="vertical-align: -5%">u = 3x + 2</math>
|-
| to find  <math>\frac{\log(u)}{3} = \frac{\log(3x + 2)}{3}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|Since this integral is an indefinite integral we have to remember to add a constant  <math style="vertical-align: 0%">C</math> at the end.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|
::<math>\int \frac{1}{3x + 2}\,dx \,=\, \frac{\ln(3x + 2)}{3} + C.</math>
|}

[[022_Exam_2_Sample_A|'''Return to Sample Exam''']]

@@ Line 16: / Line 16: @@
 |-
+|
-::<math style="vertical-align: -70%;">\int x^n dx \,=\, \frac{x^{n + 1}}{n + 1}+C,</math>&thinsp; for <math style="vertical-align: -25%;">n\neq -1</math>,
+::<math style="vertical-align: -70%;">\int x^n dx \,=\, \frac{x^{n + 1}}{n + 1}+C,</math>&thinsp; for <math style="vertical-align: -22%;">n\neq -1</math>,
 |-
 |as well as the convenient antiderivative:

@@ Line 16: / Line 16: @@
 |-
+|
-::<math style="vertical-align: -70%;">\int x^n dx \,=\, \frac{x^{n + 1}}{n + 1}+C,</math>&thinsp; for <math style="vertical-align: -23%;">n\neq 0</math>,
+::<math style="vertical-align: -70%;">\int x^n dx \,=\, \frac{x^{n + 1}}{n + 1}+C,</math>&thinsp; for <math style="vertical-align: -25%;">n\neq -1</math>,
 |-
 |as well as the convenient antiderivative:

@@ Line 36: / Line 36: @@
 !(a) Step 2: &nbsp;
 |-
-|Now, we need to substitute back into our original variables using our original substitution <math style="vertical-align: -5%">u = 3x^2 + 1</math>
+|Now, we need to substitute back into our original variable using our original substitution <math style="vertical-align: -5%">u = 3x^2 + 1</math>
 |-
 | to find&nbsp; <math style="vertical-align: -60%">\frac{1}{6}e^u = \frac{e^{3x^2 + 1}}{6}.</math>

@@ Line 16: / Line 16: @@
 |-
+|
-::<math style="vertical-align: -70%;">\int x^n dx \,=\, \frac{x^{n + 1}}{n + 1}+C,</math>&thinsp; for <math style="vertical-align: -23%;">n\neq 0</math>.
+::<math style="vertical-align: -70%;">\int x^n dx \,=\, \frac{x^{n + 1}}{n + 1}+C,</math>&thinsp; for <math style="vertical-align: -23%;">n\neq 0</math>,
 |}

@@ Line 22: / Line 22: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
+!(a) Step 1: &nbsp;
 |-
-|(a) Use a ''u''-substitution with <math style="vertical-align: -8%">u = 3x^2 + 1.</math> This means <math style="vertical-align: 0%">du = 6x\,dx</math>, or <math style="vertical-align: -20%">dx=du/6</math>. After substitution we have
+|(a) Use a ''u''-substitution with <math style="vertical-align: -8%">u = 3x^2 + 1.</math> This means <math style="vertical-align: 0%">du = 6x\,dx</math>, or <math style="vertical-align: -20%">dx=du/6</math>. Substituting, we have
-::<math>\int x e^{3x^2+1}\,dx = \frac{1}{6} \int e^{u}\, du. </math>
+::<math>\int x e^{3x^2+1}\,dx \,=\,  \int xe^{u}\cdot\frac{du}{6}\,=\,\frac{1}{6}\int e^u\,du\,=\,\frac{1}{6}u. </math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
+!(a) Step 2: &nbsp;
 |-
-|(a)
+|Now, we need to substitute back into our original variables using our original substitution <math style="vertical-align: -5%">u = 3x^2 + 1</math>
 |-
-|(b) We just need to evaluate at the endpoints to finish the problem:
+| to find&nbsp; <math style="vertical-align: -60%">\frac{1}{6}e^u = \frac{e^{3x^2 + 1}}{6}.</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 3: &nbsp;
+!(a) Step 3: &nbsp;
 |-
-|(a) Now we need to substitute back into our original variables using our original substitution <math style="vertical-align: -5%">u = 3x^2 + 1</math>
+|Since this integral is an indefinite integral, we have to remember to add a constant&thinsp; <math style="vertical-align: 0%">C</math> at the end.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 4: &nbsp;
+!(b): &nbsp;
 |-
-|Since this integral is an indefinite integral we have to remember to add a constant&thinsp; <math style="vertical-align: 0%">C</math> at the end.
+|Then, we evaluate:
 |}
@@ Line 61: / Line 63: @@
 !Final Answer: &nbsp;
 |-
-|(a)
+|'''(a)'''
-::<math>\frac{e^{3x^2 + 1}}{6} + C.</math>
+::<math>\int xe^{3x^2+1}\,dx\,=\,\frac{e^{3x^2 + 1}}{6} + C.</math>
 |-
-|(b) <math>27</math>
 |}
 [[022_Exam_2_Sample_B|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 01:49, 16 May 2015
Line 67:		Line 67:
	\|}		\|}

−	[[~~022_Exam_2_Sample_A~~\|'''<u>Return to Sample Exam</u>''']]	+	[[022_Exam_2_Sample_B\|'''<u>Return to Sample Exam</u>''']]

@@ Line 25: / Line 25: @@
 |-
 |(a) Use a ''u''-substitution with <math style="vertical-align: -8%">u = 3x^2 + 1.</math> This means <math style="vertical-align: 0%">du = 6x\,dx</math>, or <math style="vertical-align: -20%">dx=du/6</math>. After substitution we have
-::<math>\int x e^{3x^2+1} dx = \frac{1}{6} \int e^{u} du. </math>
+::<math>\int x e^{3x^2+1}\,dx = \frac{1}{6} \int e^{u}\, du. </math>
 |-
 |(b) We need to use the power rule to find that <math>\int_2^5 4x - 5 \, dx = 2x^2 - 5x \Bigr|_2^5</math>
@@ Line 64: / Line 64: @@
 ::<math>\frac{e^{3x^2 + 1}}{6} + C.</math>
 |-
-|(b) 27
+|(b) <math>27</math>
 |}
 [[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
-|This problem requires two rules of integration.  In particular, you need
+|This problem requires '''Integration by substitution (''u'' - sub):''' If <math style="vertical-align: -20%">u = g(x)</math>&thinsp; is a differentiable functions whose range is in the domain of <math style="vertical-align: -20%">f</math>, then
 |-
+|
 ::<math>\int g'(x)f(g(x)) dx \,=\, \int f(u) du.</math>
 |-
-|We also need the derivative of the natural log since we will recover natural log from integration:
+|We also need our power rule for integration:
 |-
+|
-::<math>\left(ln(x)\right)' \,=\, \frac{1}{x}.</math>
+::<math style="vertical-align: -70%;">\int x^n dx \,=\, \frac{x^{n + 1}}{n + 1}+C,</math>&thinsp; for <math style="vertical-align: -23%;">n\neq 0</math>.
 |}
@@ Line 26: / Line 24: @@
 !Step 1: &nbsp;
 |-
-|Use a ''u''-substitution with <math style="vertical-align: -8%">u = 3x + 2.</math> This means <math style="vertical-align: 0%">du = 3\,dx</math>, or <math style="vertical-align: -20%">dx=du/3</math>. After substitution we have
+|(a) Use a ''u''-substitution with <math style="vertical-align: -8%">u = 3x^2 + 1.</math> This means <math style="vertical-align: 0%">du = 6x\,dx</math>, or <math style="vertical-align: -20%">dx=du/6</math>. After substitution we have
-::<math>\int \frac{1}{3x + 2} \,=\, \int \frac{1}{u}\,\frac{du}{3}\,=\,\frac{1}{3}\int\frac{1}{u}\,du.</math>
+::<math>\int x e^{3x^2+1}\, dx &= \frac{1}{6} \int e^{u}\, du.</math>
 |}
@@ Line 33: / Line 33: @@
 !Step 2: &nbsp;
 |-
-|We can now take the integral remembering the special rule:
+|(a)
 |-
+|(b) We just need to evaluate at the endpoints to finish the problem:
-::<math>\frac{1}{3}\int\frac{1}{u}\,du\,=\, \frac{\log(u)}{3}.</math>
+<math>\begin{array}{rcl}2x^2 - 5x \Bigr|_2^5 & = & 2(5^2) - 5(5) -(2(2)^2 - 5(2)\\
 |}
@@ Line 42: / Line 47: @@
 !Step 3: &nbsp;
 |-
-| Now we need to substitute back into our original variables using our original substitution <math style="vertical-align: -5%">u = 3x + 2</math>
+|(a) Now we need to substitute back into our original variables using our original substitution <math style="vertical-align: -5%">u = 3x^2 + 1</math>
 |-
-| to find&nbsp; <math>\frac{\log(u)}{3} = \frac{\log(3x + 2)}{3}.</math>
+| to find&nbsp; <math>\frac{1}{6}e^u = \frac{e^{3x^2 + 1}}{6}.</math>
 |}
@@ Line 56: / Line 61: @@
 !Final Answer: &nbsp;
 |-
+|(a)
-::<math>\int \frac{1}{3x + 2}\,dx \,=\, \frac{\ln(3x + 2)}{3} + C.</math>
+::<math>\frac{e^{3x^2 + 1}}{6} + C.</math>
 |}
 [[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

022 Exam 2 Sample B, Problem 7 - Revision history

MathAdmin at 00:28, 18 May 2015

MathAdmin at 00:27, 18 May 2015

MathAdmin at 18:04, 17 May 2015

MathAdmin at 18:04, 17 May 2015

MathAdmin at 18:02, 17 May 2015

MathAdmin at 01:49, 16 May 2015

MathAdmin at 01:33, 16 May 2015

MathAdmin at 01:32, 16 May 2015

MathAdmin at 01:28, 16 May 2015

MathAdmin: Created page with "Find the antiderivatives: ::(a) \int xe^{3x^2+1}\,dx. ::(b) \int_2^54x - 5\,dx. {| c..."

MathAdmin: Created page with "Find the antiderivatives: ::(a) $\int xe^{3x^2+1}\,dx.$
::(b) $\int_2^54x - 5\,dx.$ {| c..."