022 Exam 2 Sample A, Problem 3 - Revision history

MathAdmin at 14:47, 16 May 2015

2015-05-16T14:47:27Z

2015-05-16T14:45:12Z

2015-05-16T14:44:45Z

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@@ Line 30: / Line 30: @@
 !Step 2: &nbsp;
 |-
-|We can now take the integral remembering the special rule:
+|We can now take the integral remembering the special rule resulting in natural log:
 |-
+|
@@ Line 39: / Line 39: @@
 !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>
+| 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
 |-
-| to find&nbsp; <math>\frac{\log(u)}{3} = \frac{\log(3x + 2)}{3}.</math>
 |}

@@ Line 47: / Line 47: @@
 !Step 4: &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.
+|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.
 |}

@@ Line 24: / Line 24: @@
 |-
 |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>
+::<math>\int \frac{1}{3x + 2}\,dx \,=\, \int \frac{1}{u}\,\frac{du}{3}\,=\,\frac{1}{3}\int\frac{1}{u}\,du.</math>
 |}

@@ Line 15: / Line 15: @@
 |-
+|
-::<math>\left(ln(x)\right)' \,=\, \frac{1}{x}</math>
+::<math>\left(ln(x)\right)' \,=\, \frac{1}{x}.</math>
 |}

@@ Line 7: / Line 7: @@
 |This problem requires two rules of integration.  In particular, you need
 |-
-|'''Integration by substitution (''u'' - sub):''' If <math style="vertical-align: -25%">u = g(x)</math> is a differentiable functions whose range is in the domain of <math style="vertical-align: -20%">f</math>, then
+|'''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
 |-
+|

@@ Line 39: / Line 39: @@
 !Step 3: &nbsp;
 |-
-| Now we need to substitute back into our original variables using our original substitution <math style="vertical-align: -10%">u = 3x + 2</math>
+| 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&thinsp; <math>\frac{\log(u)}{3} = \frac{\log(3x + 2)}{3}.</math>
+| to find&nbsp; <math>\frac{\log(u)}{3} = \frac{\log(3x + 2)}{3}.</math>
 |}
@@ Line 54: / Line 54: @@
 |-
+|
-::<math>\int \frac{1}{3x + 2} dx \,=\, \frac{\ln(3x + 2)}{3} + C.</math>
+::<math>\int \frac{1}{3x + 2}\,dx \,=\, \frac{\ln(3x + 2)}{3} + C.</math>
 |}
 [[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 32: / Line 32: @@
 |We can now take the integral remembering the special rule:
 |-
-|<math>\frac{1}{3}\int\frac{1}{u}\,du. \,=\, \frac{\log(u)}{3}.</math>
 |}