022 Sample Final A, Problem 3 - Revision history

MathAdmin at 02:10, 7 June 2015

2015-06-07T02:10:17Z

MathAdmin at 01:34, 7 June 2015

2015-06-07T01:34:15Z

MathAdmin: Created page with " Find the antiderivative: $\int \frac{6}{x^2 - x - 12}$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations: |..."

2015-06-04T23:12:32Z

Created page with "<span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}</math> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations: |..."

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<span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|1) What does the denominator factor into? What will be the form of the decomposition?
|-
|2) How do you solve for the numerators?
|-
|3) What special integral do we have to use?
|-
|Answer:
|-
|1) Since <math>x^2 - x - 12 = (x - 4)(x +3)</math>  , and each term has multiplicity one, the decomposition will be of the form: <math>\frac{A}{x - 4} + \frac{B}{x + 3}</math>
|-
|2) After writing the equality, <math>\frac{6}{x^2 -x - 12} = \frac{A}{x - 4} + \frac{B}{x + 3}</math>, clear the denominators, and evaluate both sides at x = 4, -3, Each evaluation will yield the value of one of the unknowns.
|-
|3) We have to remember that <math>\int \frac{c}{x - a} dx = c\ln(x - a)</math> , for any numbers c, a.
|}

'''Solution:'''

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we factor <math>x^2 - x - 12 = (x - 4)(x + 3)</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now we want to find the partial fraction expansion for <math>\frac{6}{(x - 4)(x + 3)}</math> , which will have the form <math>\frac{A}{x - 4} + {B}{x + 3}</math>
|-
|To do this we need to solve the equation <math>6 = A( x + 3) + B(x - 4)</math>
|-
|Plugging in -3 for x to both sides we find that <math>6 = -7B</math>   and   <math>B = -\frac{6}{7}</math>.
|-
|Now we can find A by plugging in 4 for x to both sides. This yields <math>6 = 7A</math> , so <math>A = \frac{6}{7}</math>
|-
|Finally we have the partial fraction expansion: <math>\frac{6}{x^2 -x - 12} = \frac{6}{7(x - 4)} - \frac{6}{7(x + 3)}</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now to finish the problem we integrate each fraction to get: <math>\int \frac{6}{x^2 -x -12} dx = \int \frac{6}{7(x - 4)}dx - \int \frac{6}{7(x + 3)}dx </math>  to get  <math>\frac{6}{7}\ln(x - 4) - \frac{6}{7}\ln(x + 3)</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|Now make sure you remember to add the <math> + C</math> to the integral at the end.
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|<math>\frac{6}{7}\ln(x - 4) - \frac{6}{7}\ln(x + 3) + C</math>
|}

[[022_Sample Final A|'''<u>Return to Sample Final</u>''']]

← Older revision		Revision as of 01:34, 7 June 2015
Line 1:		Line 1:
−	<span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}</math>	+	<span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}\,dx.</math>

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 10: / Line 10: @@
 |3) What special integral do we have to use?
 |-
-|Answer:
+|Answers:
 |-
-|1) Since <math>x^2 - x - 12 = (x - 4)(x +3)</math> &nbsp;, and each term has multiplicity one, the decomposition will be of the form: <math>\frac{A}{x - 4} + \frac{B}{x + 3}</math>
+|1) Since <math style="vertical-align: -5px">x^2 - x - 12 = (x - 4)(x +3)</math>&thinsp;, and each term has multiplicity one, the decomposition will be of the form:&thinsp; <math style="vertical-align: -15px">\frac{A}{x - 4} + \frac{B}{x + 3}</math>.
 |-
-|2) After writing the equality, <math>\frac{6}{x^2 -x - 12} = \frac{A}{x - 4} + \frac{B}{x + 3}</math>, clear the denominators, and evaluate both sides at x = 4, -3, Each evaluation will yield the value of one of the unknowns.
+|2) After writing the equality, <math style="vertical-align: -15px">\frac{6}{x^2 -x - 12} \,=\, \frac{A}{x - 4} + \frac{B}{x + 3}</math>, clear the denominators, and evaluate both sides at <math style="vertical-align: -4px">x = 4, -3</math>. Each evaluation will yield the value of one of the unknowns.
 |-
-|3) We have to remember that <math>\int \frac{c}{x - a} dx = c\ln(x - a)</math>&nbsp;, for any numbers c, a.
+|3) We have to remember that <math style="vertical-align: -13px">\int \frac{c}{x - a} dx = c\ln(x - a)</math>&nbsp;, for any numbers <math style="vertical-align: -4px">c, a</math>.
 |}
@@ Line 24: / Line 24: @@
 !Step 1: &nbsp;
 |-
-|First, we factor <math>x^2 - x - 12 = (x - 4)(x + 3)</math>
+|First, we factor: <math style="vertical-align: -5px">x^2 - x - 12 = (x - 4)(x +3)</math>&thinsp;.
 |}
@@ Line 30: / Line 30: @@
 !Step 2: &nbsp;
 |-
-|Now we want to find the partial fraction expansion for <math>\frac{6}{(x - 4)(x + 3)}</math>&nbsp;, which will have the form <math>\frac{A}{x - 4} + {B}{x + 3}</math>
+|Now we want to find the partial fraction expansion for <math>\frac{6}{(x - 4)(x + 3)}</math>&nbsp;, which will have the form <math style="vertical-align: -20px">\frac{A}{x - 4} + \frac{B}{x + 3}</math>.
 |-
-|To do this we need to solve the equation <math>6 = A( x + 3) + B(x - 4)</math>
+|To do this, we need to solve the equation&thinsp; <math style="vertical-align: -5px">6 \,=\, A( x + 3) + B(x - 4)</math>.
 |-
-|Plugging in -3 for x to both sides we find that <math>6 = -7B</math> &nbsp; and &nbsp; <math>B = -\frac{6}{7}</math>.
+|Plugging in&thinsp; <math style="vertical-align: 0px">-3</math> for <math style="vertical-align: 0px">x</math>, we find that&thinsp; <math style="vertical-align: -0.2px">6 \,=\, -7B</math>&thinsp;, and thus&thinsp; <math style="vertical-align: -13px">B = -\frac{6}{7}</math>&thinsp;.
 |-
-|Now we can find A by plugging in 4 for x to both sides. This yields <math>6 = 7A</math>&nbsp;, so <math>A = \frac{6}{7}</math>
+|Similarly, we can find <math style="vertical-align: -0px">A</math> by plugging in <math style="vertical-align: -1px">4</math> for <math style="vertical-align: 0px">x</math>. This yields <math style="vertical-align: -0.2px">6 = 7A</math>&nbsp;, so <math style="vertical-align: -13px">A = \frac{6}{7}</math>&thinsp;.
 |-
-|Finally we have the partial fraction expansion: <math>\frac{6}{x^2 -x - 12} = \frac{6}{7(x - 4)} - \frac{6}{7(x + 3)}</math>
+|This completes the partial fraction expansion:
 |}
@@ Line 44: / Line 45: @@
 !Step 3: &nbsp;
 |-
-|Now to finish the problem we integrate each fraction to get: <math>\int \frac{6}{x^2 -x -12} dx = \int \frac{6}{7(x - 4)}dx  - \int \frac{6}{7(x + 3)}dx </math>&nbsp; to get &nbsp;<math>\frac{6}{7}\ln(x - 4) - \frac{6}{7}\ln(x + 3)</math>
+|By the previous step, we have
 |}
@@ Line 50: / Line 55: @@
 !Step 4: &nbsp;
 |-
-|Now make sure you remember to add the <math> + C</math> to the integral at the end.
+|Now, make sure you remember to add the&thinsp; <math style="vertical-align: -2px"> + C</math>&thinsp; to the integral at the end.
 |}
@@ Line 56: / Line 61: @@
 !Final Answer: &nbsp;
 |-
 |<math>\frac{6}{7}\ln(x - 4) - \frac{6}{7}\ln(x + 3) + C</math>
 |}
 [[022_Sample Final A|'''<u>Return to Sample Final</u>''']]

022 Sample Final A, Problem 3 - Revision history

MathAdmin at 02:10, 7 June 2015

MathAdmin at 01:34, 7 June 2015

MathAdmin: Created page with " Find the antiderivative: \int \frac{6}{x^2 - x - 12} {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations: |..."

MathAdmin: Created page with " Find the antiderivative: $\int \frac{6}{x^2 - x - 12}$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations: |..."