022 Sample Final A, Problem 3 - Revision history

Matt Lee at 19:58, 30 May 2015

2015-05-30T19:58:14Z

Matt Lee at 19:56, 30 May 2015

2015-05-30T19:56:55Z

Matt Lee at 19:54, 30 May 2015

2015-05-30T19:54:50Z

Matt Lee at 19:43, 30 May 2015

2015-05-30T19:43:38Z

Matt Lee at 19:26, 30 May 2015

2015-05-30T19:26:07Z

Matt Lee: Created page with "{| 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..."

2015-05-30T19:23:04Z

Created page with "{| 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..."

New page

{| 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.
|}

← Older revision		Revision as of 19:58, 30 May 2015
Line 60:		Line 60:


−	[[022_Sample Final A\|<~~span class="biglink"~~>~~<span style="font-size:80%"> ~~Return to ~~Exam </span>~~</~~span~~>]]	+	[[022_Sample Final A\|'''<u>Return to Sample Final</u>''']]

← Older revision		Revision as of 19:56, 30 May 2015
Line 58:		Line 58:
	\|<math>\frac{6}{7}\ln(x - 4) - \frac{6}{7}\ln(x + 3) + C</math>		\|<math>\frac{6}{7}\ln(x - 4) - \frac{6}{7}\ln(x + 3) + C</math>
	\|}		\|}
		+
		+
		+	[[022_Sample Final A\|<span class="biglink"><span style="font-size:80%"> Return to Exam </span></span>]]

@@ Line 39: / Line 39: @@
 |-
 |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>
 |}

@@ Line 17: / Line 17: @@
 |-
 |3) We have to remember that <math>\int \frac{c}{x - a} dx = c\ln(x - a)</math>&nbsp;, for any numbers c, a.
 |}

@@ Line 1: / Line 1: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
@@ Line 6: / Line 8: @@
 |2) How do you solve for the numerators?
 |-
-) What special integral do we have to use?
+|3) What special integral do we have to use?
 |-
 |Answer: