Difference between revisions of "022 Exam 2 Sample A, Problem 3"

From Math Wiki
Jump to navigation Jump to search
 
(12 intermediate revisions by the same user not shown)
Line 7: Line 7:
 
|This problem requires two rules of integration.  In particular, you need
 
|This problem requires two rules of integration.  In particular, you need
 
|-
 
|-
−
|'''Integration by substitution (U - sub):''' If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, 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
−
|-
 
−
 
 
−
|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math>
 
−
|-
 
−
|<br>'''The Product Rule:''' If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
 
−
|-
 
−
|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</math>
 
−
|-
 
−
|<br>'''The Quotient Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions and <math style="vertical-align: -21%;">g(x) \neq 0</math>&thinsp;, then
 
−
|-
 
−
|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. </math>
 
−
|-
 
−
|Additionally, we will need our power rule for differentiation:
 
 
|-
 
|-
 
|
 
|
−
::<math style="vertical-align: -21%;">\left(x^n\right)'\,=\,nx^{n-1},</math> for <math style="vertical-align: -25%;">n\neq 0</math>,
+
::<math>\int g'(x)f(g(x)) dx \,=\, \int f(u) du.</math>
 
|-
 
|-
−
|as well as the derivative of natural log:
+
|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>
+
::<math>\left(ln(x)\right)' \,=\, \frac{1}{x}.</math>
−
|<br>
 
 
|}
 
|}
  
Line 37: Line 23:
 
!Step 1: &nbsp;
 
!Step 1: &nbsp;
 
|-
 
|-
−
|Use a U-substitution with <math>u = 3x + 2.</math> This means <math>du = 3 dx</math>, and after substitution we have
+
|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}{3u} 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 44: Line 30:
 
!Step 2: &nbsp;
 
!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:
 
|-
 
|-
−
|<math>\int \frac{1}{3u} du = \frac{\log(u)}{3}</math>
+
|
 +
::<math>\frac{1}{3}\int\frac{1}{u}\,du\,=\, \frac{\log(u)}{3}.</math>
 
|}
 
|}
  
Line 52: Line 39:
 
!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
|-
 
|-
−
| Now we need to substitute back into our original variables using our original substitution <math>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 get <math>\frac{\log(u)}{3} = \frac{\log(3x + 2)}{3}</math>
+
|
 +
::<math>\frac{\log(u)}{3} = \frac{\log(3x + 2)}{3}.</math>
 
|}
 
|}
  
Line 60: Line 48:
 
!Step 4: &nbsp;
 
!Step 4: &nbsp;
 
|-
 
|-
−
| Since this integral is an indefinite integral we have to remember to add ''' "+ C" ''' 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 66: Line 54:
 
!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
|<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>''']]
 
[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 06:47, 16 May 2015

Find the antiderivative of


Foundations:  
This problem requires two rules of integration. In particular, you need
Integration by substitution (u - sub): If   is a differentiable functions whose range is in the domain of , then
We also need the derivative of the natural log since we will recover natural log from integration:

 Solution:

Step 1:  
Use a u-substitution with This means , or . After substitution we have
Step 2:  
We can now take the integral remembering the special rule resulting in natural log:
Step 3:  
Now we need to substitute back into our original variables using our original substitution to find
Step 4:  
Since this integral is an indefinite integral, we have to remember to add a constant  at the end.
Final Answer:  

Return to Sample Exam

Retrieved from "https://wiki.math.ucr.edu/index.php?title=022_Exam_2_Sample_A,_Problem_3&oldid=569"