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

From Math Wiki
Jump to navigation Jump to search
(Created page with "<span class="exam"> Find the antiderivative of <math style="vertical-align: -50%">\int \frac{1}{3x+2}\,dx.</math> {| class="mw-collapsible mw-collapsed" style = "text-align:...")
 
Line 51: Line 51:
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!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>
 +
|-
 +
| to get <math>\frac{\log(u)}{3} = \frac{\log(3x + 2}{3}</math>
 +
|}
 +
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!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 C at the end.
Line 58: Line 66:
 
!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
|<math>\int \frac{1}{3x + 2} dx = \frac{\ln(3x + 2)}{3}</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>''']]

Revision as of 08:44, 15 May 2015

Find the antiderivative of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int {\frac {1}{3x+2}}\,dx.}


Foundations:  
This problem requires two rules of integration. In particular, you need
Integration by substitution (U - sub): If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are differentiable functions, then

     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (f\circ g)'(x) = f'(g(x))\cdot g'(x).}

The Product Rule: If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are differentiable functions, then

     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).}

The Quotient Rule: If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are differentiable functions and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x) \neq 0}  , then

     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. }
Additionally, we will need our power rule for differentiation:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(x^n\right)'\,=\,nx^{n-1},} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n\neq 0} ,
as well as the derivative of natural log:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\ln x\right)'\,=\,\frac{1}{x}.}

 Solution:

Step 1:  
Use a U-substitution with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = 3x + 2.} This means Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du = 3 dx} , and after substitution we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{1}{3x + 2} = \int \frac{1}{3u} du}
Step 2:  
We can now take the integral remembering the special rule:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{1}{3u} du = \frac{\log(u)}{3}}
Step 3:  
Now we need to substitute back into our original variables using our original substitution Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = 3x + 2}
to get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\log(u)}{3} = \frac{\log(3x + 2}{3}}
Step 4:  
Since this integral is an indefinite integral we have to remember to add C at the end.
Final Answer:  
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{1}{3x + 2} dx = \frac{\ln(3x + 2)}{3} + C}

Return to Sample Exam