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

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|This problem requires two rules of integration.  In particular, you need
 
|This problem requires two rules of integration.  In particular, you need
 
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|'''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
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|'''Integration by substitution (U - sub):''' If <math>u = g(x)</math> is a differentiable functions whose range is in the domain of <math>f</math>, then
 
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|<math>\int g'(x)f(g(x)) dx = \int f(u) du.</math>
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|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math>
 
 
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|<br>'''The Product Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
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|We also need the derivative of the natural log since we will recover natural log from integration:
 
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|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</math>
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|<math>\left(ln(x)\right)' = \frac{1}{x}</math>
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|<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
 
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|<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>
 
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|Additionally, we will need our power rule for differentiation:
 
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::<math style="vertical-align: -21%;">\left(x^n\right)'\,=\,nx^{n-1},</math> for <math style="vertical-align: -25%;">n\neq 0</math>,
 
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|as well as the derivative of natural log:
 
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::<math>\left(\ln x\right)'\,=\,\frac{1}{x}.</math>
 
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|<br>
 
 
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Revision as of 10:02, 15 May 2015

Find the antiderivative of 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.}


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 u = g(x)} is a differentiable functions whose range is in the domain of 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} , 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 \int g'(x)f(g(x)) dx = \int f(u) du.}
We also need the derivative of the natural log since we will recover natural log from integration:
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}

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