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%">u = g(x)</math> is a differentiable functions whose range is in the domain of <math style="vertical-align: -20%">f</math>, then
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|'''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
 
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::<math>\left(ln(x)\right)' \,=\, \frac{1}{x}</math>
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::<math>\left(ln(x)\right)' \,=\, \frac{1}{x}.</math>
 
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|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
 
|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
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::<math>\int \frac{1}{3x + 2} \,=\, \int \frac{1}{u}\,\frac{du}{3}\,=\,\frac{1}{3}\int\frac{1}{u}\,du.</math>
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::<math>\int \frac{1}{3x + 2}\,dx \,=\, \int \frac{1}{u}\,\frac{du}{3}\,=\,\frac{1}{3}\int\frac{1}{u}\,du.</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|We can now take the integral remembering the special rule:
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|We can now take the integral remembering the special rule resulting in natural log:
 
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!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
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| Now we need to substitute back into our original variables using our original substitution <math style="vertical-align: -5%">u = 3x + 2</math>
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| 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
 
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| to find&nbsp; <math>\frac{\log(u)}{3} = \frac{\log(3x + 2)}{3}.</math>
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::<math>\frac{\log(u)}{3} = \frac{\log(3x + 2)}{3}.</math>
 
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!Step 4: &nbsp;
 
!Step 4: &nbsp;
 
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|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.
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|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.
 
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Latest revision as of 07:47, 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 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} , or 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 dx=du/3} . 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}\,dx \,=\, \int \frac{1}{u}\,\frac{du}{3}\,=\,\frac{1}{3}\int\frac{1}{u}\,du.}
Step 2:  
We can now take the integral remembering the special rule resulting in 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 \frac{1}{3}\int\frac{1}{u}\,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 find
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 a constant  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 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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