Difference between revisions of "022 Exam 2 Sample A, Problem 3"
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| − | ::<math>\left(ln(x)\right)' \,=\, \frac{1}{x} | + | ::<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 | ||
| − | ::<math>\int \frac{1}{3x + 2} | + | ::<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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|We can now take the integral remembering the special rule: | |We can now take the integral remembering the special rule: | ||
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| − | | | + | |<math>\frac{1}{3}\int\frac{1}{u}\,du. \,=\, \frac{\log(u)}{3}.</math> |
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!Step 3: | !Step 3: | ||
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| − | | Now we need to substitute back into our original variables using our original substitution <math style="vertical-align: - | + | | Now we need to substitute back into our original variables using our original substitution <math style="vertical-align: -10%">u = 3x + 2</math> |
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| to find  <math>\frac{\log(u)}{3} = \frac{\log(3x + 2)}{3}.</math> | | to find  <math>\frac{\log(u)}{3} = \frac{\log(3x + 2)}{3}.</math> | ||
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!Step 4: | !Step 4: | ||
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| − | | Since this integral is an indefinite integral we have to remember to add a constant  <math style="vertical-align: 0%">C</math> at the end. | + | |Since this integral is an indefinite integral we have to remember to add a constant  <math style="vertical-align: 0%">C</math> at the end. |
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| − | ::<math>\int \frac{1}{3x + 2} | + | ::<math>\int \frac{1}{3x + 2} dx \,=\, \frac{\ln(3x + 2)}{3} + C.</math> |
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[[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 15:17, 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: |
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| 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 |
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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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Solution:
| Step 1: |
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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
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| Step 2: |
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| 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 \frac{1}{3}\int\frac{1}{u}\,du. \,=\, \frac{\log(u)}{3}.} |
| Step 3: |
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| 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: |
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| 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: |
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|