Difference between revisions of "022 Exam 2 Sample A, Problem 3"
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(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:...") |
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 3: | !Step 3: | ||
| + | |- | ||
| + | | 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: | ||
|- | |- | ||
| 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: | !Final Answer: | ||
|- | |- | ||
| − | |<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: | |
| |
| as well as the derivative of natural log: | |
|
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
|
| 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} |