Difference between revisions of "009B Sample Final 3, Problem 6"
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!Step 1: | !Step 1: | ||
|- | |- | ||
| − | | | + | |We begin by using <math>u</math>-substitution. |
| + | |- | ||
| + | |Let <math>u=\sqrt{x+1}.</math> | ||
| + | |- | ||
| + | |Then, <math>u^2=x+1</math> and <math>x=u^2-1.</math> | ||
| + | |- | ||
| + | |Also, we have | ||
| + | |- | ||
| + | | <math>\begin{array}{rcl} | ||
| + | \displaystyle{du} & = & \displaystyle{\frac{1}{2} (x+1)^{\frac{-1}{2}}dx}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\frac{1}{2\sqrt{x+1}}dx}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\frac{1}{2u}dx.} | ||
| + | \end{array}</math> | ||
| + | |- | ||
| + | |Hence, <math>dx=2udu</math>. | ||
| + | |- | ||
| + | |Using all this information, we get | ||
|- | |- | ||
| | | | ||
Revision as of 15:59, 2 March 2017
Find the following integrals
(a)
(b)
| Foundations: |
|---|
| Through partial fraction decomposition, we can write the fraction |
| for some constants |
Solution:
(a)
| Step 1: |
|---|
| First, we factor the denominator to get |
| We use the method of partial fraction decomposition. |
| We let |
| If we multiply both sides of this equation by we get |
| Step 2: |
|---|
| Now, if we let we get |
| If we let we get |
| Therefore, |
| Step 3: |
|---|
| Therefore, 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 \begin{array}{rcl} \displaystyle{\int \frac{3x-1}{2x^2-x}~dx} & = & \displaystyle{\int \frac{1}{x}+\frac{1}{2x-1}~dx}\\ &&\\ & = & \displaystyle{\int \frac{1}{x}~dx+\int \frac{1}{2x-1}~dx}\\ &&\\ & = & \displaystyle{\ln |x|+\int \frac{1}{2x-1}~dx.} \end{array}} |
| Now, we use 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} -substitution. |
| Let 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=2x-1.} |
| 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 du=2dx} 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 \frac{du}{2}=dx.} |
| Hence, 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 \begin{array}{rcl} \displaystyle{\int \frac{3x-1}{2x^2-x}~dx} & = & \displaystyle{\ln |x|+\frac{1}{2}\int \frac{1}{u}~du}\\ &&\\ & = & \displaystyle{\ln |x|+\frac{1}{2}\ln |u|+C}\\ &&\\ & = & \displaystyle{\ln |x|+\frac{1}{2}\ln |2x-1|+C.} \end{array}} |
(b)
| Step 1: |
|---|
| We begin by using 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} -substitution. |
| Let 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=\sqrt{x+1}.} |
| 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 u^2=x+1} 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 x=u^2-1.} |
| Also, 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 \begin{array}{rcl} \displaystyle{du} & = & \displaystyle{\frac{1}{2} (x+1)^{\frac{-1}{2}}dx}\\ &&\\ & = & \displaystyle{\frac{1}{2\sqrt{x+1}}dx}\\ &&\\ & = & \displaystyle{\frac{1}{2u}dx.} \end{array}} |
| Hence, 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=2udu} . |
| Using all this information, we get |
| Step 2: |
|---|
| Final Answer: |
|---|
| (a) 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 \ln |x|+\frac{1}{2}\ln |2x-1|+C} |
| (b) |