Difference between revisions of "009B Sample Final 3, Problem 2"
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!Step 1: | !Step 1: | ||
|- | |- | ||
| − | | | + | |We use <math>u</math>-substitution. |
| + | |- | ||
| + | |Let <math style="vertical-align: -5px">u=\ln(x).</math> | ||
| + | |- | ||
| + | |Then, <math style="vertical-align: -5px">du=\frac{1}{x}dx.</math> | ||
| + | |- | ||
| + | |Also, we need to change the bounds of integration. | ||
| + | |- | ||
| + | |Plugging in our values into the equation <math style="vertical-align: -5px">u=\ln(x),</math> | ||
| + | |- | ||
| + | |we get <math style="vertical-align: -15px">u_1=\ln(1)=0</math> and <math style="vertical-align: -16px">u_2=\ln(e)=1.</math> | ||
| + | |- | ||
| + | |Therefore, the integral becomes | ||
| + | |- | ||
| + | | <math style="vertical-align: -19px">\int_0^2 \cos(u)~du.</math> | ||
|- | |- | ||
| | | | ||
| Line 131: | Line 145: | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 2: | !Step 2: | ||
| + | |- | ||
| + | |We now have | ||
|- | |- | ||
| | | | ||
| + | <math>\begin{array}{rcl} | ||
| + | \displaystyle{\int_1^e \frac{\cos(\ln(x))}{x}~dx} & = & \displaystyle{\int_0^2 \cos(u)~du}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\sin(u)\bigg|_0^1}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\sin(1)-\sin(0)}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\sin(1).} | ||
| + | \end{array}</math> | ||
|- | |- | ||
| | | | ||
| Line 145: | Line 170: | ||
| '''(b)''' <math>-\frac{1}{3(1+x^3)}+C</math> | | '''(b)''' <math>-\frac{1}{3(1+x^3)}+C</math> | ||
|- | |- | ||
| − | | '''(c)''' | + | | '''(c)''' <math>\sin(1)</math> |
|} | |} | ||
[[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 15:33, 1 March 2017
Evaluate the following integrals.
(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 \int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx}
(b) 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{x^2}{(1+x^3)^2}~dx}
(c) 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_1^e \frac{\cos(\ln(x))}{x}~dx}
| Foundations: |
|---|
| 1. |
| 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}{1+x^2}~dx=\arctan(x)+C} |
| 2. How would you integrate 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{\ln x}{x}~dx?} |
|
You could use -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=\ln(x).} |
| 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=\frac{1}{x}dx.} |
|
Thus, |
|
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{\ln x}{x}~dx} & = & \displaystyle{\int u~du}\\ &&\\ & = & \displaystyle{\frac{u^2}{2}+C}\\ &&\\ & = & \displaystyle{\frac{(\ln x)^2}{2}+C.} \end{array}} |
Solution:
(a)
| Step 1: |
|---|
| First, we notice |
| 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_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx=\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+(4x)^2}~dx.} |
| 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=4x.} |
| 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=4dx} 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}{4}=dx.} |
| Also, we need to change the bounds of integration. |
| Plugging in our values into the equation 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=4x,} |
| we 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 u_1=4(0)=0} 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 u_2=4\bigg(\frac{\sqrt{3}}{4}\bigg)=\sqrt{3}.} |
| Therefore, the integral becomes |
| 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}{4}\int_0^{\sqrt{3}} \frac{1}{1+u^2}~du.} |
| Step 2: |
|---|
| We now 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_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx} & = & \displaystyle{\frac{1}{4}\int_0^{\sqrt{3}} \frac{1}{1+u^2}~du}\\ &&\\ & = & \displaystyle{\frac{1}{4}\arctan(u)\bigg|_0^{\sqrt{3}}}\\ &&\\ & = & \displaystyle{\frac{1}{4}\arctan(\sqrt{3})-\frac{1}{4}\arctan(0)}\\ &&\\ & = & \displaystyle{\frac{1}{4}\bigg(\frac{\pi}{3}\bigg)-0}\\ &&\\ & = & \displaystyle{\frac{\pi}{12}.} \end{array}} |
(b)
| Step 1: |
|---|
| 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=1+x^3.} |
| 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=3x^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}{3}=x^2dx.} |
| Therefore, the integral becomes |
| 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^2}~du.} |
| Step 2: |
|---|
| We now 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{x^2}{(1+x^3)^2}~dx} & = & \displaystyle{\frac{1}{3}\int \frac{1}{u^2}~du}\\ &&\\ & = & \displaystyle{-\frac{1}{3u}+C}\\ &&\\ & = & \displaystyle{-\frac{1}{3(1+x^3)}+C.} \end{array}} |
(c)
| Step 1: |
|---|
| 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=\ln(x).} |
| 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=\frac{1}{x}dx.} |
| Also, we need to change the bounds of integration. |
| Plugging in our values into the equation 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=\ln(x),} |
| we 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 u_1=\ln(1)=0} 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 u_2=\ln(e)=1.} |
| Therefore, the integral becomes |
| 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_0^2 \cos(u)~du.} |
| Step 2: |
|---|
| We now 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_1^e \frac{\cos(\ln(x))}{x}~dx} & = & \displaystyle{\int_0^2 \cos(u)~du}\\ &&\\ & = & \displaystyle{\sin(u)\bigg|_0^1}\\ &&\\ & = & \displaystyle{\sin(1)-\sin(0)}\\ &&\\ & = & \displaystyle{\sin(1).} \end{array}} |
| 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 \frac{\pi}{12}} |
| (b) 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(1+x^3)}+C} |
| (c) 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 \sin(1)} |