Difference between revisions of "009B Sample Midterm 1, Problem 3"

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
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|Integration by parts tells us that <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math>
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|'''1.''' Integration by parts tells us that <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math>
 
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|How would you integrate <math style="vertical-align: -12px">\int x\ln x~dx?</math>
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|'''2.''' How would you integrate <math style="vertical-align: -12px">\int x\ln x~dx?</math>
 
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|-
 
|
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|-
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|We proceed using integration by parts. Let <math style="vertical-align: 0px">u=x^2</math> and <math style="vertical-align: 0px">dv=e^xdx</math>. Then, <math style="vertical-align: 0px">du=2xdx</math> and <math style="vertical-align: 0px">v=e^x</math>.
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|We proceed using integration by parts. Let <math style="vertical-align: 0px">u=x^2</math> and <math style="vertical-align: 0px">dv=e^xdx.</math> Then, <math style="vertical-align: 0px">du=2xdx</math> and <math style="vertical-align: 0px">v=e^x.</math>
 
|-
 
|-
 
|Therefore, we have
 
|Therefore, we have
 
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|-
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| &nbsp;&nbsp; <math style="vertical-align: -12px">\int x^2 e^x~dx=x^2e^x-\int 2xe^x~dx</math>.
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|
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::<math style="vertical-align: -12px">\int x^2 e^x~dx=x^2e^x-\int 2xe^x~dx.</math>
 
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|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|Now, we need to use integration by parts again. Let <math style="vertical-align: 0px">u=2x</math> and <math style="vertical-align: 0px">dv=e^xdx</math>. Then, <math style="vertical-align: 0px">du=2dx</math> and <math style="vertical-align: 0px">v=e^x</math>.
+
|Now, we need to use integration by parts again. Let <math style="vertical-align: 0px">u=2x</math> and <math style="vertical-align: 0px">dv=e^xdx.</math> Then, <math style="vertical-align: 0px">du=2dx</math> and <math style="vertical-align: 0px">v=e^x.</math>
 
|-
 
|-
 
|Building on the previous step, we have
 
|Building on the previous step, we have
 
|-
 
|-
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| &nbsp;&nbsp; <math style="vertical-align: -15px">\int x^2 e^x~dx=x^2e^x-\bigg(2xe^x-\int 2e^x~dx\bigg)=x^2e^x-2xe^x+2e^x+C</math>.
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|
 +
::<math>\begin{array}{rcl}
 +
\displaystyle{\int x^2 e^x~dx} & = & \displaystyle{x^2e^x-\bigg(2xe^x-\int 2e^x~dx\bigg)}\\
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&&\\
 +
& = & \displaystyle{x^2e^x-2xe^x+2e^x+C.}\\
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\end{array}</math>
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|We proceed using integration by parts. Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x^3dx</math>. Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -14px">v=\frac{x^4}{4}</math>.
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|We proceed using integration by parts. Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x^3dx.</math> Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -14px">v=\frac{x^4}{4}.</math>
 
|-
 
|-
 
|Therefore, we have
 
|Therefore, we have
 
|-
 
|-
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| &nbsp;&nbsp; <math style="vertical-align: -20px">\int_{1}^{e} x^3\ln x~dx=\left.\ln x \bigg(\frac{x^4}{4}\bigg)\right|_{1}^{e}-\int_1^e \frac{x^3}{4}~dx=\left.\ln x \bigg(\frac{x^4}{4}\bigg)-\frac{x^4}{16}\right|_{1}^{e}</math>.
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|  
−
|-
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::<math>\begin{array}{rcl}
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|
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\displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\left.\ln x \bigg(\frac{x^4}{4}\bigg)\right|_{1}^{e}-\int_1^e \frac{x^3}{4}~dx }\\
 +
&&\\
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& = & \displaystyle{\left.\ln x \bigg(\frac{x^4}{4}\bigg)-\frac{x^4}{16}\right|_{1}^{e}.}\\
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\end{array}</math>
 
|}
 
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|Now, we evaluate to get  
 
|Now, we evaluate to get  
 
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|-
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| &nbsp;&nbsp; <math style="vertical-align: -17px">\int_{1}^{e} x^3\ln x~dx=\bigg((\ln e) \frac{e^4}{4}-\frac{e^4}{16}\bigg)-\bigg((\ln 1) \frac{1^4}{4}-\frac{1^4}{16}\bigg)=\frac{e^4}{4}-\frac{e^4}{16}+\frac{1}{16}=\frac{3e^4+1}{16}</math>.
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|  
−
|-
+
::<math>\begin{array}{rcl}
−
|
+
\displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\bigg((\ln e) \frac{e^4}{4}-\frac{e^4}{16}\bigg)-\bigg((\ln 1) \frac{1^4}{4}-\frac{1^4}{16}\bigg)}\\
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|-
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&&\\
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|
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& = & \displaystyle{\frac{e^4}{4}-\frac{e^4}{16}+\frac{1}{16}}\\
 +
&&\\
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& = & \displaystyle{\frac{3e^4+1}{16}.}\\
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\end{array}</math>
 
|}
 
|}
  

Revision as of 13:07, 18 April 2016

Evaluate the indefinite and definite integrals.

a) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int x^{2}e^{x}~dx}
b) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{1}^{e}x^{3}\ln x~dx}


Foundations:  
1. Integration by parts tells us that 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 u~dv=uv-\int v~du.}
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 x\ln x~dx?}
You could use integration by parts.
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} 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 dv=x~dx.} 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} 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 v=\frac{x^2}{2}.}
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 \int x\ln x~dx\,=\,\frac{x^2\ln x}{2}-\int \frac{x}{2}~dx\,=\,\frac{x^2\ln x}{2}-\frac{x^2}{4}+C.}

Solution:

(a)

Step 1:  
We proceed using integration by parts. 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=x^2} 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 dv=e^xdx.} 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=2xdx} 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 v=e^x.}
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 \int x^2 e^x~dx=x^2e^x-\int 2xe^x~dx.}
Step 2:  
Now, we need to use integration by parts again. 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} 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 dv=e^xdx.} 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 v=e^x.}
Building on the previous step, 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 x^2 e^x~dx} & = & \displaystyle{x^2e^x-\bigg(2xe^x-\int 2e^x~dx\bigg)}\\ &&\\ & = & \displaystyle{x^2e^x-2xe^x+2e^x+C.}\\ \end{array}}

(b)

Step 1:  
We proceed using integration by parts. 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} 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 dv=x^3dx.} 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} 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 v=\frac{x^4}{4}.}
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_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\left.\ln x \bigg(\frac{x^4}{4}\bigg)\right|_{1}^{e}-\int_1^e \frac{x^3}{4}~dx }\\ &&\\ & = & \displaystyle{\left.\ln x \bigg(\frac{x^4}{4}\bigg)-\frac{x^4}{16}\right|_{1}^{e}.}\\ \end{array}}
Step 2:  
Now, we evaluate 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 \begin{array}{rcl} \displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\bigg((\ln e) \frac{e^4}{4}-\frac{e^4}{16}\bigg)-\bigg((\ln 1) \frac{1^4}{4}-\frac{1^4}{16}\bigg)}\\ &&\\ & = & \displaystyle{\frac{e^4}{4}-\frac{e^4}{16}+\frac{1}{16}}\\ &&\\ & = & \displaystyle{\frac{3e^4+1}{16}.}\\ \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 x^2e^x-2xe^x+2e^x+C}
(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{3e^4+1}{16}}

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