Difference between revisions of "009B Sample Final 1, Problem 4"

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& = & \displaystyle{\int \frac{2x^2+x}{2x^2+x}~dx+\int\frac{1-x}{2x^2+x}~dx}\\
 
& = & \displaystyle{\int \frac{2x^2+x}{2x^2+x}~dx+\int\frac{1-x}{2x^2+x}~dx}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\int ~dx+\int\frac{1-x}{2x^2+x}~dx}\\
+
& = & \displaystyle{\int ~dx+\int\frac{1-x}{2x^2+x}~dx}.\\
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
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|If we let <math style="vertical-align: 0px">x=0</math>, the last equation becomes <math style="vertical-align: -1px">1=A</math>.
 
|If we let <math style="vertical-align: 0px">x=0</math>, the last equation becomes <math style="vertical-align: -1px">1=A</math>.
 
|-
 
|-
−
|If we let <math style="vertical-align: -14px">x=-\frac{1}{2}</math>, then we get <math style="vertical-align: -14px">\frac{3}{2}=-\frac{1}{2}B</math>. Thus, <math style="vertical-align: 0px">B=-3</math>.
+
|If we let <math style="vertical-align: -14px">x=-\frac{1}{2}</math>, then we get &thinsp;<math style="vertical-align: -13px">\frac{3}{2}=-\frac{1}{2}\,B</math>. Thus, <math style="vertical-align: 0px">B=-3</math>.
 
|-
 
|-
−
|So, in summation, we have <math>\frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}</math>.
+
|So, in summation, we have&thinsp; <math>\frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}</math>.
 
|}
 
|}
  
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\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int ~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\
 
\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int ~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\
+
& = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}.\\
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
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|For the final remaining integral, we use <math style="vertical-align: 0px">u</math>-substitution.  
 
|For the final remaining integral, we use <math style="vertical-align: 0px">u</math>-substitution.  
 
|-
 
|-
−
|Let <math style="vertical-align: -2px">u=2x+1</math>. Then, <math style="vertical-align: 0px">du=2dx</math> and <math style="vertical-align: -14px">\frac{du}{2}=dx</math>.
+
|Let <math style="vertical-align: -2px">u=2x+1</math>. Then, <math style="vertical-align: 0px">du=2\,dx</math> and&thinsp; <math style="vertical-align: -14px">\frac{du}{2}=dx</math>.
 
|-
 
|-
 
|Thus, our final integral becomes
 
|Thus, our final integral becomes
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& = & \displaystyle{x+\ln x+\int\frac{-3}{2u}~du}\\
 
& = & \displaystyle{x+\ln x+\int\frac{-3}{2u}~du}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{x+\ln x-\frac{3}{2}\ln u +C}\\
+
& = & \displaystyle{x+\ln x-\frac{3}{2}\ln u +C}.\\
 
\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
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|-
 
|-
 
|
 
|
−
::<math>\int \frac{2x^2+1}{2x^2+x}~dx=x+\ln x-\frac{3}{2}\ln (2x+1) +C</math>
+
::<math>\int \frac{2x^2+1}{2x^2+x}~dx\,=\,x+\ln x-\frac{3}{2}\ln (2x+1) +C.</math>
 
|}
 
|}
 +
 
== 4 ==
 
== 4 ==
 
'''(c)'''
 
'''(c)'''

Revision as of 22:55, 25 February 2016

Compute 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 e^x(x+\sin(e^x))~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{2x^2+1}{2x^2+x}~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 \sin^3x~dx}

Foundations:  
Recall:
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. Through partial fraction decomposition, we can write the fraction  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}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}}  for some constants 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 A,B} .
3. We have the Pythagorean identity 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^2(x)=1-\cos^2(x)} .

Solution:

(a)

Step 1:  
We first distribute 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 \int e^x(x+\sin(e^x))~dx\,=\,\int e^xx~dx+\int e^x\sin(e^x)~dx.}
Now, for the first integral on the right hand side of the last equation, we 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=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=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=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=e^x} .
So, 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 e^x(x+\sin(e^x))~dx} & = & \displaystyle{\bigg(xe^x-\int e^x~dx \bigg)+\int e^x\sin(e^x)~dx}\\ &&\\ & = & \displaystyle{xe^x-e^x+\int e^x\sin(e^x)~dx}.\\ \end{array}}
Step 2:  
Now, for the one remaining integral, 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=e^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=e^xdx} .
So, 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 e^x(x+\sin(e^x))~dx} & = & \displaystyle{xe^x-e^x+\int \sin(u)~du}\\ &&\\ & = & \displaystyle{xe^x-e^x-\cos(u)+C}\\ &&\\ & = & \displaystyle{xe^x-e^x-\cos(e^x)+C}.\\ \end{array}}

3

(b)

Step 1:  
First, we add and subtract 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} from the numerator.
So, 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{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int\frac{2x^2+x-x+1}{2x^2+x}~dx}\\ &&\\ & = & \displaystyle{\int \frac{2x^2+x}{2x^2+x}~dx+\int\frac{1-x}{2x^2+x}~dx}\\ &&\\ & = & \displaystyle{\int ~dx+\int\frac{1-x}{2x^2+x}~dx}.\\ \end{array}}
Step 2:  
Now, we need to use partial fraction decomposition for the second integral.
Since 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 2x^2+x=x(2x+1)} , we 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 \frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}} .
Multiplying both sides of the last equation by 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(2x+1)} ,
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 1-x=A(2x+1)+Bx} .
If we 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 x=0} , the last equation 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 1=A} .
If we 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 x=-\frac{1}{2}} , then 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 \frac{3}{2}=-\frac{1}{2}\,B} . 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 B=-3} .
So, in summation, 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 \frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}} .
Step 3:  
If we plug in the last equation from Step 2 into our final integral in Step 1, 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{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int ~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\ &&\\ & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}.\\ \end{array}}
Step 4:  
For the final remaining integral, 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=2\,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 \frac{du}{2}=dx} .
Thus, our final 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 \begin{array}{rcl} \displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\ &&\\ & = & \displaystyle{x+\ln x+\int\frac{-3}{2u}~du}\\ &&\\ & = & \displaystyle{x+\ln x-\frac{3}{2}\ln u +C}.\\ \end{array}}
Therefore, the final answer is
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{2x^2+1}{2x^2+x}~dx\,=\,x+\ln x-\frac{3}{2}\ln (2x+1) +C.}

4

(c)

Step 1:  
First, we write 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\sin^3x~dx=\int \sin^2 x \sin x~dx} .
Using the identity 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^2x+\cos^2x=1} , 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 \sin^2x=1-\cos^2x} .
If we use this identity, 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\sin^3x~dx=\int (1-\cos^2x)\sin x~dx} .
Step 2:  
Now, we proceed by 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=\cos 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=-\sin x dx} .
So 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\sin^3x~dx} & = & \displaystyle{\int -(1-u^2)~du}\\ &&\\ & = & \displaystyle{-u+\frac{u^3}{3}+C}\\ &&\\ & = & \displaystyle{-\cos x+\frac{\cos^3x}{3}+C}\\ \end{array}}

5

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 xe^x-e^x-\cos(e^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 x+\ln x-\frac{3}{2}\ln (2x+1) +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 -\cos x+\frac{\cos^3x}{3}+C}

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