Difference between revisions of "009B Sample Final 1, Problem 6"

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(Created page with "<span class="exam"> Evaluate the improper integrals: ::<span class="exam">a) <math>\int_0^{\infty} xe^{-x}~dx</math> ::<span class="exam">b) <math>\int_1^4 \frac{dx}{\sqrt{4-...")
 
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::You can write <math>\int_0^{\infty} f(x)~dx=\lim_{a\rightarrow\infty} \int_0^a f(x)~dx.</math>
 
::You can write <math>\int_0^{\infty} f(x)~dx=\lim_{a\rightarrow\infty} \int_0^a f(x)~dx.</math>
 
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|'''2.''' How could you write <math>\int_{-1}^1 \frac{1}{x}~dx</math> ?
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|'''2.''' How could you write <math>\int_{-1}^1 \frac{1}{x}~dx?</math>
 
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::The problem is that&thinsp; <math>\frac{1}{x}</math> &thinsp;is not continuous at <math style="vertical-align: 0px">x=0</math>.
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::The problem is that&thinsp; <math>\frac{1}{x}</math> &thinsp;is not continuous at <math style="vertical-align: 0px">x=0.</math>
 
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::So, you can write <math style="vertical-align: -15px">\int_{-1}^1 \frac{1}{x}~dx=\lim_{a\rightarrow 0^-} \int_{-1}^a \frac{1}{x}~dx+\lim_{a\rightarrow 0^+} \int_a^1 \frac{1}{x}~dx</math>.
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::So, you can write <math style="vertical-align: -15px">\int_{-1}^1 \frac{1}{x}~dx=\lim_{a\rightarrow 0^-} \int_{-1}^a \frac{1}{x}~dx+\lim_{a\rightarrow 0^+} \int_a^1 \frac{1}{x}~dx.</math>
 
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|'''3.''' How would you integrate <math style="vertical-align: -13px">\int xe^x\,dx</math> ?
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|'''3.''' How would you integrate <math style="vertical-align: -13px">\int xe^x\,dx?</math>
 
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::Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^xdx</math>.
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::Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^xdx.</math>
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|First, we write <math style="vertical-align: -14px">\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \int_0^a xe^{-x}~dx</math>.
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|First, we write  
 
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|Now, we proceed using integration by parts. Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^{-x}dx</math>. Then, <math style="vertical-align: 0px">du=dx</math> and <math style="vertical-align: 0px">v=-e^{-x}</math>.
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::<math style="vertical-align: -14px">\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \int_0^a xe^{-x}~dx.</math>
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 +
|Now, we proceed using integration by parts. Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^{-x}dx.</math> Then, <math style="vertical-align: 0px">du=dx</math> and <math style="vertical-align: 0px">v=-e^{-x}.</math>
 
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|Thus, the integral becomes
 
|Thus, the integral becomes
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|For the remaining integral, we need to use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: 0px">u=-x</math>. Then, <math style="vertical-align: 0px">du=-dx</math>.
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|For the remaining integral, we need to use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: 0px">u=-x.</math> Then, <math style="vertical-align: 0px">du=-dx.</math>  
 
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|Since the integral is a definite integral, we need to change the bounds of integration.
 
|Since the integral is a definite integral, we need to change the bounds of integration.
 
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|Plugging in our values into the equation <math style="vertical-align: 0px">u=-x</math>, we get <math style="vertical-align: -5px">u_1=0</math> and <math style="vertical-align: -3px">u_2=-a</math>.
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|Plugging in our values into the equation <math style="vertical-align: -4px">u=-x,</math> we get <math style="vertical-align: -5px">u_1=0</math> and <math style="vertical-align: -3px">u_2=-a.</math>
 
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|Thus, the integral becomes
 
|Thus, the integral becomes
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|First, we write <math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_1^a\frac{dx}{\sqrt{4-x}}</math>.
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|First, we write  
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::<math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_1^a\frac{dx}{\sqrt{4-x}}.</math>
 
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|Now, we proceed by <math style="vertical-align: 0px">u</math>-substitution. We let <math style="vertical-align: -1px">u=4-x</math>. Then, <math style="vertical-align: 0px">du=-dx</math>.
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|Now, we proceed by <math style="vertical-align: 0px">u</math>-substitution. We let <math style="vertical-align: -1px">u=4-x.</math> Then, <math style="vertical-align: 0px">du=-dx.</math>
 
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|-
 
|Since the integral is a definite integral, we need to change the bounds of integration.  
 
|Since the integral is a definite integral, we need to change the bounds of integration.  
 
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|-
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|Plugging in our values into the equation <math style="vertical-align: -1px">u=4-x</math>, we get <math style="vertical-align: -5px">u_1=4-1=3</math>&thinsp; and <math style="vertical-align: -3px">u_2=4-a</math>.
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|Plugging in our values into the equation <math style="vertical-align: -4px">u=4-x,</math> we get <math style="vertical-align: -5px">u_1=4-1=3</math>&thinsp; and <math style="vertical-align: -3px">u_2=4-a.</math>
 
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|Thus, the integral becomes
 
|Thus, the integral becomes

Revision as of 12:12, 18 April 2016

Evaluate the improper integrals:

a)
b)
Foundations:  
1. How could you write so that you can integrate?
You can 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_0^{\infty} f(x)~dx=\lim_{a\rightarrow\infty} \int_0^a f(x)~dx.}
2. How could you 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_{-1}^1 \frac{1}{x}~dx?}
The problem is 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 \frac{1}{x}}  is not continuous at 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.}
So, you can 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_{-1}^1 \frac{1}{x}~dx=\lim_{a\rightarrow 0^-} \int_{-1}^a \frac{1}{x}~dx+\lim_{a\rightarrow 0^+} \int_a^1 \frac{1}{x}~dx.}
3. 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 xe^x\,dx?}
You can 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.}

Solution:

(a)

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_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \int_0^a xe^{-x}~dx.}
Now, 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} 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^{-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=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}.}
Thus, 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^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \left.-xe^{-x}\right|_0^a-\int_0^a-e^{-x}\,dx.}
Step 2:  
For the remaining integral, we need to 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=-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=-dx.}
Since the integral is a definite integral, 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=-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=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=-a.}
Thus, 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 \begin{array}{rcl} \displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} -xe^{-x}\bigg|_0^a-\int_0^{-a}e^{u}~du}\\ &&\\ & = & \displaystyle{\lim_{a\rightarrow \infty} -xe^{-x}\bigg|_0^a-e^{u}\bigg|_0^{-a}}\\ &&\\ & = & \displaystyle{\lim_{a\rightarrow \infty} -ae^{-a}-(e^{-a}-1)}.\\ \end{array}}
Step 3:  
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_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} -ae^{-a}-(e^{-a}-1)}\\ &&\\ & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-a}{e^a}-\frac{1}{e^a}+1}\\ &&\\ & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-a-1}{e^a}+1}.\\ \end{array}}
Using L'Hôpital's Rule, 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 \begin{array}{rcl} \displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-1}{e^a}+1}\\ &&\\ & = & \displaystyle{0+1}\\ &&\\ & = & \displaystyle{1}.\\ \end{array}}

(b)

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_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_1^a\frac{dx}{\sqrt{4-x}}.}
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. 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 u=4-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=-dx.}
Since the integral is a definite integral, 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=4-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=4-1=3}   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-a.}
Thus, 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_1^4 \frac{dx}{\sqrt{4-x}}\,=\,\lim_{a\rightarrow 4} \int_3^{4-a}\frac{-1}{\sqrt{u}}~du.}
Step 2:  
We integrate 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_1^4 \frac{dx}{\sqrt{4-x}}} & = & \displaystyle{\lim_{a\rightarrow 4} -2u^{\frac{1}{2}}\bigg|_{3}^{4-a}}\\ &&\\ & = & \displaystyle{\lim_{a\rightarrow 4}-2\sqrt{4-a}+2\sqrt{3}}\\ &&\\ & = & \displaystyle{2\sqrt{3}}.\\ \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 1}
(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 2\sqrt{3}}

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