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

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!Step 1:    
 
!Step 1:    
 
|-
 
|-
|First, we write <math>\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \int_0^a xe^{-x}~dx</math>.
+
|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>.
 
|-
 
|-
|Now, we proceed using integration by parts. Let <math>u=x</math> and <math>dv=e^{-x}dx</math>. Then, <math>du=dx</math> and <math>v=-e^{-x}</math>.
+
|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>.
 
|-
 
|-
 
|Thus, the integral becomes
 
|Thus, the integral becomes
 
|-
 
|-
|<math>\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \left.-xe^{-x}\right|_0^a-\int_0^a-e^{-x}dx</math>
+
|
 +
::<math>\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \left.-xe^{-x}\right|_0^a-\int_0^a-e^{-x}dx</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|For the remaining integral, we need to use <math>u</math>-substitution. Let <math>u=-x</math>. Then, <math>du=-dx</math>.  
+
|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>.  
 
|-
 
|-
 
|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.
 
|-
 
|-
|Plugging in our values into the equation <math>u=-x</math>, we get <math>u_1=0</math> and <math>u_2=-a</math>.
+
|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>.
 
|-
 
|-
 
|Thus, the integral becomes
 
|Thus, the integral becomes
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|
 
|
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
\displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{}\\
+
\displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-1}{e^a}+1}\\
&&\\
 
& = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-1}{e^a}+1}\\
 
 
&&\\
 
&&\\
 
& = & \displaystyle{0+1}\\
 
& = & \displaystyle{0+1}\\

Revision as of 10:22, 22 February 2016

Evaluate the improper integrals:

a)
b)
Foundations:  
Review integration by parts

Solution:

(a)

Step 1:  
First, we write .
Now, we proceed using integration by parts. Let and . Then, and .
Thus, the integral becomes
Step 2:  
For the remaining integral, we need to use -substitution. Let . Then, .
Since the integral is a definite integral, we need to change the bounds of integration.
Plugging in our values into the equation , we get and .
Thus, the integral becomes
Step 3:  
Now, we evaluate to get
Using L'Hopital's Rule, we get

(b)

Step 1:  
First, we write .
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{-du}{\sqrt{u}}} .
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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