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

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<span class="exam"> Evaluate the improper integrals:
 
<span class="exam"> Evaluate the improper integrals:
  
::<span class="exam">a) <math>\int_0^{\infty} xe^{-x}~dx</math>
+
<span class="exam">(a) &nbsp; <math>\int_0^{\infty} xe^{-x}~dx</math>
::<span class="exam">b) <math>\int_1^4 \frac{dx}{\sqrt{4-x}}</math>
+
 
 +
<span class="exam">(b) &nbsp; <math>\int_1^4 \frac{dx}{\sqrt{4-x}}</math>
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
|'''1.''' How could you write <math style="vertical-align: -14px">\int_0^{\infty} f(x)~dx</math> so that you can integrate?
+
|'''1.''' How could you write &nbsp; <math style="vertical-align: -14px">\int_0^{\infty} f(x)~dx</math> so that you can integrate?
 
|-
 
|-
 
|  
 
|  
::You can write <math>\int_0^{\infty} f(x)~dx=\lim_{a\rightarrow\infty} \int_0^a f(x)~dx.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; You can write &nbsp; <math>\int_0^{\infty} f(x)~dx=\lim_{a\rightarrow\infty} \int_0^a f(x)~dx.</math>
 
|-
 
|-
|'''2.''' How could you write <math>\int_{-1}^1 \frac{1}{x}~dx?</math>
+
|'''2.''' How could you write &nbsp; <math>\int_{-1}^1 \frac{1}{x}~dx?</math>
 
|-
 
|-
 
|
 
|
::The problem is that&thinsp; <math>\frac{1}{x}</math> &thinsp;is not continuous at <math style="vertical-align: 0px">x=0.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; The problem is that &nbsp;<math>\frac{1}{x}</math>&nbsp; is not continuous at &nbsp;<math style="vertical-align: 0px">x=0.</math>
 
|-
 
|-
 
|
 
|
::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>
+
&nbsp; &nbsp; &nbsp; &nbsp; So, you can write &nbsp;<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>
 
|-
 
|-
|'''3.''' How would you integrate <math style="vertical-align: -13px">\int xe^x\,dx?</math>
+
|'''3.''' How would you integrate &nbsp;<math style="vertical-align: -13px">\int xe^x\,dx?</math>
 
|-
 
|-
 
|
 
|
::You can use integration by parts.  
+
&nbsp; &nbsp; &nbsp; &nbsp;You can use integration by parts.  
 
|-
 
|-
 
|
 
|
::Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^xdx.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;Let &nbsp;<math style="vertical-align: 0px">u=x</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^xdx.</math>
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|First, we write  
+
|First, we write &nbsp;<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.
::<math style="vertical-align: -14px">\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \int_0^a xe^{-x}~dx.</math>
+
|-
 +
|Let &nbsp;<math style="vertical-align: 0px">u=x</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-x}dx.</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>
+
|Then, &nbsp;<math style="vertical-align: 0px">du=dx</math>&nbsp; and &nbsp;<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>
+
&nbsp; &nbsp; &nbsp; &nbsp;<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 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>  
+
|For the remaining integral, we need to use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.  
 +
|-
 +
|Let &nbsp;<math style="vertical-align: 0px">u=-x.</math>&nbsp; Then, &nbsp;<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 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>
+
|Plugging in our values into the equation &nbsp;<math style="vertical-align: -4px">u=-x,</math>&nbsp; we get
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">u_1=0</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">u_2=-a.</math>
 
|-
 
|-
 
|Thus, the integral becomes
 
|Thus, the integral becomes
 
|-
 
|-
 
|
 
|
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\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{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} -xe^{-x}\bigg|_0^a-\int_0^{-a}e^{u}~du}\\
 
&&\\
 
&&\\
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|-
 
|-
 
|
 
|
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} -ae^{-a}-(e^{-a}-1)}\\
 
\displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} -ae^{-a}-(e^{-a}-1)}\\
 
&&\\
 
&&\\
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|-
 
|-
 
|
 
|
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-1}{e^a}+1}\\
 
\displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-1}{e^a}+1}\\
 
&&\\
 
&&\\
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|First, we write  
+
|First, we write &nbsp;<math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_1^a\frac{dx}{\sqrt{4-x}}.</math>
 
|-
 
|-
|
+
|Now, we proceed by &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
::<math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_1^a\frac{dx}{\sqrt{4-x}}.</math>
 
 
|-
 
|-
|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>
+
|We let &nbsp;<math style="vertical-align: -1px">u=4-x.</math>&nbsp; Then, &nbsp;<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 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>
+
|Plugging in our values into the equation &nbsp;<math style="vertical-align: -4px">u=4-x,</math>&nbsp; we get  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">u_1=4-1=3</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">u_2=4-a.</math>
 
|-
 
|-
 
|Thus, the integral becomes
 
|Thus, the integral becomes
 
|-
 
|-
 
|
 
|
::<math>\int_1^4 \frac{dx}{\sqrt{4-x}}\,=\,\lim_{a\rightarrow 4} \int_3^{4-a}\frac{-1}{\sqrt{u}}~du.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\int_1^4 \frac{dx}{\sqrt{4-x}}\,=\,\lim_{a\rightarrow 4} \int_3^{4-a}\frac{-1}{\sqrt{u}}~du.</math>
 
|}
 
|}
  
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|-
 
|-
 
|
 
|
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\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{\int_1^4 \frac{dx}{\sqrt{4-x}}} & = & \displaystyle{\lim_{a\rightarrow 4} -2u^{\frac{1}{2}}\bigg|_{3}^{4-a}}\\
 
&&\\
 
&&\\
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\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|&nbsp;&nbsp; '''(a)''' &nbsp;<math style="vertical-align: -3px">1</math>
+
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;<math style="vertical-align: -3px">1</math>
 
|-
 
|-
|&nbsp;&nbsp; '''(b)''' &nbsp;<math style="vertical-align: -4px">2\sqrt{3}</math>
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;<math style="vertical-align: -4px">2\sqrt{3}</math>
 
|}
 
|}
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 08:45, 10 April 2017

Evaluate the improper integrals:

(a)  

(b)  

Foundations:  
1. How could you write   so that you can integrate?

        You can write  

2. How could you write  

        The problem is that    is not continuous at  

        So, you can write  

3. How would you integrate  

       You can use integration by parts.

       Let    and  


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  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
         and  
Thus, the integral becomes

       

Step 3:  
Now, we evaluate to get

       

Using L'Hôpital's Rule, we get

       

(b)

Step 1:  
First, we write  
Now, we proceed by  -substitution.
We 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 2:  
We integrate to get

       


Final Answer:  
    (a)    
    (b)    

Return to Sample Exam