009B Sample Final 2, Problem 7 - Revision history

Kayla Murray: Replaced content with "Evaluate the following integrals or show that they are divergent: (a) $\int_1^\infty \frac{\ln x}{x^4}~dx$
2017-12-03T00:33:02Z

Replaced content with "Evaluate the following integrals or show that they are divergent: (a) <math>\int_1^\infty \frac{\ln x}{x^4}~dx</math> (b)  <math> \int_0^1 \frac{3\ln x}{\sqrt{x}}~dx</math>

(b)  <math> \int_0^1 \frac{3\ln x}{\sqrt{x}}~dx</math>

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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<hr>

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!Foundations:  
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[[009B Sample Final 2, Problem 7 Solution|'''Solution''']]

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|'''1.''' How could you write   <math style="vertical-align: -14px">~~\int_0^{\infty} f(x)~dx</math> so that you can integrate?

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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>

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|'''2.''' How could you write   <math>\int_{0}^1 \frac{1}{x}~dx?</math>

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        The problem is that  <math>\frac{1}{x}</math>  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_{0}^1 \frac{1}{x}~dx=\lim_{a\rightarrow 0} \int_{a}^1 \frac{1}{x}~dx.</math>

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|}

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'''Solution:'''
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[[009B Sample Final 2, Problem 7 Detailed Solution|'''Detailed Solution''']]

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'''(a)'''

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

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!Step 1:  

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|-

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|First, we write

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|-

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|       <math>\int_1^\infty \frac{\ln x}{x^4}~dx=\lim_{a\rightarrow \infty} \int_1^a \frac{\ln x}{x^4}~dx.</math>

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|-

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|Now, we use integration by parts.

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|-

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|Let  <math style="vertical-align: -2px">u=\ln x</math>  and  <math style="vertical-align: -13px">dv=\frac{1}{x^4}dx.</math>

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|-

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|Then,  <math style="vertical-align: -13px">du=\frac{1}{x}dx</math>  and  <math style="vertical-align: -13px">v=\frac{1}{-3x^3}.</math>

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|-

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|Using integration by parts, we get

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|-

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|        <math>\begin{array}{rcl}

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\displaystyle{\int_1^\infty \frac{\ln x}{x^4}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{\ln x}{-3x^3}\bigg|_1^a+\int_1^a \frac{1}{3x^4}~dx}\\

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&&\\

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& = & \displaystyle{\lim_{a\rightarrow \infty} \frac{\ln x}{-3x^3}-\frac{1}{9x^3}\bigg|_1^a.}

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\end{array}</math>

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|}

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

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!Step 2:  

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|-

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|Now, using L'Hopital's Rule, we get

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|-

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|        <math>\begin{array}{rcl}

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\displaystyle{\int_1^\infty \frac{\ln x}{x^4}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{\ln a}{-3a^3}-\frac{1}{9a^3}-\bigg(\frac{\ln 1}{-3}-\frac{1}{9}\bigg)}\\

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&&\\

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& = & \displaystyle{\lim_{a\rightarrow \infty} \frac{\ln(a)}{-3a^3}+0+0+\frac{1}{9}}\\

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&&\\

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& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln(x)}{-3x^3}+\frac{1}{9}}\\

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&&\\

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& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{x}}{-9x^2}+\frac{1}{9}}\\

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&&\\

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& = & \displaystyle{\frac{1}{9}.}

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\end{array}</math>

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|}

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'''(b)'''

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

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!Step 1:  

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|-

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|First, we write

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|       <math>\int_0^1 \frac{3\ln x}{\sqrt{x}}~dx=\lim_{a\rightarrow 0^+} \int_a^1 \frac{3\ln x}{\sqrt{x}}~dx.</math>

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|-

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|Now, we use integration by parts.

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|-

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|Let  <math style="vertical-align: -2px">u=3\ln x</math>  and  <math style="vertical-align: -19px">dv=\frac{1}{\sqrt{x}}dx.</math>

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|-

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|Then,  <math style="vertical-align: -13px">du=\frac{3}{x}dx</math>  and  <math style="vertical-align: -5px">v=2\sqrt{x}.</math>

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|-

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|Using integration by parts, we get

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|-

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|        <math>\begin{array}{rcl}

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\displaystyle{\int_0^1 \frac{3\ln x}{\sqrt{x}}~dx} & = & \displaystyle{\lim_{a\rightarrow 0^+} (3\ln x)(2\sqrt{x})\bigg|_a^1-\int_a^1 \frac{6}{\sqrt{x}}~dx}\\

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&&\\

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& = & \displaystyle{\lim_{a\rightarrow 0^+} 6\sqrt{x}\ln(x)-12\sqrt{x}\bigg|_a^1.}

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\end{array}</math>

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|}

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

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!Step 2:  

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|-

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|Now, using L'Hopital's Rule, we get

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|-

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|        <math>\begin{array}{rcl}

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\displaystyle{\int_0^1 \frac{3\ln x}{\sqrt{x}}~dx} & = & \displaystyle{\lim_{a\rightarrow 0^+} (6\sqrt{1}\ln(1)-12\sqrt{1})-(6\sqrt{a}\ln(a)-12\sqrt{a})}\\

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&&\\

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& = & \displaystyle{\lim_{a\rightarrow 0^+} -12 -6\sqrt{a}\ln(a) +12\sqrt{a}}\\

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&&\\

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& = & \displaystyle{\lim_{a\rightarrow 0^+} -12 -6\sqrt{a}\ln(a)+0}\\

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&&\\

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& = & \displaystyle{\lim_{x\rightarrow 0^+} -12-6\sqrt{x}\ln(x)}\\

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&&\\

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& = & \displaystyle{-12-\lim_{x\rightarrow 0^+} \frac{6\ln(x)}{\frac{1}{\sqrt{x}}}}\\

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&&\\

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& \overset{L'H}{=} & \displaystyle{-12-\lim_{x\rightarrow 0^+} \frac{\frac{6}{x}}{-\frac{1}{2x^{3/2}}}}\\

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&&\\

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& = & \displaystyle{-12+\lim_{x\rightarrow 0^+} 12\sqrt{x}}\\

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&&\\

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& = & \displaystyle{-12.}

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\end{array}</math>

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

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!Final Answer:  

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|   '''(a)'''    <math>\frac{1}{9}</math>

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|   '''(b)'''    <math>-12</math>~~

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|}

[[009B_Sample_Final_2|'''Return to Sample Exam''']]

[[009B_Sample_Final_2|'''Return to Sample Exam''']]

Kayla Murray at 21:19, 20 May 2017

2017-05-20T21:19:39Z

Kayla Murray at 21:31, 3 March 2017

2017-03-03T21:31:08Z

Kayla Murray at 21:28, 3 March 2017

2017-03-03T21:28:56Z

Kayla Murray at 21:11, 3 March 2017

2017-03-03T21:11:38Z

Kayla Murray at 21:08, 3 March 2017

2017-03-03T21:08:08Z

Kayla Murray at 20:56, 3 March 2017

2017-03-03T20:56:49Z

Kayla Murray at 20:51, 3 March 2017

2017-03-03T20:51:02Z

Kayla Murray at 02:20, 19 February 2017

2017-02-19T02:20:03Z

Kayla Murray at 17:42, 18 February 2017

2017-02-18T17:42:29Z

@@ Line 1: / Line 1: @@
 <span class="exam">Evaluate the following integrals or show that they are divergent:
-::<span class="exam">a) <math>\int_1^\infty \frac{\ln x}{x^4}~dx</math>
+<span class="exam">(a) <math>\int_1^\infty \frac{\ln x}{x^4}~dx</math>
-::<span class="exam">b) <math> \int_0^1 \frac{3\ln x}{\sqrt{x}}~dx</math>
+<span class="exam">(b) <math> \int_0^1 \frac{3\ln x}{\sqrt{x}}~dx</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 16: / Line 16: @@
+|
 |}
 '''Solution:'''
@@ Line 80: / Line 81: @@
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 1: / Line 1: @@
-<span class="exam">Consider the region bounded by the following two functions:
+<span class="exam">Evaluate the following integrals or show that they are divergent:
-<span class="exam">a) Using the lower sum with three rectangles having equal width, approximate the area.
+::<span class="exam">a) <math>\int_1^\infty \frac{\ln x}{x^4}~dx</math>
-<span class="exam">b) Using the upper sum with three rectangles having equal width, approximate the area.
+::<span class="exam">b) <math> \int_0^1 \frac{3\ln x}{\sqrt{x}}~dx</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp;<math> \int_0^1 \frac{3\ln x}{\sqrt{x}}~dx</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009B Sample Final 2, Problem 7 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009B Sample Final 2, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 74: / Line 74: @@
 |First, we write
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\int_0^1 \frac{3\ln x}{\sqrt{x}}~dx=\lim_{a\rightarrow 0} \int_a^1 \frac{3\ln x}{\sqrt{x}}~dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\int_0^1 \frac{3\ln x}{\sqrt{x}}~dx=\lim_{a\rightarrow 0^+} \int_a^1 \frac{3\ln x}{\sqrt{x}}~dx.</math>
 |-
 |Now, we use integration by parts.
@@ Line 85: / Line 85: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int_0^1 \frac{3\ln x}{\sqrt{x}}~dx} & = & \displaystyle{\lim_{a\rightarrow 0} (3\ln x)(2\sqrt{x})\bigg|_a^1-\int_a^1 \frac{6}{\sqrt{x}}~dx}\\
+\displaystyle{\int_0^1 \frac{3\ln x}{\sqrt{x}}~dx} & = & \displaystyle{\lim_{a\rightarrow 0^+} (3\ln x)(2\sqrt{x})\bigg|_a^1-\int_a^1 \frac{6}{\sqrt{x}}~dx}\\
 &&\\
-& = & \displaystyle{\lim_{a\rightarrow 0} 6\sqrt{x}\ln(x)-12\sqrt{x}\bigg|_a^1.}
+& = & \displaystyle{\lim_{a\rightarrow 0^+} 6\sqrt{x}\ln(x)-12\sqrt{x}\bigg|_a^1.}
 \end{array}</math>
 |}
@@ Line 97: / Line 97: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int_0^1 \frac{3\ln x}{\sqrt{x}}~dx} & = & \displaystyle{\lim_{a\rightarrow 0} (6\sqrt{1}\ln(1)-12\sqrt{1})-(6\sqrt{a}\ln(a)-12\sqrt{a})}\\
+\displaystyle{\int_0^1 \frac{3\ln x}{\sqrt{x}}~dx} & = & \displaystyle{\lim_{a\rightarrow 0^+} (6\sqrt{1}\ln(1)-12\sqrt{1})-(6\sqrt{a}\ln(a)-12\sqrt{a})}\\
 &&\\
-& = & \displaystyle{\lim_{a\rightarrow 0} -12 -6\sqrt{a}\ln(a) +12\sqrt{a}}\\
+& = & \displaystyle{\lim_{a\rightarrow 0^+} -12 -6\sqrt{a}\ln(a) +12\sqrt{a}}\\
 &&\\
-& = & \displaystyle{\lim_{a\rightarrow 0} -12 -6\sqrt{a}\ln(a)+0}\\
+& = & \displaystyle{\lim_{a\rightarrow 0^+} -12 -6\sqrt{a}\ln(a)+0}\\
 &&\\
-& = & \displaystyle{\lim_{x\rightarrow 0} -12-6\sqrt{x}\ln(x)}\\
+& = & \displaystyle{\lim_{x\rightarrow 0^+} -12-6\sqrt{x}\ln(x)}\\
 &&\\
-& = & \displaystyle{-12-\lim_{x\rightarrow 0} \frac{6\ln(x)}{\frac{1}{\sqrt{x}}}}\\
+& = & \displaystyle{-12-\lim_{x\rightarrow 0^+} \frac{6\ln(x)}{\frac{1}{\sqrt{x}}}}\\
 &&\\
-& \overset{L'H}{=} & \displaystyle{-12-\lim_{x\rightarrow 0} \frac{\frac{6}{x}}{-\frac{1}{2x^{3/2}}}}\\
+& \overset{L'H}{=} & \displaystyle{-12-\lim_{x\rightarrow 0^+} \frac{\frac{6}{x}}{-\frac{1}{2x^{3/2}}}}\\
 &&\\
-& = & \displaystyle{-12+\lim_{x\rightarrow 0} 12\sqrt{x}}\\
+& = & \displaystyle{-12+\lim_{x\rightarrow 0^+} 12\sqrt{x}}\\
 &&\\
 & = & \displaystyle{-12.}

@@ Line 78: / Line 78: @@
 |Now, we use integration by parts.
 |-
-|Let &nbsp;<math style="vertical-align: -2px">u=3\ln x</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">dv=\frac{1}{\sqrt{x}}dx.</math>
+|Let &nbsp;<math style="vertical-align: -2px">u=3\ln x</math>&nbsp; and &nbsp;<math style="vertical-align: -19px">dv=\frac{1}{\sqrt{x}}dx.</math>
 |-
-|Then, &nbsp;<math style="vertical-align: -13px">du=\frac{3}{x}dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">v=2\sqrt{x}.</math>
+|Then, &nbsp;<math style="vertical-align: -13px">du=\frac{3}{x}dx</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">v=2\sqrt{x}.</math>
 |-
 |Using integration by parts, we get

@@ Line 72: / Line 72: @@
 !Step 1: &nbsp;
 |-
+|First, we write
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 80: / Line 94: @@
 !Step 2: &nbsp;
 |-
+|Now, using L'Hopital's Rule, we get
 |-
+|
@@ Line 91: / Line 123: @@
 |&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>\frac{1}{9}</math>
 |-
-|'''(b)'''
+|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>-12</math>
 |}
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 36: / Line 36: @@
 |Now, we use integration by parts.
 |-
-|Let <math>u=\ln x</math> and <math>dv=\frac{1}{x^4}dx.</math>
+|Let &nbsp;<math style="vertical-align: -2px">u=\ln x</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">dv=\frac{1}{x^4}dx.</math>
 |-
-|Then, <math>du=\frac{1}{x}dx</math> and <math>v=\frac{1}{-3x^3}.</math>
+|Then, &nbsp;<math style="vertical-align: -13px">du=\frac{1}{x}dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">v=\frac{1}{-3x^3}.</math>
 |-
 |Using integration by parts, we get

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
+|'''1.''' How could you write &nbsp; <math style="vertical-align: -14px">\int_0^{\infty} f(x)~dx</math> so that you can integrate?
 |-
+|'''2.''' How could you write &nbsp; <math>\int_{0}^1 \frac{1}{x}~dx?</math>
 |-
+|
 |-
+|
 |}