MathAdmin at 01:08, 21 May 2017

2017-05-21T01:08:56Z

MathAdmin: Created page with "Evaluate the following integrals or show that they are divergent: (a) $\int_1^\infty \frac{\ln x}{x^4}~dx$
2017-04-10T16:50:58Z

Created page with "Evaluate the following integrals or show that they are divergent: (a) <math>\int_1^\infty \frac{\ln x}{x^4}~dx</math> 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>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' How could you write   <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>
|-
|'''2.''' How could you write   <math>\int_{0}^1 \frac{1}{x}~dx?</math>
|-
|
        The problem is that  <math>\frac{1}{x}</math>  is not continuous at  <math style="vertical-align: 0px">x=0.</math>
|-
|
        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>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write
|-
|       <math>\int_1^\infty \frac{\ln x}{x^4}~dx=\lim_{a\rightarrow \infty} \int_1^a \frac{\ln x}{x^4}~dx.</math>
|-
|Now, we use integration by parts.
|-
|Let  <math style="vertical-align: -2px">u=\ln x</math>  and  <math style="vertical-align: -13px">dv=\frac{1}{x^4}dx.</math>
|-
|Then,  <math style="vertical-align: -13px">du=\frac{1}{x}dx</math>  and  <math style="vertical-align: -13px">v=\frac{1}{-3x^3}.</math>
|-
|Using integration by parts, we get
|-
|        <math>\begin{array}{rcl}
\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}\\
&&\\
& = & \displaystyle{\lim_{a\rightarrow \infty} \frac{\ln x}{-3x^3}-\frac{1}{9x^3}\bigg|_1^a.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using L'Hopital's Rule, we get
|-
|        <math>\begin{array}{rcl}
\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)}\\
&&\\
& = & \displaystyle{\lim_{a\rightarrow \infty} \frac{\ln(a)}{-3a^3}+0+0+\frac{1}{9}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln(x)}{-3x^3}+\frac{1}{9}}\\
&&\\
& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{x}}{-9x^2}+\frac{1}{9}}\\
&&\\
& = & \displaystyle{\frac{1}{9}.}
\end{array}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write
|-
|       <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.
|-
|Let  <math style="vertical-align: -2px">u=3\ln x</math>  and  <math style="vertical-align: -19px">dv=\frac{1}{\sqrt{x}}dx.</math>
|-
|Then,  <math style="vertical-align: -13px">du=\frac{3}{x}dx</math>  and  <math style="vertical-align: -5px">v=2\sqrt{x}.</math>
|-
|Using integration by parts, we get
|-
|        <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{\lim_{a\rightarrow 0} 6\sqrt{x}\ln(x)-12\sqrt{x}\bigg|_a^1.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using L'Hopital's Rule, we get
|-
|        <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{\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_{x\rightarrow 0} -12-6\sqrt{x}\ln(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}}}}\\
&&\\
& = & \displaystyle{-12+\lim_{x\rightarrow 0} 12\sqrt{x}}\\
&&\\
& = & \displaystyle{-12.}
\end{array}</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>\frac{1}{9}</math>
|-
|   '''(b)'''    <math>-12</math>
|}
[[009B_Sample_Final_2|'''Return to Sample Exam''']]

@@ 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.}

009B Sample Final 2, Problem 7 - Revision history

MathAdmin at 01:08, 21 May 2017