Difference between revisions of "009B Sample Final 2, Problem 7"

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!Step 1:    
 
!Step 1:    
 
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|First, we write
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|&nbsp; &nbsp; &nbsp; &nbsp;<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>u=\ln x</math> and <math>dv=\frac{1}{x^4}dx.</math>
 
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|Then, <math>du=\frac{1}{x}dx</math> and <math>v=\frac{1}{-3x^3}.</math>
 
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|Using integration by parts, we get
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <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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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|-
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|Now, using L'Hopital's Rule, we get
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|&nbsp; &nbsp; &nbsp; &nbsp; <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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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|'''(a)'''  
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|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>\frac{1}{9}</math>
 
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|'''(b)'''  
 
|'''(b)'''  
 
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[[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 13:08, 3 March 2017

Evaluate the following integrals or show that they are divergent:

(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 \int_1^\infty \frac{\ln x}{x^4}~dx}

(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 \int_0^1 \frac{3\ln x}{\sqrt{x}}~dx}

Foundations:  
1. 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_0^{\infty} f(x)~dx} 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_{0}^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_{0}^1 \frac{1}{x}~dx=\lim_{a\rightarrow 0} \int_{a}^1 \frac{1}{x}~dx.}


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_1^\infty \frac{\ln x}{x^4}~dx=\lim_{a\rightarrow \infty} \int_1^a \frac{\ln x}{x^4}~dx.}
Now, we 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=\ln 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=\frac{1}{x^4}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=\frac{1}{x}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=\frac{1}{-3x^3}.}
Using integration by parts, 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_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}}
Step 2:  
Now, using L'Hopital'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_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}}

(b)

Step 1:  
Step 2:  


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 \frac{1}{9}}
(b)

Return to Sample Exam

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