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

Revision as of 14:08, 3 March 2017

Evaluate the following integrals or show that they are divergent:

(a) $\int _{1}^{\infty }{\frac {\ln x}{x^{4}}}~dx$

(b) $\int _{0}^{1}{\frac {3\ln x}{\sqrt {x}}}~dx$

Foundations:
1. How could you write $\int _{0}^{\infty }f(x)~dx$ so that you can integrate?
You can write $\int _{0}^{\infty }f(x)~dx=\lim _{a\rightarrow \infty }\int _{0}^{a}f(x)~dx.$
2. How could you write $\int _{0}^{1}{\frac {1}{x}}~dx?$
The problem is that ${\frac {1}{x}}$ 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

@@ Line 30: / Line 30: @@
 !Step 1: &nbsp;
 |-
-|
+|First, we write
+|-
+|&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>
+|-
+|Now, we use integration by parts.
+|-
+|Let <math>u=\ln x</math> and <math>dv=\frac{1}{x^4}dx.</math>
 |-
-|
+|Then, <math>du=\frac{1}{x}dx</math> and <math>v=\frac{1}{-3x^3}.</math>
 |-
-|
+|Using integration by parts, we get
 |-
-|
+|&nbsp; &nbsp; &nbsp; &nbsp; <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>
 |}
@@ Line 42: / Line 52: @@
 !Step 2: &nbsp;
 |-
-|
+|Now, using L'Hopital's Rule, we get
-|-
-|
-|-
-|
 |-
-|
+|&nbsp; &nbsp; &nbsp; &nbsp; <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>
 |}
@@ Line 73: / Line 89: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>\frac{1}{9}</math>
 |-
 |'''(b)'''
 |}
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

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

Revision as of 14:08, 3 March 2017

Navigation menu

Search