Difference between revisions of "009B Sample Final 2, Problem 7"
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|First, we write | |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> | + | | <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. | |Now, we use integration by parts. | ||
| Line 85: | Line 85: | ||
|- | |- | ||
| <math>\begin{array}{rcl} | | <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> | \end{array}</math> | ||
|} | |} | ||
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|- | |- | ||
| <math>\begin{array}{rcl} | | <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.} | & = & \displaystyle{-12.} | ||
Revision as of 13:19, 20 May 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 |
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: |
|---|
| 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_0^1 \frac{3\ln x}{\sqrt{x}}~dx=\lim_{a\rightarrow 0^+} \int_a^1 \frac{3\ln x}{\sqrt{x}}~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=3\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}{\sqrt{x}}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{3}{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=2\sqrt{x}.} |
| 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_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}} |
| 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_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}} |
| 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) 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 -12} |