anonymous
  • anonymous
Determine whether the improper integral converges or diverges Integral from 1 to infinity of [sqrt(9x+4)]/x^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\int\limits_{1}^{\infty} \frac{ \sqrt{9x+4} }{ x ^{2} } dx\]
SolomonZelman
  • SolomonZelman
u=9x+4 seems to be bad
anonymous
  • anonymous
could I use trig substitution for the numerator?

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SolomonZelman
  • SolomonZelman
\(\large\color{#000000}{\displaystyle\int\limits_{~}^{~}\frac{\sqrt{9x+4}}{ x^2 }~dx}\) \(\large\color{#000000}{\displaystyle x=(4/9)\tan^2\theta}\) \(\large\color{#000000}{\displaystyle x^2=(16/81)\tan^4\theta}\) \(\large\color{#000000}{\displaystyle dx=2(4/9)\tan\theta\sec^2\theta~d\theta}\) \(\large\color{#000000}{\displaystyle dx=(8/9)\tan\theta\sec^2\theta~d\theta}\)
SolomonZelman
  • SolomonZelman
let's see where this will get us
SolomonZelman
  • SolomonZelman
\(\large\color{#000000}{\displaystyle\int\limits_{~}^{~}\frac{\sqrt{9(4/9)\tan^2\theta +4}}{ (16/81)\tan^4\theta }~(8/9)\tan\theta\sec^2\theta~d\theta}\) \(\large\color{#000000}{\displaystyle\int\limits_{~}^{~}\frac{\sqrt{4\tan^2\theta +4}}{ (2/9)\tan^3\theta }\sec^2\theta~d\theta}\) \(\large\color{#000000}{\displaystyle\int\limits_{~}^{~}\frac{4\sec\theta}{ (2/9)\tan^3\theta }\sec^2\theta~d\theta}\) \(\large\color{#000000}{\displaystyle\frac{8}{ 9}\int\limits_{~}^{~}\csc^3 \theta~d\theta}\)
SolomonZelman
  • SolomonZelman
this is what I got so far... I have to go right now, but I hope this is not too dfiffficult, but you have a lot of dirty work left...
anonymous
  • anonymous
ok, thank you!
SolomonZelman
  • SolomonZelman
I am sorry, we have to determine convergence, I thought we had to evaluate
SolomonZelman
  • SolomonZelman
The convergence goes much easily... For this integral to converge, a series \(\large\color{#000000 }{ \displaystyle \sum_{n=1 }^\infty \frac{\sqrt{9n+4}}{n^2} }\) has to converge
SolomonZelman
  • SolomonZelman
We can do this in a snap shot. Intuitively, your power is n^1.5/n^2=1/n^1.5 And any series in a form of 1/n^p where p>1, will converge.
SolomonZelman
  • SolomonZelman
So the only thing we need to do is to actually show that \(\large\color{#000000 }{ \displaystyle \sum_{n=1 }^\infty \frac{\sqrt{9n+4}}{n^2} }\) actually converges, SO, we will COMPARE this series to a larger series If the larger series (we compare to) converges, then our series converges (and thus the initial integral converges).
SolomonZelman
  • SolomonZelman
We will compare \(\color{#000000 }{ \displaystyle \sum_{n=1 }^\infty \frac{\sqrt{9n+4}}{n^2} {\bf \quad~~to~~\quad }\sum_{n=1 }^\infty \frac{\sqrt{9n+9n}}{n^2} }\) Note that, \(\color{#000000 }{ \displaystyle \sum_{n=1 }^\infty \frac{\sqrt{9n+4}}{n^2} <\sum_{n=1 }^\infty \frac{\sqrt{9n+9n}}{n^2} }\) \(\color{#000000 }{ \displaystyle \sum_{n=1 }^\infty \frac{\sqrt{9n+4}}{n^2} <\sum_{n=1 }^\infty \frac{\sqrt{18n}}{n^2} }\) \(\color{#000000 }{ \displaystyle \sum_{n=1 }^\infty \frac{\sqrt{9n+4}}{n^2} <\sqrt{18}\cdot \sum_{n=1 }^\infty \frac{n^{0.5} }{n^2} }\) \(\color{#000000 }{ \displaystyle \sum_{n=1 }^\infty \frac{\sqrt{9n+4}}{n^2} <\sqrt{18}\cdot \sum_{n=1 }^\infty \frac{1 }{n^{1.5}}\Longleftarrow {\rm Converges} }\)
SolomonZelman
  • SolomonZelman
Since the right side that is larger converges, therefore the series converges (and again- thus the integral also converges).
SolomonZelman
  • SolomonZelman
The basic concepts that relate to these techniques of series convergence, that you have to know are as follows: \(\color{#000000 }{ \displaystyle {\bf Geometric~Series}}\) In the form; \(\color{#000000 }{ \displaystyle {\bf \sum_{n=j}^{\infty} a(r)^n}}\) Will only converge for \(\color{#000000 }{ \bf \displaystyle |r|<1}\) and diverges if \(\color{#000000 }{ \bf \displaystyle |r|\ge1}\) \(\color{#000000 }{ \displaystyle \LARGE {\bf ^\text{_________________}}}\) \(\color{#000000 }{ \displaystyle {\bf P-Series}}\) In the form; \(\color{#000000 }{ \displaystyle {\bf \sum_{n=j}^{\infty} a\cdot\frac{1}{n^p} }}\) Will only converge for \(\color{#000000 }{ \bf \displaystyle p>1}\) and diverges if \(\color{#000000 }{ \bf \displaystyle p\le1}\) \(\color{#000000 }{ \displaystyle \LARGE {\bf ^\text{_________________}}}\) \(\color{#000000 }{ \displaystyle {\bf Also,}}\) \(\color{#000000 }{ \displaystyle {\bf Comparison~Test}}\) \(\color{#000000 }{ \displaystyle {\bf Integral~Test}}\) (relates to what I did just now) \(\color{#000000 }{ \displaystyle {\bf Limit~Test}}\) \(\color{#000000 }{ \displaystyle {\bf Ratio~Test}}\) \(\color{#000000 }{ \displaystyle {\bf Nth~Root~Test}}\) Also, \(\color{#000000 }{ \displaystyle {\bf Alternating~Series~Test}}\) and perhaps some other concepts/terms... THE \(\color{#000000 }{ \displaystyle {\bf subject}}\) OF THIS IS; \(\color{#ff6433 }{ \displaystyle {\bf Determining~Series~Convergence}}\)

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