Question

Determine whether the improper integral converges or diverges

Integral from 1 to infinity of [sqrt(9x+4)]/x^2

Answer 1

\[\int\limits_{1}^{\infty} \frac{ \sqrt{9x+4} }{ x ^{2} } dx\]

Answer 2

u=9x+4 seems to be bad

Answer 3

could I use trig substitution for the numerator?

Answer 4

\(\large\color{#000000}{\displaystyle\int\limits_{~}^{~}\frac{\sqrt{9x+4}}{ x^2 }~dx}\)

\(\large\color{#000000}{\displaystyle x=(4/9)\tan^2\theta}\)

\(\large\color{#000000}{\displaystyle x^2=(16/81)\tan^4\theta}\)

\(\large\color{#000000}{\displaystyle dx=2(4/9)\tan\theta\sec^2\theta~d\theta}\)
\(\large\color{#000000}{\displaystyle dx=(8/9)\tan\theta\sec^2\theta~d\theta}\)

Answer 5

let's see where this will get us

Answer 6

\(\large\color{#000000}{\displaystyle\int\limits_{~}^{~}\frac{\sqrt{9(4/9)\tan^2\theta +4}}{ (16/81)\tan^4\theta }~(8/9)\tan\theta\sec^2\theta~d\theta}\)

\(\large\color{#000000}{\displaystyle\int\limits_{~}^{~}\frac{\sqrt{4\tan^2\theta +4}}{ (2/9)\tan^3\theta }\sec^2\theta~d\theta}\)

\(\large\color{#000000}{\displaystyle\int\limits_{~}^{~}\frac{4\sec\theta}{ (2/9)\tan^3\theta }\sec^2\theta~d\theta}\)

\(\large\color{#000000}{\displaystyle\frac{8}{ 9}\int\limits_{~}^{~}\csc^3 \theta~d\theta}\)

Answer 7

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...

Answer 8

ok, thank you!

Answer 9

I am sorry, we have to determine convergence, I thought we had to evaluate

Answer 10

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

Answer 11

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.

Answer 12

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).

Answer 13

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,

\(\color{#000000 }{ \displaystyle \sum_{n=1 }^\infty \frac{\sqrt{9n+4}}{n^2} <\sum_{n=1 }^\infty \frac{\sqrt{9n+9n}}{n^2} }\)

\(\color{#000000 }{ \displaystyle \sum_{n=1 }^\infty \frac{\sqrt{9n+4}}{n^2} <\sum_{n=1 }^\infty \frac{\sqrt{18n}}{n^2} }\)

\(\color{#000000 }{ \displaystyle \sum_{n=1 }^\infty \frac{\sqrt{9n+4}}{n^2} <\sqrt{18}\cdot \sum_{n=1 }^\infty \frac{n^{0.5} }{n^2} }\)

\(\color{#000000 }{ \displaystyle \sum_{n=1 }^\infty \frac{\sqrt{9n+4}}{n^2} <\sqrt{18}\cdot \sum_{n=1 }^\infty \frac{1 }{n^{1.5}}\Longleftarrow {\rm Converges} }\)

Answer 14

Since the right side that is larger converges, therefore the series converges (and again- thus the integral also converges).

Answer 15

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;
\(\color{#ff6433 }{ \displaystyle {\bf Determining~Series~Convergence}}\)