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sjg13e
 one year ago
Improper Integral, (2 to infinity)
sjg13e
 one year ago
Improper Integral, (2 to infinity)

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sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{2}^{\infty}\frac{ 1 }{ v^2 + 8v  9 }dv\] Does it converge or diverge?

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\frac{ 1 }{ v^2 + 8v  9 }dv = \frac{ 1 }{ v } + \frac{ 1 }{ 8 }\lnv  \frac{ 1 }{ 9 }v\] (this is what I got)

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{t \rightarrow \infty}[\frac{ 1 }{ v }+\frac{ 1 }{ 8 }\lnv\frac{ 1 }{ 9 }v]\] where a = 2 and b = t

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0I got \[\frac{ 1 }{ 3 }\frac{ 1 }{ 8 }\ln2\] as my answer. But it's wrong. I know it converges, maybe I messed up with inputting the values?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Suppose you want to know whether \(\large\color{black}{\displaystyle\sum_{n=7}^{\infty}\frac{1}{n^2} }\) converges or not. you know that converges, (to π²/6 by Euler) And a bigger series \(\large\color{black}{\displaystyle\sum_{n=7}^{\infty}\frac{1}{n^2+8v9} }\) would all the more so converge

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1So if the first series converges (and must be  so does its integral) THEN, all the more so the second series (and must be  so does the integral).

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1That is ian observation if you studied about the *integral test*... have you?

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0Okay, yes. That is also expressed by the comparison test, right?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Comparison test, to a larger series.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1It should really start from n=2, not n=7, but that doesn't matter.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1If you want, we can solve the integral tho

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0okay thank you, id be awesome if you helped me solve it

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{v^2+8v9}{~\rm dv} }\) \(\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{v^2+8v+16169}{~\rm dv} }\) \(\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{(v+4)^225}{~\rm dv} }\) set u=v+4 this is where I would start

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I think I see what you did... you think \[\frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \text{ but this is not true }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you could use trig sub or do partial fractions

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes, partical fractions from the beginning is better I guess, because with my way you still have partial fractions

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{(v+9)(v1)}{~\rm dv} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1and do the partial fractions...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1@SolomonZelman I think you could have done trig sub without partial fractions I was just offering another route

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1trig after u sub?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\int\limits \frac{1}{(v+4)^225} dv \\ \int\limits \frac{1}{25[\frac{1}{25}(v+4)^21]} dv \\ \frac{1}{25} \int\limits \frac{1}{(\frac{v+4}{5})^21} dv \\ \text{ recall } \tan^2(x)=\sec^2(x)1 \\ \text{ so use sub } \frac{v+4}{5}=\sec(x) \\ \frac{1}{5} dv=\sec(x) \tan(x) dx \\ \frac{1}{25} \int\limits \frac{1}{\sec^2(x)1} 5 \sec(x) \tan(x) dx \\ \frac{1}{25} \int\limits \frac{5 \sec(x)\tan(x)}{\tan^2(x)}dx\] unless... I did something wrong we don't need partial fractions for the trig route here

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, my wifi was spotty. I'll read through your explanation now

freckles
 one year ago
Best ResponseYou've already chosen the best response.1though I much prefer the partial fractions

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, definitely... dw:1443654159980:dw

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0okay thank you! i messed up in the beginning with integration. i'll probably used partial fractions to solve it then

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes, just partial fractions for 1/((v1)(v+9))

freckles
 one year ago
Best ResponseYou've already chosen the best response.1@sjg13e well you messed up before the integration you used that 1/(a+b+c)=1/a+1/b+1/c which is not true

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0yes, that's what i meant.

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0thank you all very much! i appreciate your help!
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