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sjg13e

  • one year ago

Improper Integral, (2 to infinity)

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  1. sjg13e
    • one year ago
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    \[\int\limits_{2}^{\infty}\frac{ 1 }{ v^2 + 8v - 9 }dv\] Does it converge or diverge?

  2. sjg13e
    • one year ago
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    \[\int\limits_{}^{}\frac{ 1 }{ v^2 + 8v - 9 }dv = \frac{ -1 }{ v } + \frac{ 1 }{ 8 }\ln|v| - \frac{ 1 }{ 9 }v\] (this is what I got)

  3. sjg13e
    • one year ago
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    \[\lim_{t \rightarrow \infty}[\frac{ -1 }{ v }+\frac{ 1 }{ 8 }\ln|v|-\frac{ 1 }{ 9 }v]\] where a = 2 and b = t

  4. sjg13e
    • one year ago
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    I got \[\frac{ 1 }{ 3 }-\frac{ 1 }{ 8 }\ln|2|\] as my answer. But it's wrong. I know it converges, maybe I messed up with inputting the values?

  5. SolomonZelman
    • one year ago
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    Suppose 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 \(\large\color{black}{\displaystyle\sum_{n=7}^{\infty}\frac{1}{n^2+8v-9} }\) would all the more so converge

  6. SolomonZelman
    • one year ago
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    So 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).

  7. SolomonZelman
    • one year ago
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    That is ian observation if you studied about the *integral test*... have you?

  8. sjg13e
    • one year ago
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    Okay, yes. That is also expressed by the comparison test, right?

  9. SolomonZelman
    • one year ago
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    Yes

  10. SolomonZelman
    • one year ago
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    Comparison test, to a larger series.

  11. SolomonZelman
    • one year ago
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    It should really start from n=2, not n=7, but that doesn't matter.

  12. SolomonZelman
    • one year ago
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    If you want, we can solve the integral tho

  13. sjg13e
    • one year ago
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    okay thank you, id be awesome if you helped me solve it

  14. SolomonZelman
    • one year ago
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    \(\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{v^2+8v-9}{~\rm dv} }\) \(\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{v^2+8v+16-16-9}{~\rm dv} }\) \(\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{(v+4)^2-25}{~\rm dv} }\) set u=v+4 this is where I would start

  15. freckles
    • one year ago
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    I think I see what you did... you think \[\frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \text{ but this is not true }\]

  16. freckles
    • one year ago
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    you could use trig sub or do partial fractions

  17. SolomonZelman
    • one year ago
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    yes, partical fractions from the beginning is better I guess, because with my way you still have partial fractions

  18. SolomonZelman
    • one year ago
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    \(\large\color{black}{\displaystyle\int\limits_{2}^{\infty }\frac{1}{(v+9)(v-1)}{~\rm dv} }\)

  19. SolomonZelman
    • one year ago
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    and do the partial fractions...

  20. freckles
    • one year ago
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    @SolomonZelman I think you could have done trig sub without partial fractions I was just offering another route

  21. SolomonZelman
    • one year ago
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    trig after u sub?

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

  23. sjg13e
    • one year ago
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    Sorry, my wifi was spotty. I'll read through your explanation now

  24. freckles
    • one year ago
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    though I much prefer the partial fractions

  25. SolomonZelman
    • one year ago
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    Yeah, definitely... |dw:1443654159980:dw|

  26. sjg13e
    • one year ago
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    okay thank you! i messed up in the beginning with integration. i'll probably used partial fractions to solve it then

  27. SolomonZelman
    • one year ago
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    yes, just partial fractions for 1/((v-1)(v+9))

  28. freckles
    • one year ago
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    @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

  29. sjg13e
    • one year ago
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    yes, that's what i meant.

  30. sjg13e
    • one year ago
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    thank you all very much! i appreciate your help!

  31. freckles
    • one year ago
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    np :)

  32. SolomonZelman
    • one year ago
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    Σ = np × 2

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