anonymous
  • anonymous
Integrate this pls
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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freckles
  • freckles
integration!!!!!
freckles
  • freckles
looks like we going to need some completing the square and some trig sub action
welshfella
  • welshfella
yes

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freckles
  • freckles
\[as^2+bs+c \\ a(s^2+\frac{b}{a}s)+c \\ a(s^2+\frac{b}{a}s+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2 \\ a(s+\frac{b}{2a})^2+c-\frac{ab^2}{4a^2} \\ a(s+\frac{b}{2a})^2+c-\frac{b^2}{4a} \\ a(s+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}\]
freckles
  • freckles
that is generally how I complete the square
freckles
  • freckles
if you want you can use that as a formula too
freckles
  • freckles
you know of course assuming a isn't 0
SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle\int\limits_{0}^{2000}\frac{1}{12s+0.025^2}~ds}\) it is like this?!
SolomonZelman
  • SolomonZelman
or what is exactly the integral?
freckles
  • freckles
there is a square root over the bottom part
anonymous
  • anonymous
Yes @SolomonZelman
freckles
  • freckles
oh no square root?
anonymous
  • anonymous
Trying to find the value of s
freckles
  • freckles
wait that 5 is a s?
freckles
  • freckles
I'm no confused
SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle\int\limits_{0}^{2000}\frac{1}{\sqrt{12s+0.025}^2}~ds}\) ??? I am also confused
anonymous
  • anonymous
that's s
SolomonZelman
  • SolomonZelman
the last thing I wrote
SolomonZelman
  • SolomonZelman
?
SolomonZelman
  • SolomonZelman
so square root is raised to the second power?
freckles
  • freckles
I thought it was: \[\int\limits_{0}^{2000}\frac{ds}{\sqrt{12s+0.02s^2}}\]
SolomonZelman
  • SolomonZelman
is that 5 or s at the end of the root?
SolomonZelman
  • SolomonZelman
can you draw it pliz? sliver x?
anonymous
  • anonymous
@freckles was right. It isnt 5..
SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle\int\limits_{0}^{2000}\frac{1}{\sqrt{12s+0.02s^2}}~ds}\)
anonymous
  • anonymous
yes yes
SolomonZelman
  • SolomonZelman
I ended up with partial fractions which are solvable, but it took too many steps, I think it is not correct to post it here.
freckles
  • freckles
you show eventually wind up with this (if you do the completing the square and you aim for a trig sub) \[\frac{1}{30 \sqrt{2}}\int\limits_{0}^{2000} \frac{ds}{\sqrt{(\frac{s+300}{300})^2-1} }\]
freckles
  • freckles
should (not show)
freckles
  • freckles
\[\sec^2(\theta)-1=\tan^2(\theta) \\ \text{ so } \sec(\theta)=\frac{s+300}{300} \\ \]
anonymous
  • anonymous
so its sec^2 theta
SolomonZelman
  • SolomonZelman
this is what I ended up after I have written all possible latex :o-:()
SolomonZelman
  • SolomonZelman
I completed the square in the root, and factored out of 1/ √(0.02) then I did u=s+300
SolomonZelman
  • SolomonZelman
then partial fractions decomposition/expansion
SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle\frac{\sqrt{2}}{120} \int\limits_{30}^{2300}\frac{\sqrt{u^2-90,000}}{u-300}-\frac{\sqrt{u^2-90,000}}{u+300}~du}\)
SolomonZelman
  • SolomonZelman
(also factored out of 600)
SolomonZelman
  • SolomonZelman
won't disturb tho'
anonymous
  • anonymous
And I know its a long long process..
SolomonZelman
  • SolomonZelman
I hate bad integrals like yours....
SolomonZelman
  • SolomonZelman
I would use geom. shapes if it was my choice. To many equations, sick of them:o-:(
anonymous
  • anonymous
Well actually this was derived in a dynamics problem. :(
SolomonZelman
  • SolomonZelman
oh
SolomonZelman
  • SolomonZelman
if you sure u derived it correctly.... (don't think you would make a mistake on that... i have seen you solve DE)
freckles
  • freckles
I'm not really sure if you need anymore help me but let me know
freckles
  • freckles
also @SolomonZelman your way is equivalent but I think it still needs trig sub
anonymous
  • anonymous
@freckles after noted that it was sectheta.. whats next?
freckles
  • freckles
differentiate
SolomonZelman
  • SolomonZelman
perhaps, I was thinking of separating them into two integrals and in one go w=u-300 and in other p=u+300
freckles
  • freckles
\[\frac{1}{30 \sqrt{2}}\int\limits_{0}^{2000} \frac{ds}{\sqrt{(\frac{s+300}{300})^2-1} } \] \[\sec(\theta)=\frac{s+300}{300} \\ \sec(\theta) \tan(\theta) d \theta=\frac{1}{300} ds \\ 300 \sec(\theta) \tan(\theta) d \theta =ds\]
freckles
  • freckles
\[\frac{10}{\sqrt{2}} \int\limits_0^{arcsec(\frac{23}{3})} \sec(\theta) d \theta \]
anonymous
  • anonymous
okay @freckles im following =)
SolomonZelman
  • SolomonZelman
((( When I did those in my class I had a nice math professor who would let use simpson's trapezoidal and midpoint rules, but on a condition that we find the max possible error, and use a bigger n. ))) not a fan of these:) bye
freckles
  • freckles
should finally get: \[\frac{10}{\sqrt{2}} \ln|\frac{23}{3}+\frac{1}{3} \sqrt{23^2-3^2}|-\frac{10}{\sqrt{2}}\ln|1+0|\]
anonymous
  • anonymous
lol @SolomonZelman thank you so much =)
freckles
  • freckles
the second term is just 0
freckles
  • freckles
\[\frac{10}{\sqrt{2}} \ln|\frac{23}{3}+\frac{2 \sqrt{130}}{3}|\]
freckles
  • freckles
I will let you play with that and dress it however you want
anonymous
  • anonymous
Got 17.78
SolomonZelman
  • SolomonZelman
didn't do much.. yw
freckles
  • freckles
by the way you can drop the | |
SolomonZelman
  • SolomonZelman
close
SolomonZelman
  • SolomonZelman
wolfram gives 19.274
anonymous
  • anonymous
Ops. Its 22.004
freckles
  • freckles
yeah you are a little bit off by the way since this was an ugly beast and you probably will come up again a lot more ugly foe I suggest wolfram for checking your answers
freckles
  • freckles
oops and solomon already said something about wolfram
freckles
  • freckles
http://www.wolframalpha.com/input/?i=10%2Fsqrt%282%29*ln%7C23%2F3%2B2%28sqrt%28130%29%2F3%29%7C
freckles
  • freckles
which you could also put the initial integral in there and compare what I got with that should be same thing
freckles
  • freckles
http://www.wolframalpha.com/input/?i=integrate%281%2Fsqrt%280.02s%5E2%2B12s%29%2Cs%3D0..2000%29 they actually round theirs to the nearest thousandth and don't off any more digits
anonymous
  • anonymous
Aaah yes yes..
anonymous
  • anonymous
Its 19..
anonymous
  • anonymous
Thank you so much @freckles
anonymous
  • anonymous
I still need to digest this..
freckles
  • freckles
if you want to be exact the answer is: \[\frac{10}{\sqrt{2}} \ln(\frac{23+2 \sqrt{130}}{3}) \\ \text{ or if you just want to give a better answer than wolfram then you can say something like } \\ 19.273952\]
anonymous
  • anonymous
Thank you so much =)
freckles
  • freckles
np
anonymous
  • anonymous
@freckles I wanna ask something.. Why did you use the completing the square?
freckles
  • freckles
I did the completing the square to do the trig sub that usually works for things in this form: \[\int\limits_{}^{}\frac{1}{\sqrt{(ax^2+bx+c)^n}}dx\]
anonymous
  • anonymous
Oh yes thank you!
anonymous
  • anonymous
@freckles how did you get the 300?
anonymous
  • anonymous
and also the 1/30 sqrt of 2?
freckles
  • freckles
I followed what I said about completing the square: \[as^2+bs+c \\ a(s^2+\frac{b}{a}s)+c \\ a(s^2+\frac{b}{a}s+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2 \\ a(s+\frac{b}{2a})^2+c-\frac{ab^2}{4a^2} \\ a(s+\frac{b}{2a})^2+c-\frac{b^2}{4a} \\ a(s+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}\] You can use this as a formula but I will go through the steps with the numbers instead just for another example of how to complete the square: \[0.02s^2+12s+0 \\ 0.02s^2+12s\\0.02(s^2+\frac{12}{0.02}s) \\ 0.02(s^2+\frac{12}{0.02}s+(\frac{12}{2(0.02)})^2)-0.02(\frac{12}{2(0.02)})^2 \\ 0.02(s+\frac{12}{2(0.02)})^2-0.02 \frac{12^2}{2^2(0.02^2)} \\ 0.02(s+\frac{12}{2} \frac{1}{0.02})^2-0.02 \frac{(2 \cdot 6)^2}{2^2(0.02)^2} \\ 0.02(s+6 \frac{1}{0.02})^2-0.02 \frac{2^2 6^2}{2^2(0.02)^2} \\ 0.02(s+6 \frac{100}{2})^2-0.02 \frac{ \cancel{2^2}6^2}{\cancel{2^2}(0.02)^2} \\ 0.02(s+3(100))^2-\cancel{0.02} \frac{6^2}{0.02^{\cancel{2}}} \\ 0.02(s+300)^2-\frac{36}{0.02} \\ 0.02(s+300)^2-\frac{3600}{2} \\ 0.02(s+300)^2-1800 \\ \frac{2}{100}(s+300)^2-1800 \\ 1800( \frac{2}{100(1800)}(s+300)^2-1) \\ 1800(\frac{2}{(100)18(100)}(s+300)^2-1) \\ 1800(\frac{2}{18(100)^2}(s+300)^2-1) \\ 1800(\frac{1}{9(100)^2}(s+300)^2-1) \\ \\ 1800(\frac{1}{(300)^2}(s+300)^2-1) \\ 1800((\frac{s+300}{300})^2-1)\]
freckles
  • freckles
\[\sqrt{1800}=\sqrt{900 \cdot 2}=\sqrt{900} \sqrt{2} =30 \sqrt{2}\]
freckles
  • freckles
is how I came to: \[\frac{1}{30 \sqrt{2}}\int\limits_{0}^{2000} \frac{1}{\sqrt{(\frac{s+300}{300})^2-1}} ds\]
anonymous
  • anonymous
Thank you thank you soo much
anonymous
  • anonymous
@freckles how did you get the upper limit (arc sec 23/3)? =)
freckles
  • freckles
s=2000 and \[\sec(\theta)=\frac{s+300}{300}\] plug in old limit into sub to find new limit
anonymous
  • anonymous
Ahh thank you so much @freckles =)

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