Integrate this pls

- anonymous

Integrate this pls

- Stacey Warren - Expert brainly.com

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- schrodinger

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- freckles

integration!!!!!

- freckles

looks like we going to need some completing the square and some trig sub action

- welshfella

yes

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## More answers

- 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

that is generally how I complete the square

- freckles

if you want you can use that as a formula too

- freckles

you know of course assuming a isn't 0

- SolomonZelman

\(\large\color{slate}{\displaystyle\int\limits_{0}^{2000}\frac{1}{12s+0.025^2}~ds}\) it is like this?!

- SolomonZelman

or what is exactly the integral?

- freckles

there is a square root over the bottom part

- anonymous

Yes @SolomonZelman

- freckles

oh no square root?

- anonymous

Trying to find the value of s

- freckles

wait that 5 is a s?

- freckles

I'm no confused

- SolomonZelman

\(\large\color{slate}{\displaystyle\int\limits_{0}^{2000}\frac{1}{\sqrt{12s+0.025}^2}~ds}\) ???
I am also confused

- anonymous

that's s

- SolomonZelman

the last thing I wrote

- SolomonZelman

?

- SolomonZelman

so square root is raised to the second power?

- freckles

I thought it was:
\[\int\limits_{0}^{2000}\frac{ds}{\sqrt{12s+0.02s^2}}\]

- SolomonZelman

is that 5 or s at the end of the root?

- SolomonZelman

can you draw it pliz? sliver x?

- anonymous

@freckles was right. It isnt 5..

- SolomonZelman

\(\large\color{slate}{\displaystyle\int\limits_{0}^{2000}\frac{1}{\sqrt{12s+0.02s^2}}~ds}\)

- anonymous

yes yes

- 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

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

should (not show)

- freckles

\[\sec^2(\theta)-1=\tan^2(\theta) \\ \text{ so } \sec(\theta)=\frac{s+300}{300} \\ \]

- anonymous

so its sec^2 theta

- SolomonZelman

this is what I ended up after I have written all possible latex :o-:()

- SolomonZelman

I completed the square in the root, and factored out of 1/ √(0.02)
then I did u=s+300

- SolomonZelman

then partial fractions decomposition/expansion

- 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

(also factored out of 600)

- SolomonZelman

won't disturb tho'

- anonymous

And I know its a long long process..

- SolomonZelman

I hate bad integrals like yours....

- SolomonZelman

I would use geom. shapes if it was my choice.
To many equations, sick of them:o-:(

- anonymous

Well actually this was derived in a dynamics problem. :(

- SolomonZelman

oh

- 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

I'm not really sure if you need anymore help me but let me know

- freckles

also @SolomonZelman your way is equivalent
but I think it still needs trig sub

- anonymous

@freckles after noted that it was sectheta.. whats next?

- freckles

differentiate

- 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

\[\frac{1}{30 \sqrt{2}}\int\limits_{0}^{2000} \frac{ds}{\sqrt{(\frac{s+300}{300})^2-1} }
\]
\[\sec(\theta)=\frac{s+300}{300} \\ \sec(\theta) \tan(\theta) d \theta=\frac{1}{300} ds \\ 300 \sec(\theta) \tan(\theta) d \theta =ds\]

- freckles

\[\frac{10}{\sqrt{2}} \int\limits_0^{arcsec(\frac{23}{3})} \sec(\theta) d \theta \]

- anonymous

okay @freckles im following =)

- 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

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

lol @SolomonZelman thank you so much =)

- freckles

the second term is just 0

- freckles

\[\frac{10}{\sqrt{2}} \ln|\frac{23}{3}+\frac{2 \sqrt{130}}{3}|\]

- freckles

I will let you play with that and dress it however you want

- anonymous

Got 17.78

- SolomonZelman

didn't do much.. yw

- freckles

by the way you can drop the | |

- SolomonZelman

close

- SolomonZelman

wolfram gives 19.274

- anonymous

Ops. Its 22.004

- 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

oops and solomon already said something about wolfram

- freckles

http://www.wolframalpha.com/input/?i=10%2Fsqrt%282%29*ln%7C23%2F3%2B2%28sqrt%28130%29%2F3%29%7C

- freckles

which you could also put the initial integral in there
and compare what I got with that
should be same thing

- 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

Aaah yes yes..

- anonymous

Its 19..

- anonymous

Thank you so much @freckles

- anonymous

I still need to digest this..

- 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

Thank you so much =)

- freckles

np

- anonymous

@freckles I wanna ask something.. Why did you use the completing the square?

- 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

Oh yes thank you!

- anonymous

@freckles how did you get the 300?

- anonymous

and also the 1/30 sqrt of 2?

- 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

\[\sqrt{1800}=\sqrt{900 \cdot 2}=\sqrt{900} \sqrt{2} =30 \sqrt{2}\]

- 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

Thank you thank you soo much

- anonymous

@freckles how did you get the upper limit (arc sec 23/3)? =)

- freckles

s=2000 and \[\sec(\theta)=\frac{s+300}{300}\] plug in old limit into sub to find new limit

- anonymous

Ahh thank you so much @freckles =)

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