A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Integrate this pls
anonymous
 one year ago
Integrate this pls

This Question is Closed

freckles
 one year ago
Best ResponseYou've already chosen the best response.2looks like we going to need some completing the square and some trig sub action

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[as^2+bs+c \\ a(s^2+\frac{b}{a}s)+c \\ a(s^2+\frac{b}{a}s+(\frac{b}{2a})^2)+ca(\frac{b}{2a})^2 \\ a(s+\frac{b}{2a})^2+c\frac{ab^2}{4a^2} \\ a(s+\frac{b}{2a})^2+c\frac{b^2}{4a} \\ a(s+\frac{b}{2a})^2+\frac{4acb^2}{4a}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2that is generally how I complete the square

freckles
 one year ago
Best ResponseYou've already chosen the best response.2if you want you can use that as a formula too

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you know of course assuming a isn't 0

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{slate}{\displaystyle\int\limits_{0}^{2000}\frac{1}{12s+0.025^2}~ds}\) it is like this?!

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1or what is exactly the integral?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2there is a square root over the bottom part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Trying to find the value of s

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{slate}{\displaystyle\int\limits_{0}^{2000}\frac{1}{\sqrt{12s+0.025}^2}~ds}\) ??? I am also confused

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1the last thing I wrote

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1so square root is raised to the second power?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I thought it was: \[\int\limits_{0}^{2000}\frac{ds}{\sqrt{12s+0.02s^2}}\]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1is that 5 or s at the end of the root?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1can you draw it pliz? sliver x?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@freckles was right. It isnt 5..

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{slate}{\displaystyle\int\limits_{0}^{2000}\frac{1}{\sqrt{12s+0.02s^2}}~ds}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.2you show eventually wind up with this (if you do the completing the square and you aim for a trig sub) \[\frac{1}{30 \sqrt{2}}\int\limits_{0}^{2000} \frac{ds}{\sqrt{(\frac{s+300}{300})^21} }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\sec^2(\theta)1=\tan^2(\theta) \\ \text{ so } \sec(\theta)=\frac{s+300}{300} \\ \]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1this is what I ended up after I have written all possible latex :o:()

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I completed the square in the root, and factored out of 1/ √(0.02) then I did u=s+300

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1then partial fractions decomposition/expansion

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{slate}{\displaystyle\frac{\sqrt{2}}{120} \int\limits_{30}^{2300}\frac{\sqrt{u^290,000}}{u300}\frac{\sqrt{u^290,000}}{u+300}~du}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1(also factored out of 600)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1won't disturb tho'

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And I know its a long long process..

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I hate bad integrals like yours....

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I would use geom. shapes if it was my choice. To many equations, sick of them:o:(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well actually this was derived in a dynamics problem. :(

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1if you sure u derived it correctly.... (don't think you would make a mistake on that... i have seen you solve DE)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I'm not really sure if you need anymore help me but let me know

freckles
 one year ago
Best ResponseYou've already chosen the best response.2also @SolomonZelman your way is equivalent but I think it still needs trig sub

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@freckles after noted that it was sectheta.. whats next?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1perhaps, I was thinking of separating them into two integrals and in one go w=u300 and in other p=u+300

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{1}{30 \sqrt{2}}\int\limits_{0}^{2000} \frac{ds}{\sqrt{(\frac{s+300}{300})^21} } \] \[\sec(\theta)=\frac{s+300}{300} \\ \sec(\theta) \tan(\theta) d \theta=\frac{1}{300} ds \\ 300 \sec(\theta) \tan(\theta) d \theta =ds\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{10}{\sqrt{2}} \int\limits_0^{arcsec(\frac{23}{3})} \sec(\theta) d \theta \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay @freckles im following =)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1((( 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
 one year ago
Best ResponseYou've already chosen the best response.2should finally get: \[\frac{10}{\sqrt{2}} \ln\frac{23}{3}+\frac{1}{3} \sqrt{23^23^2}\frac{10}{\sqrt{2}}\ln1+0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol @SolomonZelman thank you so much =)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the second term is just 0

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{10}{\sqrt{2}} \ln\frac{23}{3}+\frac{2 \sqrt{130}}{3}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I will let you play with that and dress it however you want

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1didn't do much.. yw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2by the way you can drop the  

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1wolfram gives 19.274

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yeah 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
 one year ago
Best ResponseYou've already chosen the best response.2oops and solomon already said something about wolfram

freckles
 one year ago
Best ResponseYou've already chosen the best response.2http://www.wolframalpha.com/input/?i=10%2Fsqrt%282%29*ln%7C23%2F3%2B2%28sqrt%28130%29%2F3%29%7C

freckles
 one year ago
Best ResponseYou've already chosen the best response.2which you could also put the initial integral in there and compare what I got with that should be same thing

freckles
 one year ago
Best ResponseYou've already chosen the best response.2http://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
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much @freckles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I still need to digest this..

freckles
 one year ago
Best ResponseYou've already chosen the best response.2if 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
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much =)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@freckles I wanna ask something.. Why did you use the completing the square?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.0@freckles how did you get the 300?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and also the 1/30 sqrt of 2?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I followed what I said about completing the square: \[as^2+bs+c \\ a(s^2+\frac{b}{a}s)+c \\ a(s^2+\frac{b}{a}s+(\frac{b}{2a})^2)+ca(\frac{b}{2a})^2 \\ a(s+\frac{b}{2a})^2+c\frac{ab^2}{4a^2} \\ a(s+\frac{b}{2a})^2+c\frac{b^2}{4a} \\ a(s+\frac{b}{2a})^2+\frac{4acb^2}{4a}\] 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: \[0.02s^2+12s+0 \\ 0.02s^2+12s\\0.02(s^2+\frac{12}{0.02}s) \\ 0.02(s^2+\frac{12}{0.02}s+(\frac{12}{2(0.02)})^2)0.02(\frac{12}{2(0.02)})^2 \\ 0.02(s+\frac{12}{2(0.02)})^20.02 \frac{12^2}{2^2(0.02^2)} \\ 0.02(s+\frac{12}{2} \frac{1}{0.02})^20.02 \frac{(2 \cdot 6)^2}{2^2(0.02)^2} \\ 0.02(s+6 \frac{1}{0.02})^20.02 \frac{2^2 6^2}{2^2(0.02)^2} \\ 0.02(s+6 \frac{100}{2})^20.02 \frac{ \cancel{2^2}6^2}{\cancel{2^2}(0.02)^2} \\ 0.02(s+3(100))^2\cancel{0.02} \frac{6^2}{0.02^{\cancel{2}}} \\ 0.02(s+300)^2\frac{36}{0.02} \\ 0.02(s+300)^2\frac{3600}{2} \\ 0.02(s+300)^21800 \\ \frac{2}{100}(s+300)^21800 \\ 1800( \frac{2}{100(1800)}(s+300)^21) \\ 1800(\frac{2}{(100)18(100)}(s+300)^21) \\ 1800(\frac{2}{18(100)^2}(s+300)^21) \\ 1800(\frac{1}{9(100)^2}(s+300)^21) \\ \\ 1800(\frac{1}{(300)^2}(s+300)^21) \\ 1800((\frac{s+300}{300})^21)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\sqrt{1800}=\sqrt{900 \cdot 2}=\sqrt{900} \sqrt{2} =30 \sqrt{2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2is how I came to: \[\frac{1}{30 \sqrt{2}}\int\limits_{0}^{2000} \frac{1}{\sqrt{(\frac{s+300}{300})^21}} ds\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you thank you soo much

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@freckles how did you get the upper limit (arc sec 23/3)? =)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2s=2000 and \[\sec(\theta)=\frac{s+300}{300}\] plug in old limit into sub to find new limit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahh thank you so much @freckles =)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.