Question

Integrate this pls

Answer 1

integration!!!!!

Answer 2

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

Answer 3

yes

Answer 4

\[as^2+bs+c \\ a(s^2+\frac{b}{a}s)+c \\ a(s^2+\frac{b}{a}s+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2 \\ a(s+\frac{b}{2a})^2+c-\frac{ab^2}{4a^2} \\ a(s+\frac{b}{2a})^2+c-\frac{b^2}{4a} \\ a(s+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}\]

Answer 5

that is generally how I complete the square

Answer 6

if you want you can use that as a formula too

Answer 7

you know of course assuming a isn't 0

Answer 8

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

Answer 9

or what is exactly the integral?

Answer 10

there is a square root over the bottom part

Answer 11

Yes @SolomonZelman

Answer 12

oh no square root?

Answer 13

Trying to find the value of s

Answer 14

wait that 5 is a s?

Answer 15

I'm no confused

Answer 16

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

I am also confused

Answer 17

that's s

Answer 18

the last thing I wrote

Answer 19

?

Answer 20

so square root is raised to the second power?

Answer 21

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

Answer 22

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

Answer 23

can you draw it pliz? sliver x?

Answer 24

@freckles was right. It isnt 5..

Answer 25

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

Answer 26

yes yes

Answer 27

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.

Answer 28

you show eventually wind up with
this (if you do the completing the square and you aim for a trig sub)
\[\frac{1}{30 \sqrt{2}}\int\limits_{0}^{2000} \frac{ds}{\sqrt{(\frac{s+300}{300})^2-1} }\]

Answer 29

should (not show)

Answer 30

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

Answer 31

so its sec^2 theta

Answer 32

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

Answer 33

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

Answer 34

then partial fractions decomposition/expansion

Answer 35

\(\large\color{slate}{\displaystyle\frac{\sqrt{2}}{120} \int\limits_{30}^{2300}\frac{\sqrt{u^2-90,000}}{u-300}-\frac{\sqrt{u^2-90,000}}{u+300}~du}\)

Answer 36

(also factored out of 600)

Answer 37

won't disturb tho'

Answer 38

And I know its a long long process..

Answer 39

I hate bad integrals like yours....

Answer 40

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

Answer 41

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

Answer 42

oh

Answer 43

if you sure u derived it correctly.... (don't think you would make a mistake on that... i have seen you solve DE)

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

differentiate

Answer 48

perhaps, I was thinking of separating them into two integrals and in one go w=u-300
and in other p=u+300

Answer 49

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

Answer 50

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

Answer 51

okay @freckles im following =)

Answer 52

((( 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

Answer 53

should finally get:
\[\frac{10}{\sqrt{2}} \ln|\frac{23}{3}+\frac{1}{3} \sqrt{23^2-3^2}|-\frac{10}{\sqrt{2}}\ln|1+0|\]

Answer 54

lol @SolomonZelman thank you so much =)

Answer 55

the second term is just 0

Answer 56

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

Answer 57

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

Answer 58

Got 17.78

Answer 59

didn't do much.. yw

Answer 60

by the way you can drop the | |

Answer 61

close

Answer 62

wolfram gives 19.274

Answer 63

Ops. Its 22.004

Answer 64

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

Answer 65

oops and solomon already said something about wolfram

Answer 66

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

Answer 67

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

Answer 68

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

Answer 69

Aaah yes yes..

Answer 70

Its 19..

Answer 71

Thank you so much @freckles

Answer 72

I still need to digest this..

Answer 73

if you want to be exact
the answer is:
\[\frac{10}{\sqrt{2}} \ln(\frac{23+2 \sqrt{130}}{3}) \\ \text{ or if you just want to give a better answer than wolfram then you can say something like } \\ 19.273952\]

Answer 74

Thank you so much =)

Answer 75

np

Answer 76

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

Answer 77

I did the completing the square
to do the trig sub
that usually works for things in this form:
\[\int\limits_{}^{}\frac{1}{\sqrt{(ax^2+bx+c)^n}}dx\]

Answer 78

Oh yes thank you!

Answer 79

@freckles how did you get the 300?

Answer 80

and also the 1/30 sqrt of 2?

Answer 81

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:
\[0.02s^2+12s+0 \\ 0.02s^2+12s\\0.02(s^2+\frac{12}{0.02}s) \\ 0.02(s^2+\frac{12}{0.02}s+(\frac{12}{2(0.02)})^2)-0.02(\frac{12}{2(0.02)})^2 \\ 0.02(s+\frac{12}{2(0.02)})^2-0.02 \frac{12^2}{2^2(0.02^2)} \\ 0.02(s+\frac{12}{2} \frac{1}{0.02})^2-0.02 \frac{(2 \cdot 6)^2}{2^2(0.02)^2} \\ 0.02(s+6 \frac{1}{0.02})^2-0.02 \frac{2^2 6^2}{2^2(0.02)^2} \\ 0.02(s+6 \frac{100}{2})^2-0.02 \frac{ \cancel{2^2}6^2}{\cancel{2^2}(0.02)^2} \\ 0.02(s+3(100))^2-\cancel{0.02} \frac{6^2}{0.02^{\cancel{2}}} \\ 0.02(s+300)^2-\frac{36}{0.02} \\ 0.02(s+300)^2-\frac{3600}{2} \\ 0.02(s+300)^2-1800 \\ \frac{2}{100}(s+300)^2-1800 \\ 1800( \frac{2}{100(1800)}(s+300)^2-1) \\ 1800(\frac{2}{(100)18(100)}(s+300)^2-1) \\ 1800(\frac{2}{18(100)^2}(s+300)^2-1) \\ 1800(\frac{1}{9(100)^2}(s+300)^2-1) \\ \\ 1800(\frac{1}{(300)^2}(s+300)^2-1) \\ 1800((\frac{s+300}{300})^2-1)\]

Answer 82

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

Answer 83

is how I came to:
\[\frac{1}{30 \sqrt{2}}\int\limits_{0}^{2000} \frac{1}{\sqrt{(\frac{s+300}{300})^2-1}} ds\]

Answer 84

Thank you thank you soo much

Answer 85

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

Answer 86

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

Answer 87

Ahh thank you so much @freckles =)