## anonymous one year ago Integrate this pls

1. freckles

integration!!!!!

2. freckles

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

3. welshfella

yes

4. freckles

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

5. freckles

that is generally how I complete the square

6. freckles

if you want you can use that as a formula too

7. freckles

you know of course assuming a isn't 0

8. SolomonZelman

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

9. SolomonZelman

or what is exactly the integral?

10. freckles

there is a square root over the bottom part

11. anonymous

Yes @SolomonZelman

12. freckles

oh no square root?

13. anonymous

Trying to find the value of s

14. freckles

wait that 5 is a s?

15. freckles

I'm no confused

16. SolomonZelman

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

17. anonymous

that's s

18. SolomonZelman

the last thing I wrote

19. SolomonZelman

?

20. SolomonZelman

so square root is raised to the second power?

21. freckles

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

22. SolomonZelman

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

23. SolomonZelman

can you draw it pliz? sliver x?

24. anonymous

@freckles was right. It isnt 5..

25. SolomonZelman

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

26. anonymous

yes yes

27. 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.

28. freckles

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

29. freckles

should (not show)

30. freckles

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

31. anonymous

so its sec^2 theta

32. SolomonZelman

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

33. SolomonZelman

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

34. SolomonZelman

then partial fractions decomposition/expansion

35. SolomonZelman

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

36. SolomonZelman

(also factored out of 600)

37. SolomonZelman

won't disturb tho'

38. anonymous

And I know its a long long process..

39. SolomonZelman

I hate bad integrals like yours....

40. SolomonZelman

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

41. anonymous

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

42. SolomonZelman

oh

43. SolomonZelman

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

44. freckles

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

45. freckles

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

46. anonymous

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

47. freckles

differentiate

48. SolomonZelman

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

49. freckles

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

50. freckles

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

51. anonymous

okay @freckles im following =)

52. 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

53. freckles

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

54. anonymous

lol @SolomonZelman thank you so much =)

55. freckles

the second term is just 0

56. freckles

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

57. freckles

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

58. anonymous

Got 17.78

59. SolomonZelman

didn't do much.. yw

60. freckles

by the way you can drop the | |

61. SolomonZelman

close

62. SolomonZelman

wolfram gives 19.274

63. anonymous

Ops. Its 22.004

64. 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

65. freckles

66. freckles
67. freckles

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

68. 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

69. anonymous

Aaah yes yes..

70. anonymous

Its 19..

71. anonymous

Thank you so much @freckles

72. anonymous

I still need to digest this..

73. freckles

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

74. anonymous

Thank you so much =)

75. freckles

np

76. anonymous

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

77. freckles

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

78. anonymous

Oh yes thank you!

79. anonymous

@freckles how did you get the 300?

80. anonymous

and also the 1/30 sqrt of 2?

81. 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: $0.02s^2+12s+0 \\ 0.02s^2+12s\\0.02(s^2+\frac{12}{0.02}s) \\ 0.02(s^2+\frac{12}{0.02}s+(\frac{12}{2(0.02)})^2)-0.02(\frac{12}{2(0.02)})^2 \\ 0.02(s+\frac{12}{2(0.02)})^2-0.02 \frac{12^2}{2^2(0.02^2)} \\ 0.02(s+\frac{12}{2} \frac{1}{0.02})^2-0.02 \frac{(2 \cdot 6)^2}{2^2(0.02)^2} \\ 0.02(s+6 \frac{1}{0.02})^2-0.02 \frac{2^2 6^2}{2^2(0.02)^2} \\ 0.02(s+6 \frac{100}{2})^2-0.02 \frac{ \cancel{2^2}6^2}{\cancel{2^2}(0.02)^2} \\ 0.02(s+3(100))^2-\cancel{0.02} \frac{6^2}{0.02^{\cancel{2}}} \\ 0.02(s+300)^2-\frac{36}{0.02} \\ 0.02(s+300)^2-\frac{3600}{2} \\ 0.02(s+300)^2-1800 \\ \frac{2}{100}(s+300)^2-1800 \\ 1800( \frac{2}{100(1800)}(s+300)^2-1) \\ 1800(\frac{2}{(100)18(100)}(s+300)^2-1) \\ 1800(\frac{2}{18(100)^2}(s+300)^2-1) \\ 1800(\frac{1}{9(100)^2}(s+300)^2-1) \\ \\ 1800(\frac{1}{(300)^2}(s+300)^2-1) \\ 1800((\frac{s+300}{300})^2-1)$

82. freckles

$\sqrt{1800}=\sqrt{900 \cdot 2}=\sqrt{900} \sqrt{2} =30 \sqrt{2}$

83. freckles

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

84. anonymous

Thank you thank you soo much

85. anonymous

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

86. freckles

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

87. anonymous

Ahh thank you so much @freckles =)