## anonymous 5 years ago 1/(6t+1)^1/2 dt evaluated from 0 to 4

1. anonymous

The integral of that you mean?

2. anonymous

Yes, sorry I wasn't clear

3. anonymous

Definite Integral

4. myininaya

let u=6t+1

5. anonymous

Use a u substation as myininaya has just suggested ;p

6. anonymous

substitution I mean.

7. anonymous

I have to use the Fundamental Theorem of Calculus

8. myininaya

we will

9. anonymous

Ok

10. myininaya

there is no other way unless you used riemann sums but lets not mention him anymore

11. anonymous

haha

12. myininaya

using riemann sums can also be very difficult to evaluate the limit

13. myininaya

or maybe impossible

14. anonymous

Well I'm teaching myself Integrals so I'm not quite that far yet.

15. anonymous

16. myininaya

do you know substation?

17. anonymous

substitution ;p

18. myininaya

stupid thingy i hit whatever popped up because i hate spelling substitution

19. anonymous

You have to learn u substitutions to do these they allow you to undo the chain rule.

20. anonymous

Ok, So after the substitution, What's my next step?

21. myininaya

if u=6t+1 then du/dt=.....

22. anonymous

6

23. myininaya

right do du=....

24. anonymous

6 dt?

25. myininaya

right

26. anonymous

Then I hit both sides with integrals, correct?

27. anonymous

No. Then you substitute 6t +1 as u, and your dt as 1/6 du.

28. anonymous

And take the integral with respect to u (though you also have to change your limits)

29. anonymous

Oh, ok, that's where I get lost.

30. myininaya

polpak divided both sides by 6 of that equation du=6dt

31. anonymous

Well 6t + 1 = u the 1/(6t+1) = 1/u

32. myininaya

you dont have to change the limits you can just leave it blank and come back after finding the antiderivative in terms of x in using the limits that we started with

33. anonymous

and dt = 1/6 du so $\int_0^4 \frac{1}{6t+1}dt = \int_1^{25} \frac{1}{u}(\frac{1}{6} du)$

34. anonymous

Since 1/6 is just a multiplicative constant you can pull it out front.

35. myininaya

when i say x i meant t lol

36. anonymous

Hmmmm...

37. anonymous

Ok, let's try this one step at a time. Let u = 6t +1 $\implies du = 6dt \implies dt = \frac{1}{6}du$ $t = 0 \implies u = 0 + 1 = 1$ $t = 4 \implies u = 6(4) + 1 = 25$ Therefore $\int_0^4 \frac{1}{6t+1} dt = \int_1^{25} \frac{1}{u}(\frac{1}{6}du)$ $= \frac{1}{6} \int_1^{25} \frac{1}{u}du$

38. anonymous

Make more sense?

39. anonymous

The only thing I'm confused about is where $\int\limits_{1}^{25}$ comes from

40. anonymous

See the t=0 and t=4 lines? If we are evaluating t from 0 to 4, that means u goes from 1 to 25.

41. anonymous

Ohhh, okay, I see

42. myininaya

43. anonymous

But, the original equation it's $\sqrt{6t+1}$ What does that change in your answer?

44. myininaya

i just posted it

45. myininaya

1 over that right?

46. anonymous

Yes

47. anonymous

Wow, thanks for that effort in explaining, that helps so much!

48. anonymous

Oh, sorry didn't see the exponent. It changes things some in that it's $$1/u^2$$ instead of 1/u, but that just changes the end result, not too much in what I posted.

49. anonymous

Ok, thank you both, very much appreciated guys, I have another question as well if you want to answer it.

50. myininaya

51. anonymous

The population of a certain community (t) years after the year 2000 is given by $P(t)= e ^{0.2t}\div4+e ^{0.2t} million people$ what was the average population from 2000-2010?

52. myininaya

for 2000, t=0 is that right? t=1?

53. anonymous

I'm not sure, that's all I was given and I'm not too sharp in this subject of integrals yet, sorry.

54. myininaya

polpak come back? lol

55. anonymous

Haha

56. myininaya

we can work on integrating P(t) together let me look at it, and I will scan it in k?

57. anonymous

Thank you so much! Please! lol

58. myininaya

hey so we have [e^(0.2t)]/4 + e^(0.2t) ={5e^(0.2t)}/4?

59. anonymous

I'm so stressed out. Lol

60. myininaya

61. anonymous

Where did the 5 come from?

62. myininaya

which 5?

63. anonymous

Oh duh, I was looking at it too hard.

64. myininaya

so you are teaching yourself?

65. anonymous

Yes

66. myininaya

thats cool. we need to figure out what limits to use? its ether 1 to 11 or 0 to 10 depending if we start 2000 with t=1 or t=0 i will get polpak and see if he knows

67. anonymous

Ok, so how do you determine what limits to use?

68. myininaya

well it wants to know the average population from 2000 to 2010 so from there we can figure out the limits

69. anonymous

Ok, so the t=1 or t=0 would represent the year 2000, as the 10/11 would represent the 10 year time frame from 2000.

70. myininaya

right

71. anonymous

Alrighty, so what difference would it make to use either one of the limits?

72. anonymous

Sorry , what's up?

73. anonymous

Oh, 2000 is t=0

74. myininaya

ok cool lol

75. anonymous

"(t) years after the year 2000" so if t = 0 the year is 2000.

76. anonymous

t=10 would be 2010

77. myininaya

ok so 2010 would be t=10 so we have our upper and lower limit just use the fundamental thm of calculus on the antiderivative we found

78. myininaya

right i was not sure at first if it was 1 to 11 or 0 to 10 thanks polpak

79. anonymous

np =)

80. anonymous

Thank you polpak!

81. myininaya

so you do [25e^(0.2*10)]/4 - [25e^(0*10)]/4

82. anonymous

What did you get after evaluation?

83. anonymous

I got 40 million people