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anonymous

  • 5 years ago

1/(6t+1)^1/2 dt evaluated from 0 to 4

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  1. anonymous
    • 5 years ago
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    The integral of that you mean?

  2. anonymous
    • 5 years ago
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    Yes, sorry I wasn't clear

  3. anonymous
    • 5 years ago
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    Definite Integral

  4. myininaya
    • 5 years ago
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    let u=6t+1

  5. anonymous
    • 5 years ago
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    Use a u substation as myininaya has just suggested ;p

  6. anonymous
    • 5 years ago
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    substitution I mean.

  7. anonymous
    • 5 years ago
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    I have to use the Fundamental Theorem of Calculus

  8. myininaya
    • 5 years ago
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    we will

  9. anonymous
    • 5 years ago
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    Ok

  10. myininaya
    • 5 years ago
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    there is no other way unless you used riemann sums but lets not mention him anymore

  11. anonymous
    • 5 years ago
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    haha

  12. myininaya
    • 5 years ago
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    using riemann sums can also be very difficult to evaluate the limit

  13. myininaya
    • 5 years ago
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    or maybe impossible

  14. anonymous
    • 5 years ago
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    Well I'm teaching myself Integrals so I'm not quite that far yet.

  15. anonymous
    • 5 years ago
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    I'm just having trouble finding where to start with these functions.

  16. myininaya
    • 5 years ago
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    do you know substation?

  17. anonymous
    • 5 years ago
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    substitution ;p

  18. myininaya
    • 5 years ago
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    stupid thingy i hit whatever popped up because i hate spelling substitution

  19. anonymous
    • 5 years ago
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    You have to learn u substitutions to do these they allow you to undo the chain rule.

  20. anonymous
    • 5 years ago
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    Ok, So after the substitution, What's my next step?

  21. myininaya
    • 5 years ago
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    if u=6t+1 then du/dt=.....

  22. anonymous
    • 5 years ago
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    6

  23. myininaya
    • 5 years ago
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    right do du=....

  24. anonymous
    • 5 years ago
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    6 dt?

  25. myininaya
    • 5 years ago
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    right

  26. anonymous
    • 5 years ago
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    Then I hit both sides with integrals, correct?

  27. anonymous
    • 5 years ago
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    No. Then you substitute 6t +1 as u, and your dt as 1/6 du.

  28. anonymous
    • 5 years ago
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    And take the integral with respect to u (though you also have to change your limits)

  29. anonymous
    • 5 years ago
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    Oh, ok, that's where I get lost.

  30. myininaya
    • 5 years ago
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    polpak divided both sides by 6 of that equation du=6dt

  31. anonymous
    • 5 years ago
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    Well 6t + 1 = u the 1/(6t+1) = 1/u

  32. myininaya
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    Since 1/6 is just a multiplicative constant you can pull it out front.

  35. myininaya
    • 5 years ago
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    when i say x i meant t lol

  36. anonymous
    • 5 years ago
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    Hmmmm...

  37. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Make more sense?

  39. anonymous
    • 5 years ago
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    The only thing I'm confused about is where \[\int\limits_{1}^{25}\] comes from

  40. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Ohhh, okay, I see

  42. myininaya
    • 5 years ago
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  43. anonymous
    • 5 years ago
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    But, the original equation it's \[\sqrt{6t+1}\] What does that change in your answer?

  44. myininaya
    • 5 years ago
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    i just posted it

  45. myininaya
    • 5 years ago
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    1 over that right?

  46. anonymous
    • 5 years ago
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    Yes

  47. anonymous
    • 5 years ago
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    Wow, thanks for that effort in explaining, that helps so much!

  48. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Ok, thank you both, very much appreciated guys, I have another question as well if you want to answer it.

  50. myininaya
    • 5 years ago
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  51. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    for 2000, t=0 is that right? t=1?

  53. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    polpak come back? lol

  55. anonymous
    • 5 years ago
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    Haha

  56. myininaya
    • 5 years ago
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    we can work on integrating P(t) together let me look at it, and I will scan it in k?

  57. anonymous
    • 5 years ago
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    Thank you so much! Please! lol

  58. myininaya
    • 5 years ago
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    hey so we have [e^(0.2t)]/4 + e^(0.2t) ={5e^(0.2t)}/4?

  59. anonymous
    • 5 years ago
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    I'm so stressed out. Lol

  60. myininaya
    • 5 years ago
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  61. anonymous
    • 5 years ago
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    Where did the 5 come from?

  62. myininaya
    • 5 years ago
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    which 5?

  63. anonymous
    • 5 years ago
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    Oh duh, I was looking at it too hard.

  64. myininaya
    • 5 years ago
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    so you are teaching yourself?

  65. anonymous
    • 5 years ago
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    Yes

  66. myininaya
    • 5 years ago
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    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
    • 5 years ago
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    Ok, so how do you determine what limits to use?

  68. myininaya
    • 5 years ago
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    well it wants to know the average population from 2000 to 2010 so from there we can figure out the limits

  69. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    right

  71. anonymous
    • 5 years ago
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    Alrighty, so what difference would it make to use either one of the limits?

  72. anonymous
    • 5 years ago
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    Sorry , what's up?

  73. anonymous
    • 5 years ago
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    Oh, 2000 is t=0

  74. myininaya
    • 5 years ago
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    ok cool lol

  75. anonymous
    • 5 years ago
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    "(t) years after the year 2000" so if t = 0 the year is 2000.

  76. anonymous
    • 5 years ago
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    t=10 would be 2010

  77. myininaya
    • 5 years ago
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    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
    • 5 years ago
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    right i was not sure at first if it was 1 to 11 or 0 to 10 thanks polpak

  79. anonymous
    • 5 years ago
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    np =)

  80. anonymous
    • 5 years ago
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    Thank you polpak!

  81. myininaya
    • 5 years ago
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    so you do [25e^(0.2*10)]/4 - [25e^(0*10)]/4

  82. anonymous
    • 5 years ago
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    What did you get after evaluation?

  83. anonymous
    • 5 years ago
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    I got 40 million people

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