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Ldaniel

  • 4 years ago

Consider the function f(x) = 2sin(x2) on the interval 0 ≤ x ≤ 3. (a) Find the exact value in the given interval where an antiderivative, F, reaches its maximum. x = (b) If F(1) = 9, estimate the maximum value attained by F. (Round your answer to three decimal places.) y ≈

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  1. inkyvoyd
    • 4 years ago
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    set 2sin(x^2)=0

  2. inkyvoyd
    • 4 years ago
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    Wait, over an interval -_-

  3. inkyvoyd
    • 4 years ago
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    Ignore the interval, and just solve that trig equation.

  4. Ldaniel
    • 4 years ago
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    i know that the answer for part "a" is

  5. Ldaniel
    • 4 years ago
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    square(pi)

  6. inkyvoyd
    • 4 years ago
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    square or sqrt?

  7. Ldaniel
    • 4 years ago
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    i dont know how to do part b

  8. Ldaniel
    • 4 years ago
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    sqrt

  9. inkyvoyd
    • 4 years ago
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    Well, your interval s defined only up to 3. Is that supposed to be that way, or is the three suposed to be pi?

  10. Ldaniel
    • 4 years ago
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    up to 3

  11. RaphaelFilgueiras
    • 4 years ago
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    a)\[2*\sin(2*x)=0\rightarrow x=k*pi/2\] In this interval x=pi/2

  12. inkyvoyd
    • 4 years ago
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    Do you understand a though?

  13. Ldaniel
    • 4 years ago
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    well i got x= |dw:1334814197335:dw|

  14. inkyvoyd
    • 4 years ago
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    |dw:1334814169184:dw|

  15. Ldaniel
    • 4 years ago
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    and is the right answer

  16. inkyvoyd
    • 4 years ago
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    |dw:1334814236680:dw|

  17. inkyvoyd
    • 4 years ago
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    Naturally, sqrt pi would be the largest value up to 3, because it squares out to pi.

  18. Ldaniel
    • 4 years ago
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    the equation is f(x) = 2sin(x^2)

  19. Ldaniel
    • 4 years ago
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    sorry about that

  20. RaphaelFilgueiras
    • 4 years ago
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    \[F(x)=-\cos(2*x)+K'\] \[F(1)=-\cos(2*1)+K'=9\]

  21. inkyvoyd
    • 4 years ago
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    I know - That is sin x

  22. inkyvoyd
    • 4 years ago
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    but, sin x^2 is similar to the graph in that it has its highest values at sqrtpi

  23. Ldaniel
    • 4 years ago
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    sqrt(pi) is the right answer for part a is marking it right

  24. Ldaniel
    • 4 years ago
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    the part that i dont get how to do is part b

  25. inkyvoyd
    • 4 years ago
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    Yes. Can you tell me what calc course you are enrolled in?

  26. Ldaniel
    • 4 years ago
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    calculus I

  27. inkyvoyd
    • 4 years ago
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    Alright, give me a second.

  28. Ldaniel
    • 4 years ago
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    2sin(x^2) = 0 x^2 = pi x = sqrt(pi)

  29. Ldaniel
    • 4 years ago
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    thats how i got my answer

  30. inkyvoyd
    • 4 years ago
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    Have you learned how to approximate integrals?

  31. Ldaniel
    • 4 years ago
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    kinda

  32. Ldaniel
    • 4 years ago
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    thats what we are learning

  33. inkyvoyd
    • 4 years ago
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    What's it called?

  34. Ldaniel
    • 4 years ago
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    antiderivatives

  35. inkyvoyd
    • 4 years ago
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    No, what kind of approximations have you learned?

  36. inkyvoyd
    • 4 years ago
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    I just figured out how I can solve the problem.

  37. inkyvoyd
    • 4 years ago
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    Here's the graph, btw

  38. Ldaniel
    • 4 years ago
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    i dont know

  39. inkyvoyd
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=integrate+2+sin%28x%5E2%29

  40. Ldaniel
    • 4 years ago
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    the tittle of this chapter is antiderivatives

  41. inkyvoyd
    • 4 years ago
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    Note how the function uses something called the fresnal S, which I don't know either. But mor eimportantly, notice how it gets to its peak at about 1,7

  42. Ldaniel
    • 4 years ago
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    so how do i do part b?

  43. inkyvoyd
    • 4 years ago
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    I'm thinking ;)

  44. inkyvoyd
    • 4 years ago
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    Have you learned trapozoidal approximation?

  45. Ldaniel
    • 4 years ago
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    no

  46. noginogi
    • 4 years ago
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    I also have a question for part b... help us figure this out plz

  47. inkyvoyd
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=+2+sin%28x%5E2%29

  48. inkyvoyd
    • 4 years ago
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    Alright. that is the graph of f(x)=2 sin(x^2) right?

  49. inkyvoyd
    • 4 years ago
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    Notice how the waves become wider and wider close to 0.

  50. inkyvoyd
    • 4 years ago
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    after that, they become thinner and thinner.

  51. inkyvoyd
    • 4 years ago
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    Now, give me a second to explain something.

  52. inkyvoyd
    • 4 years ago
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    Let me get htis plot into mathematica, and show you what I'm talking about.

  53. Ldaniel
    • 4 years ago
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    k

  54. inkyvoyd
    • 4 years ago
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    1 Attachment
  55. inkyvoyd
    • 4 years ago
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    Ok. The area of a>b>c>d Tell me if you understand.

  56. inkyvoyd
    • 4 years ago
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    @Ldaniel ?

  57. Ldaniel
    • 4 years ago
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    yes i do

  58. inkyvoyd
    • 4 years ago
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    So, because of this, we know that from 0 to 3, the area enclosed in a is the greatest

  59. Ldaniel
    • 4 years ago
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    yes

  60. inkyvoyd
    • 4 years ago
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    So we are essentially looking for the area there.

  61. inkyvoyd
    • 4 years ago
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    @FoolForMath , I give up/

  62. Ldaniel
    • 4 years ago
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    ;/

  63. inkyvoyd
    • 4 years ago
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    I can give you the answer though, me being a cheater cheater.

  64. noginogi
    • 4 years ago
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    what's the answer?

  65. mr.luna
    • 4 years ago
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    i wish i could do this math. -_-

  66. inkyvoyd
    • 4 years ago
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    Give me a second.

  67. inkyvoyd
    • 4 years ago
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    I'm going to numerically evaluate the integral with wolfram Mathematica

  68. inkyvoyd
    • 4 years ago
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    1.78966

  69. inkyvoyd
    • 4 years ago
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    @lgbasallote , here you go.

  70. Ldaniel
    • 4 years ago
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    the answer is wrong

  71. inkyvoyd
    • 4 years ago
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    WHAT?

  72. inkyvoyd
    • 4 years ago
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    Try the square root of pi then.

  73. Ldaniel
    • 4 years ago
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    as the answer?

  74. inkyvoyd
    • 4 years ago
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    Yes.

  75. Ldaniel
    • 4 years ago
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    wrong

  76. inkyvoyd
    • 4 years ago
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    btw, try 1.790 if you haven't already.

  77. Ldaniel
    • 4 years ago
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    i did its wrong

  78. inkyvoyd
    • 4 years ago
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    Ahem.

  79. Ldaniel
    • 4 years ago
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    ;(

  80. lgbasallote
    • 4 years ago
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    lol why call me =)) i only know algebra :P i dont like these stuffs

  81. Ldaniel
    • 4 years ago
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    :((

  82. inkyvoyd
    • 4 years ago
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    Wrong, call >.<

  83. inkyvoyd
    • 4 years ago
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    So, you are supposed to use an estimation method that isn't accurate, nor do I know. Or, the problem is set up funy.

  84. inkyvoyd
    • 4 years ago
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    Got any more info about what you are learning right now?

  85. inkyvoyd
    • 4 years ago
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    If I used the methods they gave, I might just get the answer they are looking for.

  86. Ldaniel
    • 4 years ago
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    i got the answer is y = 10.16912...

  87. inkyvoyd
    • 4 years ago
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    That's th correct answer?

  88. Ldaniel
    • 4 years ago
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    |dw:1334818371099:dw|

  89. Ldaniel
    • 4 years ago
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    yes

  90. dumbcow
    • 4 years ago
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    so\[\int\limits_{1}^{\sqrt{\pi}}f(x) = F(\sqrt{\pi}) - F(1)\] all you need to do is find approximation for definite integral

  91. inkyvoyd
    • 4 years ago
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    Well, then I'm completely useless.

  92. Ldaniel
    • 4 years ago
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    10.170 is the answer

  93. inkyvoyd
    • 4 years ago
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    Nice job.

  94. Ldaniel
    • 4 years ago
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    thanks anyways

  95. Ldaniel
    • 4 years ago
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    how old are you?

  96. inkyvoyd
    • 4 years ago
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    15

  97. inkyvoyd
    • 4 years ago
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    I'm still not sure how I missed that problem so badly. :/

  98. Ldaniel
    • 4 years ago
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    15? really?

  99. dumbcow
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=integrate+2sin%28x^2%29+dx+from+1+to+sqrt%28pi%29 add 9 to get answer

  100. Ldaniel
    • 4 years ago
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    yes dumbcow

  101. dumbcow
    • 4 years ago
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    sorry for being redundant, i know you already got the answer just putting it out there for everyones benefit

  102. apoorvk
    • 4 years ago
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    I believe graph would go like this. (the pitch decreasing with increase in x) |dw:1334819574505:dw|

  103. inkyvoyd
    • 4 years ago
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    btw, yes, I'm trying to jump the gun and finish calculus and some math afterwards before I enter college.

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