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jennychan12

  • 3 years ago

Find an antiderivative F of f(x) = x^2*sin(x^2) such that F(1) = 0.

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  1. jennychan12
    • 3 years ago
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    I don't think you can use u-sub here. cuz then it'd be f(u) = u*sinu..

  2. jennychan12
    • 3 years ago
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    and i haven't learned trig sub or integration by parts yet.

  3. Argos
    • 3 years ago
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    jenny you are a liar i think

  4. jennychan12
    • 3 years ago
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    -_- i haven't. but i can do integration by parts. but my teacher never taught it.

  5. Argos
    • 3 years ago
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    becos that problem required fresnal integral

  6. jennychan12
    • 3 years ago
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    what?

  7. wio
    • 3 years ago
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    Use \(u = x^2\). Then \(x = \sqrt{u}\)

  8. Argos
    • 3 years ago
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    u can use u sub on that

  9. jennychan12
    • 3 years ago
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    \[f(x) = x^2\sin(x^2) \]

  10. jennychan12
    • 3 years ago
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    if u = x^2 then du = 2xdx

  11. Argos
    • 3 years ago
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    well what method are you allowed to use if you can't use trig or parts

  12. jennychan12
    • 3 years ago
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    that's what i don't understand. u-sub doesn't really work...

  13. Argos
    • 3 years ago
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    so ur on u-sub section of the book?

  14. jennychan12
    • 3 years ago
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    no it's a review packet. i learned area, u-sub, trapezoid rule, and simpson's rule

  15. igotzbeard
    • 3 years ago
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    thats the way it is

  16. igotzbeard
    • 3 years ago
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    not an exponent

  17. jennychan12
    • 3 years ago
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    no it's \[f(x) = x^2\sin(x^2)\]

  18. wio
    • 3 years ago
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    This isn't a candidate for u sub.

  19. igotzbeard
    • 3 years ago
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    u = x^2 du = 2x dx cant use u du sub

  20. igotzbeard
    • 3 years ago
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    no place to apply a substitution

  21. wio
    • 3 years ago
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    \[ \frac{1}{2}\int \sqrt{u}\sin(u)du \]Isn't helpful.

  22. k.rajabhishek
    • 3 years ago
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    |dw:1357267172296:dw|

  23. jennychan12
    • 3 years ago
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    -_- it was in the textbook in the chapter 5. we don't do trig sub/parts until chapter 8...

  24. LogicalApple
    • 3 years ago
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    whats the name of the textbook

  25. jennychan12
    • 3 years ago
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    although i know how to do parts but my teacher never taught it.

  26. jennychan12
    • 3 years ago
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    stewart

  27. LogicalApple
    • 3 years ago
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    7th edition?

  28. jennychan12
    • 3 years ago
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    5th

  29. LogicalApple
    • 3 years ago
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    which chapter section?

  30. jennychan12
    • 3 years ago
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    chapter 5 review #49

  31. Argos
    • 3 years ago
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    integral x^2 sin(x^2) dx = 1/4 (sqrt(2 pi) C(sqrt(2/pi) x)-2 x cos(x^2))+constant C(x) is the fresnal C integral Medal pls

  32. Argos
    • 3 years ago
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    fresnal function is introduced in section 5.3

  33. igotzbeard
    • 3 years ago
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    lol wolframalpha? xD

  34. jennychan12
    • 3 years ago
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    haha same 5.3 is fundemental theorum

  35. Argos
    • 3 years ago
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    go to 5.3 examlpe number 3

  36. jennychan12
    • 3 years ago
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    we don't use stewart. we use larson. my teacher just photocopied this from the stewart textbook

  37. Argos
    • 3 years ago
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    thats illegail

  38. Argos
    • 3 years ago
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    but u gonna have to look up fresnal function elsewhere then

  39. jennychan12
    • 3 years ago
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    she owns like ten stewart textbooks. -_- i'll just use parts then...

  40. LogicalApple
    • 3 years ago
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    I guess one way to approach it is: \[F(t) = \int\limits_{1}^{x} t ^{2} \sin(t ^{2})\] This way the integral evaluated at x = 1 results in 0, and its derivative is just the original function

  41. LogicalApple
    • 3 years ago
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    wait let me rewrite that

  42. LogicalApple
    • 3 years ago
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    \[F(x) = \int\limits\limits_{1}^{x} t ^{2} \sin(t ^{2}) dt\]

  43. LogicalApple
    • 3 years ago
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    That way when it is evaluated at F(1), you still get 0. And the derivative is the original function.

  44. wio
    • 3 years ago
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    If you used Simpson's rule or Midpoint rule on that, you'd get a nasty summation but it is within the rules.

  45. jennychan12
    • 3 years ago
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    oh wow. thanks @LogicalApple that's the answer in the book -___-

  46. LogicalApple
    • 3 years ago
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    Well it makes sense, right? We don't actually have to take an integration. We know from the fundamental theorem of calculus (either part 1 or 2, i forget) that the derivative of F(x) here is just the inside of the integral evaluated at x. And for the limits of integration, we know that if we integrate from a to a we always get 0. So let a = 1, that way F(1) is an integral evaluated from 1 to 1.

  47. jennychan12
    • 3 years ago
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    yes but i never thought of it that way... *sigh*

  48. LogicalApple
    • 3 years ago
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    Stewart doesn't mess around.. I'm in chapter 5 on the 7th edition.

  49. jennychan12
    • 3 years ago
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    oh same here i think. my teacher uses both stewart and larson. we teach from larson.

  50. LogicalApple
    • 3 years ago
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    I will have to check that one out. Anyway good luck with your studies!

  51. jennychan12
    • 3 years ago
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    thanks :)

  52. wio
    • 3 years ago
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    I think it's sort of an unfair question. Should have specified a bit more the rigor.

  53. jennychan12
    • 3 years ago
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    same. my teacher likes stewart because they have "more pictures" and "good problems". which means that she puts them on quizzes and tests. :[

  54. jennychan12
    • 3 years ago
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    i guess it just makes u think about the concept more?

  55. wio
    • 3 years ago
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    If they said something like "in terms of an indefinite integral" then it would have been fair.

  56. LogicalApple
    • 3 years ago
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    Yeah sometimes Stewart's questions are very very chapter-specific That's why I asked what chapter the problem was in to get an idea of the context of the question.

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