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ksaimouli Group Title

http://online.math.uh.edu/apcalculus/exams/images/AP_AB_version1__37.gif

  • one year ago
  • one year ago

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  1. ksaimouli Group Title
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    @mathmate

    • one year ago
  2. ksaimouli Group Title
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    i did this with u substituion it did not work

    • one year ago
  3. ksaimouli Group Title
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    i am getting something like ln2/2

    • one year ago
  4. mathmate Group Title
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    It looks to me more an atan(2x)

    • one year ago
  5. ksaimouli Group Title
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    how

    • one year ago
  6. ksaimouli Group Title
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    i took u=1+4t^2

    • one year ago
  7. ksaimouli Group Title
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    du/8t=dt

    • one year ago
  8. mathmate Group Title
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    The standard integral is \( \int \frac{dx}{1+x^2} = atan(x) + C \)

    • one year ago
  9. ksaimouli Group Title
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    |dw:1358025284718:dw|

    • one year ago
  10. mathmate Group Title
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    So you'd be integrating \( \int \frac{du}{8u(1+4t^2)} \)

    • one year ago
  11. ksaimouli Group Title
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    use the equation button

    • one year ago
  12. mathmate Group Title
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    The result is not ln(u)

    • one year ago
  13. ksaimouli Group Title
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    ok i should remember that formula right?

    • one year ago
  14. mathmate Group Title
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    I suggest you use y=2t

    • one year ago
  15. mathmate Group Title
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    Yes, it's on the first page of standard integrals. Check it out.

    • one year ago
  16. ksaimouli Group Title
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    ok

    • one year ago
  17. mathmate Group Title
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    \( \large \int_0^{1/2} \frac{dt}{1+4t^2}= \int_0^{1} \frac{2dy}{1+y^2} = [atan(y)]_0^{1}= 2\pi/4=\pi/2 \)

    • one year ago
  18. ksaimouli Group Title
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    @mathmate i found the formula to be \[\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a }+c\]

    • one year ago
  19. mathmate Group Title
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    Right, so a=1/2.

    • one year ago
  20. ksaimouli Group Title
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    what is x then

    • one year ago
  21. ksaimouli Group Title
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    \[\int\limits_{}^{}\frac{ 1 }{ a^2+x^2}=\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a}+c\]

    • one year ago
  22. mathmate Group Title
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    x is your t. you can replace the identity with \( \large \int \frac{dx}{a^2+x^2} = \int \frac{dt}{a^2(1+(t/a)^2)} \)

    • one year ago
  23. mathmate Group Title
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    and a=1/2

    • one year ago
  24. ksaimouli Group Title
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    can i do \[\frac{ 1 }{ 1 }\tan^{-1} \frac{ 2t }{ 1}\]

    • one year ago
  25. ksaimouli Group Title
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    and t=1/2

    • one year ago
  26. mathmate Group Title
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    t is a variable, and you cannot equate it to a constant. Use the formula you found, change t for x, and 1/2 for a, then the left-hand side is exactly your question. All you need to do after that is to evaluate the right-hand side.

    • one year ago
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