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Math Help - [SOLVED] Integration?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Integration?

    If ω = 2π/T then 1/T ∫ from 0 to T of 1/2*m*ω^2*A^2*cos^2 ωt dt=?

    A)1/2 m*ω^2*A^2
    B) zero
    C) 1/4 mω^2 A^2
    D) 1

    I got B) zero as the answer.
    Is it correct?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    If ω = 2π/T then 1/T ∫ from 0 to T of 1/2*m*ω^2*A^2*cos^2 ωt dt=?

    A)1/2 m*ω^2*A^2
    B) zero
    C) 1/4 mω^2 A^2
    D) 1

    I got B) zero as the answer.
    Is it correct?
    i am pretty sure B) is wrong. running it through in my head (not sure how confident you want to be in that ) i got C)

    do you get zero when you do \int_0^T \cos^2 \omega t~dt ? nope. so the answer can't be zero
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    If ω = 2π/T then 1/T ∫ from 0 to T of 1/2*m*ω^2*A^2*cos^2 ωt dt=?

    A)1/2 m*ω^2*A^2
    B) zero
    C) 1/4 mω^2 A^2
    D) 1

    I got B) zero as the answer.
    Is it correct?
    I get c, I am not sure how you would get zero for this answer.
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  4. #4
    Super Member fardeen_gen's Avatar
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    What is ? I dunno...
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    What is ? I dunno...
    \cos(\omega{t})^2=\frac{1+\cos(2\omega{t})}{2}

    So we have

    \frac{1}{2}\int\bigg[1+\cos(2\omega{t})\bigg]dt=\frac{1}{2}\bigg[t+\frac{1}{2\omega}\sin(2\omega{t})\bigg]+\bold{\color{red}C}
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    ... +\bold{\color{red}C}
    Aha! you remembered!
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Jhevon View Post
    Aha! you remembered!
    I can't believe you got that
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