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hba

  • 3 years ago

Integrate

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  1. hba
    • 3 years ago
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    \[\int\limits _{0}^{\infty} \lfloor x \rfloor e^{-x} dx\]

  2. hba
    • 3 years ago
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    I am getting 1/(e-1)

  3. Frostbite
    • 3 years ago
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    absolute value of x right?

  4. hba
    • 3 years ago
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    Right.

  5. Mathmuse
    • 3 years ago
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    I get 1, but i'll check again

  6. Frostbite
    • 3 years ago
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    Hmm can't do the integral but the aproximation I do get to 1.

  7. hartnn
    • 3 years ago
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    that initially seemed like floor value of x..... :P

  8. Frostbite
    • 3 years ago
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    trid partial integration?

  9. hartnn
    • 3 years ago
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    i also get 1.

  10. hba
    • 3 years ago
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    Yeah one more which i was doing was Integral (x^2/x^2+1) and i got x- arctanx by trig sub.

  11. hartnn
    • 3 years ago
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    thats correct....and can be done without substitution...

  12. hartnn
    • 3 years ago
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    if u can use this integral as standard int 1/(1+x^2) dx = arctan x +c

  13. Frostbite
    • 3 years ago
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    btw forget what I said about partial integration... looks like geting no where.

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  14. Mathmuse
    • 3 years ago
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    after IBP, i get the integral of: \[\large [-e^{-x}(x+1)]^\infty_0\] i made the leap that -e^(-x) tends to zero much faster than x+1 tends to infinity as x goes to infinity, but can't give a real justification right now

  15. Frostbite
    • 3 years ago
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    Looks just about right Mathmuse. Seems like I made a bad choise when I did IBP.

  16. Mathmuse
    • 3 years ago
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    looks the same, wots 'sign(x)'?

  17. Frostbite
    • 3 years ago
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    The sign function?

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