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mariomintchev Group Title

i need to see how you solve problem #18

  • 2 years ago
  • 2 years ago

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  1. mariomintchev Group Title
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    • 2 years ago
  2. KingGeorge Group Title
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    \[\left({1 \over t}\right)^2 ={1 \over t^2}=t^{-2}\]From here, you can use the reverse power rule to integrate, and then evaluate.

    • 2 years ago
  3. iHelp Group Title
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    answer is D, 1/t^2 = t^-2, integral of that is -1/t

    • 2 years ago
  4. mariomintchev Group Title
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    i know the answer. they give it to me. lol i want to see how you get the answer.

    • 2 years ago
  5. KingGeorge Group Title
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    \[\large \int\limits_3^5 t^{-2}\;\; dt =-\,{1 \over t} \;\;\Big|^5_3=-\,{1 \over 5} + {1 \over 3}\]From this, you get that the answer is \(2/15\)

    • 2 years ago
  6. mariomintchev Group Title
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    i dont see how you got -1/t

    • 2 years ago
  7. KingGeorge Group Title
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    \[\large \int\limits t^{-2} dt = {1 \over -2+1}\cdot t^{-2+1}=-1 t^{-1}=-\,{1 \over t}\]

    • 2 years ago
  8. mariomintchev Group Title
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    ah this is too complex. no easier way to solve this?

    • 2 years ago
  9. KingGeorge Group Title
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    This is just the reverse of the power rule. In general, \[\large \int\limits x^n\;\; dx ={1 \over n+1}\cdot x^{n+1}\]Where I just used \(n=-2\)

    • 2 years ago
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