anonymous
  • anonymous
i need to see how you solve problem #18
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
KingGeorge
  • KingGeorge
\[\left({1 \over t}\right)^2 ={1 \over t^2}=t^{-2}\]From here, you can use the reverse power rule to integrate, and then evaluate.
anonymous
  • anonymous
answer is D, 1/t^2 = t^-2, integral of that is -1/t

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anonymous
  • anonymous
i know the answer. they give it to me. lol i want to see how you get the answer.
KingGeorge
  • KingGeorge
\[\large \int\limits_3^5 t^{-2}\;\; dt =-\,{1 \over t} \;\;\Big|^5_3=-\,{1 \over 5} + {1 \over 3}\]From this, you get that the answer is \(2/15\)
anonymous
  • anonymous
i dont see how you got -1/t
KingGeorge
  • KingGeorge
\[\large \int\limits t^{-2} dt = {1 \over -2+1}\cdot t^{-2+1}=-1 t^{-1}=-\,{1 \over t}\]
anonymous
  • anonymous
ah this is too complex. no easier way to solve this?
KingGeorge
  • KingGeorge
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 \(n=-2\)

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