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KingGeorgeBest ResponseYou've already chosen the best response.1
\[\left({1 \over t}\right)^2 ={1 \over t^2}=t^{2}\]From here, you can use the reverse power rule to integrate, and then evaluate.
 one year ago

iHelpBest ResponseYou've already chosen the best response.0
answer is D, 1/t^2 = t^2, integral of that is 1/t
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
i know the answer. they give it to me. lol i want to see how you get the answer.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
\[\large \int\limits_3^5 t^{2}\;\; dt =\,{1 \over t} \;\;\Big^5_3=\,{1 \over 5} + {1 \over 3}\]From this, you get that the answer is \(2/15\)
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
i dont see how you got 1/t
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
\[\large \int\limits t^{2} dt = {1 \over 2+1}\cdot t^{2+1}=1 t^{1}=\,{1 \over t}\]
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
ah this is too complex. no easier way to solve this?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
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 \(n=2\)
 one year ago
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