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KingGeorge Group TitleBest 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.
 2 years ago

iHelp Group TitleBest ResponseYou've already chosen the best response.0
answer is D, 1/t^2 = t^2, integral of that is 1/t
 2 years ago

mariomintchev Group TitleBest 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.
 2 years ago

KingGeorge Group TitleBest 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 \(2/15\)
 2 years ago

mariomintchev Group TitleBest ResponseYou've already chosen the best response.0
i dont see how you got 1/t
 2 years ago

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

mariomintchev Group TitleBest ResponseYou've already chosen the best response.0
ah this is too complex. no easier way to solve this?
 2 years ago

KingGeorge Group TitleBest 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 \(n=2\)
 2 years ago
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