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## mariomintchev 3 years ago i need to see how you solve problem #18

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1. mariomintchev

2. 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.

3. iHelp

answer is D, 1/t^2 = t^-2, integral of that is -1/t

4. mariomintchev

i know the answer. they give it to me. lol i want to see how you get the answer.

5. 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 $$2/15$$

6. mariomintchev

i dont see how you got -1/t

7. KingGeorge

$\large \int\limits t^{-2} dt = {1 \over -2+1}\cdot t^{-2+1}=-1 t^{-1}=-\,{1 \over t}$

8. mariomintchev

ah this is too complex. no easier way to solve this?

9. 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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