## mariomintchev Group Title i need to see how you solve problem #18 2 years ago 2 years ago

1. mariomintchev Group Title

2. KingGeorge Group Title

$\left({1 \over t}\right)^2 ={1 \over t^2}=t^{-2}$From here, you can use the reverse power rule to integrate, and then evaluate.

3. iHelp Group Title

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

4. mariomintchev Group Title

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

5. KingGeorge Group Title

$\large \int\limits_3^5 t^{-2}\;\; dt =-\,{1 \over t} \;\;\Big|^5_3=-\,{1 \over 5} + {1 \over 3}$From this, you get that the answer is $$2/15$$

6. mariomintchev Group Title

i dont see how you got -1/t

7. KingGeorge Group Title

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

8. mariomintchev Group Title

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

9. KingGeorge Group Title

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 $$n=-2$$