Question

Really quick help with ln and integrals!

Answer 1

\[\int\limits_{1}^{2}\frac{ dt }{ 8-3t }\]

Answer 2

so using the
ln(x)=1/x
then I would just get ln(8-3t) evaluated at 1,2
but that doesn't work

Answer 3

@rvc @inkyvoyd

Answer 4

I end up with something like ln(2/5)
when it's meant to be
(1/3)ln(5/2)

Answer 5

Oh I figured it out
\[u=8-3t\]\[du=-3dt = \frac{ du }{ -3 }\]\[\frac{ -1 }{ 3 }\int\limits_{1}^{2}\frac{ du }{ u }\]\[\frac{ 1 }{ 3 }\int\limits_{2}^{1}\frac{ du }{ u }\]\[\frac{ 1 }{ 3 }\ln(u)|^1_2\]\[\frac{ 1 }{ 3 }\ln \frac{ 5 }{ 2 }\]

Answer 6

Nice :)