if anyone can help me with this, that would be great!
this is from ln3 to ln2
∫ 3/(e^(3t)) dt=???
If you are not getting what the above post says:
let $\displaystyle u = 3t$
then, $\displaystyle du = 3 dt $
so,$\displaystyle dt= \frac{du}{3}$
so you have
$\displaystyle \int_{ln3}^{ln2} {3 e^{-u}}\frac{du}{3} $
= $\displaystyle [-e^{-u}]_{ln3}^{ln2} $
substitute u = 3t, you get
$\displaystyle [-e^{-3t}]_{ln3}^{ln2}$
calculate the value for the given upper and lower limits