$t = 0.2u \implies dt = 0.2 \, du$
substitute and reset the limits of integration
$\displaystyle \lim_{b \to \infty} \int_{2}^b e^{-t} \, dt$
$\displaystyle \lim_{b \to \infty} \bigg[-e^{-t} \bigg]_2^b$
$\displaystyle -\lim_{b \to \infty} \bigg[e^{-b} - e^{-2} \bigg] = -\bigg[0 - \dfrac{1}{e^2} \bigg] = \dfrac{1}{e^2}$