I'm assuming "actual" means the total probability of the PDF (the integral from $-\infty to \infty$) must be 1, so
$$\int\limits_{-\infty}^{\infty} ke^{-0.1t}dt = 1$$
Wolfram Alpha seems to be ambiguous regarding the antiderivative of this function, so can somene write it down for me? I'm sure I can complete this as soon as I have that.
We must have $$\int_{-\infty}^\infty f_T(t)\,dt=1.$$ Note that in the material you quoted, it says that the density function is $0$ for $t\lt 0$. Thus $$\int_{-\infty}^\infty f_T(t)\,dt=\int_{-\infty}^0 0\,dt+\int_{0}^\infty ke^{-0.1t}\,dt.$$ So we want to choose $k$ such that $$\int_{0}^\infty ke^{-0.1t}\,dt=1.$$
An antiderivative of $ke^{-0.1 t}$ is $-\frac{k}{0.1}e^{-0.1t}$.