Answer :
The average value of a continuous function f(x) over an interval [a, b] is given by the integral,
[tex]\displaystyle \frac1{b-a}\int_a^b f(x)\,\mathrm dx[/tex]
Compute the integral for f(x) = e ⁻ˣ over [0, 3] :
[tex]\displaystyle \frac1{3-0}\int_0^3e^{-x}\,\mathrm dx=\frac13(-e^{-x})\bigg|_0^3=\frac13(-e^{-3}-(-e^{-0})) = \frac13 \left(1-\frac1{e^3}\right) = \boxed{\frac{e^3-1}{3e^3}}[/tex]