Answer
${f_{avg}} = \frac{1}{3}\left( {e - \root 4 \of e } \right)$
Work Step by Step
$$\eqalign{
& f\left( z \right) = \frac{{{e^{1/z}}}}{{{z^2}}},{\text{ }}\left[ {1,4} \right] \cr
& {\text{Use The Mean Value Theorem for Integrals }} \cr
& {f_{avg}} = \frac{1}{{b - a}}\int_a^b {f\left( x \right)} dx \cr
& {\text{Therefore}}{\text{,}} \cr
& {f_{avg}} = \frac{1}{{4 - 1}}\int_1^4 {\left( {\frac{{{e^{1/z}}}}{{{z^2}}}} \right)} dz \cr
& {\text{Rewrite the integrand}} \cr
& {f_{avg}} = - \frac{1}{{4 - 1}}\int_1^4 {{e^{1/z}}\left( { - \frac{1}{{{z^2}}}} \right)} dz \cr
& {\text{Integrate using }}\int {{e^u}} du = {e^u} + C,{\text{ }}u = \frac{1}{z},{\text{ }}du = - \frac{1}{{{z^2}}}du,{\text{ so}} \cr
& {f_{avg}} = - \frac{1}{3}\left[ {{e^{1/z}}} \right]_1^4 \cr
& {\text{Evaluate and simplify}} \cr
& {f_{avg}} = - \frac{1}{3}\left[ {{e^{1/4}} - {e^1}} \right] \cr
& {f_{avg}} = \frac{1}{3}\left( {e - \root 4 \of e } \right) \cr} $$
