Answer
$$e - 2$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^2}{e^x}{\text{ over }}\left[ {0,1} \right] \cr
& {\text{The average value of the function is given by}} \cr
& {f_{avg}} = \frac{1}{{b - a}}\int_a^b {f\left( x \right)} dx \cr
& {f_{avg}} = \frac{1}{{1 - 0}}\int_0^1 {{x^2}{e^x}} dx \cr
& {f_{avg}} = \int_0^1 {{x^2}{e^x}} dx \cr
& {\text{Integrate }}\int {{x^2}{e^x}dx} {\text{ by parts}} \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx \cr
& dv = {e^x}dx,{\text{ }}v = {e^x} \cr
& \int u dv = uv - \int v du \cr
& \int {{x^2}{e^x}dx} = {x^2}{e^x} - \int {2x{e^x}} dx \cr
& {\text{Let }}u = 2x,{\text{ }}du = 2dx \cr
& dv = {e^x}dx,{\text{ }}v = {e^x} \cr
& \int {{x^2}{e^x}dx} = {x^2}{e^x} - \left( {2x{e^x} - \int {2{e^x}dx} } \right) \cr
& \int {{x^2}{e^x}dx} = {x^2}{e^x} - 2x{e^x} + \int {2{e^x}dx} \cr
& \int {{x^2}{e^x}dx} = {x^2}{e^x} - 2x{e^x} + 2{e^x} + C \cr
& {\text{Therefore,}} \cr
& {f_{avg}} = \int_0^1 {{x^2}{e^x}} dx = \left[ {{x^2}{e^x} - 2x{e^x} + 2e} \right]_0^1 \cr
& = \left[ {e - 2e + 2e} \right] - \left[ {2{e^0}} \right] \cr
& = e - 2 \cr} $$
