Answer
$$\frac{1}{{3{e^3}}} + \frac{2}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^3 {\frac{{3 - x}}{{3{e^x}}}} dx \cr
& {\text{use property of exponents }}\frac{1}{{{e^a}}} = {e^{ - a}} \cr
& = \frac{1}{3}\int {\left( {3 - x} \right){e^{ - x}}} dx \cr
& {\text{setting }}\,\,\,\,\,\,u = 3 - x{\text{ then }}du = - dx\,\,\,\,\,\,\,\,\,\, \cr
& {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^{ - x}}dx{\text{ then }}v = - {e^{ - x}} \cr
& {\text{Substituting these values into the formula for integration by parts}} \cr
& \int u dv = uv - \int {vdu} \cr
& \frac{1}{3}\int {\left( {3 - x} \right){e^{ - x}}} dx = \frac{1}{3}\left( {3 - x} \right)\left( { - {e^{ - x}}} \right) - \frac{1}{3}\int {\left( { - {e^{ - x}}} \right)\left( { - dx} \right)} \cr
& \frac{1}{3}\int {\left( {3 - x} \right){e^{ - x}}} dx = \frac{{\left( {x - 3} \right){e^{ - x}}}}{3} - \frac{1}{3}\int {{e^{ - x}}dx} \cr
& {\text{integrate using }}\int {{e^{kx}}dx} = \frac{{{e^{kx}}}}{k} + C \cr
& \frac{1}{3}\int {\left( {3 - x} \right){e^{ - x}}} dx = \frac{{\left( {x - 3} \right){e^{ - x}}}}{3} - \frac{1}{3}\left( { - {e^{ - x}}} \right) + C \cr
& \frac{1}{3}\int {\left( {3 - x} \right){e^{ - x}}} dx = \frac{{\left( {x - 3} \right){e^{ - x}}}}{3} + \frac{1}{3}{e^{ - x}} + C \cr
& \cr
& {\text{Now find the definite integral}} \cr
& \int_0^3 {\frac{{3 - x}}{{3{e^x}}}} dx = \left. {\left( {\frac{{\left( {x - 3} \right){e^{ - x}}}}{3} + \frac{1}{3}{e^{ - x}}} \right)} \right|_0^3 \cr
& {\text{evaluating the limits}} \cr
& = \left( {\frac{{\left( {3 - 3} \right){e^{ - 3}}}}{3} + \frac{1}{3}{e^{ - 3}}} \right) - \left( {\frac{{\left( {0 - 3} \right){e^{ - 0}}}}{3} + \frac{1}{3}{e^{ - 0}}} \right) \cr
& {\text{simplifying}} \cr
& = \left( {0 + \frac{1}{3}{e^{ - 3}}} \right) - \left( {\frac{{ - 3}}{3} + \frac{1}{3}{e^{ - 0}}} \right) \cr
& = \frac{1}{{3{e^3}}} + \frac{2}{3} \cr} $$
