Answer
$$\frac{1}{4}\left( {\frac{{15}}{{16}} - {e^4} + {e^2}} \right)$$
Work Step by Step
$$\eqalign{
& \int_{0.5}^1 {\left( {{p^3} - {e^{4p}}} \right)} dp \cr
& {\text{integrate by using }}\int_a^b {{e^{kx}}} dx = \left( {\frac{{{e^{kx}}}}{k}} \right)_a^b{\text{ and }}\int_a^b {{x^n}} dx = \left( {\frac{{{x^{n + 1}}}}{{n + 1}}} \right)_a^b{\text{ }} \cr
& = \left( {\frac{{{p^{3 + 1}}}}{{3 + 1}} - \frac{{{e^{4p}}}}{4}} \right)_{0.5}^1 \cr
& = \left( {\frac{{{p^4}}}{4} - \frac{{{e^{4p}}}}{4}} \right)_{0.5}^1 \cr
& {\text{factor out 4}} \cr
& = \frac{1}{4}\left( {{p^4} - {e^{4p}}} \right)_{0.5}^1 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \frac{1}{4}\left( {{{\left( 1 \right)}^4} - {e^{4\left( 1 \right)}}} \right) - \frac{1}{4}\left( {{{\left( {0.5} \right)}^4} - {e^{4\left( {0.5} \right)}}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{4}\left( {1 - {e^4}} \right) - \frac{1}{4}\left( {\frac{1}{{16}} - {e^2}} \right) \cr
& = \frac{1}{4}\left( {1 - {e^4} - \frac{1}{{16}} + {e^2}} \right) \cr
& = \frac{1}{4}\left( {\frac{{15}}{{16}} - {e^4} + {e^2}} \right) \cr} $$
