Answer
$$ - \frac{{16}}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^4 {2\left( {{t^{1/2}} - t} \right)} dt \cr
& = \int_0^4 {\left( {2{t^{1/2}} - 2t} \right)} dt \cr
& {\text{use }}\int {{t^n}dt = \frac{{{t^{n + 1}}}}{{n + 1}} + C,{\text{ so}}} \cr
& = \left( {2\left( {\frac{{{t^{1/2 + 1}}}}{{1/2 + 1}}} \right) - 2\left( {\frac{{{t^2}}}{2}} \right)} \right)_0^4 \cr
& = \left( {2\left( {\frac{{{t^{3/2}}}}{{3/2}}} \right) - {t^2}} \right)_0^4 \cr
& = \left( {\frac{4}{3}{t^{3/2}} - {t^2}} \right)_0^4 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \left( {\frac{4}{3}{{\left( 4 \right)}^{3/2}} - {{\left( 4 \right)}^2}} \right) - \left( {\frac{4}{3}{{\left( 0 \right)}^{3/2}} - {{\left( 0 \right)}^2}} \right) \cr
& {\text{simplifying}} \cr
& = \left( {\frac{{32}}{3} - 16} \right) - \left( 0 \right) \cr
& = - \frac{{16}}{3} \cr} $$