Answer
$$ - \frac{{3038}}{{15}}$$
Work Step by Step
$$\eqalign{
& \int_4^9 {\left( {4\sqrt r - 3r\sqrt r } \right)} dr \cr
& {\text{use the radical property }}\sqrt r = {r^{1/2}} \cr
& = \int_4^9 {\left( {4{r^{1/2}} - 3r \cdot {r^{1/2}}} \right)} dr \cr
& = \int_4^9 {\left( {4{r^{1/2}} - 3{r^{3/2}}} \right)} dr \cr
& {\text{integrate by using }}\int_a^b {{r^n}dr} = \left( {\frac{{{r^{n + 1}}}}{{n + 1}}} \right)_a^b \cr
& = \left( {4\left( {\frac{{{r^{3/2}}}}{{3/2}}} \right) - 3\left( {\frac{{{r^{5/2}}}}{{5/2}}} \right)} \right)_4^9 \cr
& = \left( {\frac{8}{3}{r^{3/2}} - \frac{6}{5}{r^{5/2}}} \right)_4^9 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \left( {\frac{8}{3}{{\left( 9 \right)}^{3/2}} - \frac{6}{5}{{\left( 9 \right)}^{5/2}}} \right) - \left( {\frac{8}{3}{{\left( 4 \right)}^{3/2}} - \frac{6}{5}{{\left( 4 \right)}^{5/2}}} \right) \cr
& {\text{simplifying}} \cr
& = \left( {72 - \frac{{1458}}{5}} \right) - \left( {\frac{{64}}{3} - \frac{{192}}{5}} \right) \cr
& = - \frac{{1098}}{5} - \left( { - \frac{{256}}{{15}}} \right) \cr
& = - \frac{{1098}}{5} + \frac{{256}}{{15}} \cr
& = - \frac{{3038}}{{15}} \cr} $$
