Answer
$$13$$
Work Step by Step
$$\eqalign{
& \int_0^2 {3\sqrt {4u + 1} } du \cr
& {\text{use substitution}}{\text{. Let }}t = 4u + 1,{\text{ so that }}dt = 4du \cr
& {\text{the new limits on }}t{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}u = 2,{\text{ then }}t = 4\left( 2 \right) + 1 = 9 \cr
& \,\,\,\,\,\,{\text{If }}u = 0,{\text{ then }}t = 4\left( 0 \right) + 1 = 1 \cr
& {\text{Then}} \cr
& \int_0^2 {3\sqrt {4u + 1} } du = \int_1^9 {3\sqrt t } \left( {\frac{1}{4}dt} \right) \cr
& = \frac{3}{4}\int_1^9 {\sqrt t } dt \cr
& = \frac{3}{4}\int_1^9 {{t^{1/2}}} dt \cr
& {\text{integrate}} \cr
& = \frac{3}{4}\left( {\frac{{{t^{3/2}}}}{{3/2}}} \right)_1^9 \cr
& = \frac{1}{2}\left( {{t^{3/2}}} \right)_1^9 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \frac{1}{2}\left( {{{\left( 9 \right)}^{3/2}} - {{\left( 1 \right)}^{3/2}}} \right) \cr
& = \frac{1}{2}\left( {27 - 1} \right) \cr
& = 13 \cr} $$
