Answer
$$\frac{{56}}{3}$$
Work Step by Step
$$\eqalign{
& \int_3^9 {\sqrt {2r - 2} } dr \cr
& {\text{use substitution}}{\text{. Let }}u = 2r - 2,{\text{ so that }}du = 2dr,\,\,\,\,\,\,dr = \frac{1}{2}du \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}r = 9,{\text{ then }}u = 2\left( 9 \right) - 2 = 16 \cr
& \,\,\,\,\,\,{\text{If }}r = 3,{\text{ then }}u = 2\left( 3 \right) - 2 = 4 \cr
& {\text{Then}} \cr
& \int_3^9 {\sqrt {2r - 2} } dr = \int_4^{16} {\sqrt u } \left( {\frac{1}{2}du} \right) \cr
& = \frac{1}{2}\int_4^{16} {\sqrt u } du \cr
& = \frac{1}{2}\int_4^{16} {{u^{1/2}}} du \cr
& {\text{use }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C,{\text{ so}}} \cr
& = \frac{1}{2}\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right)_4^{16} \cr
& = \frac{1}{3}\left( {{u^{3/2}}} \right)_4^{16} \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \frac{1}{3}\left( {{{\left( {16} \right)}^{3/2}} - {{\left( 4 \right)}^{3/2}}} \right) \cr
& = \frac{1}{3}\left( {64 - 8} \right) \cr
& = \frac{{56}}{3} \cr} $$
