Answer
$$\frac{{19}}{{15}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {x\sqrt {5{x^2} + 4} } dx \cr
& {\text{use substitution}}{\text{. Let }}u = 5{x^2} + 4,{\text{ so that }}du = 10xdx,\,\,\,\,\frac{1}{{10}}du = xdx \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 1,{\text{ then }}u = 5{\left( 1 \right)^2} + 4 = 9 \cr
& \,\,\,\,\,\,{\text{If }}x = 0,{\text{ then }}u = 5{\left( 0 \right)^2} + 4 = 4 \cr
& {\text{Then}} \cr
& \int_0^1 {x\sqrt {5{x^2} + 4} } dx = \int_4^9 {\sqrt u \left( {\frac{1}{{10}}du} \right)} \cr
& = \frac{1}{{10}}\int_4^9 {{u^{1/2}}du} \cr
& {\text{use }}\int_a^b {{u^n}} du = \left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right)_a^b \cr
& = \frac{1}{{10}}\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right)_4^9 \cr
& = \frac{1}{{15}}\left( {{u^{3/2}}} \right)_4^9 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \frac{1}{{15}}\left( {{9^{3/2}} - {4^{3/2}}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{{15}}\left( {27 - 8} \right) \cr
& = \frac{{19}}{{15}} \cr} $$
