Answer
$$\frac{{447}}{7}$$
Work Step by Step
$$\eqalign{
& \int_1^{64} {\frac{{\sqrt z - 2}}{{\root 3 \of z }}} dz \cr
& {\text{rewrite the radicals using the property }}\root n \of a = {a^{1/n}} \cr
& = \int_1^{64} {\left( {\frac{{{z^{1/2}} - 2}}{{{z^{1/3}}}}} \right)} dz \cr
& {\text{distributive property}} \cr
& = \int_1^{64} {\left( {\frac{{{z^{1/2}}}}{{{z^{1/3}}}} - \frac{2}{{{z^{1/3}}}}} \right)} dz \cr
& = \int_1^{64} {\left( {{z^{1/6}} - 2{z^{ - 1/3}}} \right)} dz \cr
& {\text{integrate by using }}\int_a^b {{u^n}} du = \left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right)_a^b \cr
& = \left( {\frac{{{z^{7/6}}}}{{7/6}} - 2\left( {\frac{{{z^{2/3}}}}{{2/3}}} \right)} \right)_1^{64} \cr
& = \left( {\frac{{6{z^{7/6}}}}{7} - 3{z^{2/3}}} \right)_1^{64} \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \left( {\frac{{6{{\left( {64} \right)}^{7/6}}}}{7} - 3{{\left( {64} \right)}^{2/3}}} \right) - \left( {\frac{{6{{\left( 1 \right)}^{7/6}}}}{7} - 3{{\left( 1 \right)}^{2/3}}} \right) \cr
& {\text{simplifying}} \cr
& = \left( {\frac{{6\left( {128} \right)}}{7} - 3\left( {16} \right)} \right) - \left( {\frac{6}{7} - 3} \right) \cr
& = \frac{{432}}{7} + \frac{{15}}{7} \cr
& = \frac{{432 + 15}}{7} \cr
& = \frac{{447}}{7} \cr} $$
