Answer
$$\frac{9}{2}$$
Work Step by Step
$$\eqalign{
& \int_1^8 {\frac{{3 - {y^{1/3}}}}{{{y^{2/3}}}}} dy \cr
& {\text{distributive property}} \cr
& = \int_1^8 {\left( {\frac{3}{{{y^{2/3}}}} - \frac{{{y^{1/3}}}}{{{y^{2/3}}}}} \right)} dy \cr
& {\text{simplify using the property of exponents}} \cr
& = \int_1^8 {\left( {3{y^{ - 2/3}} - {y^{ - 1/3}}} \right)} dy \cr
& {\text{integrate by using }}\int_a^b {{u^n}} du = \left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right)_a^b \cr
& = \left( {3\left( {\frac{{{y^{1/3}}}}{{1/3}}} \right) - \frac{{{y^{2/3}}}}{{2/3}}} \right)_1^8 \cr
& = \left( {9{y^{1/3}} - \frac{{3{y^{2/3}}}}{2}} \right)_1^8 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \left( {9{{\left( 8 \right)}^{1/3}} - \frac{{3{{\left( 8 \right)}^{2/3}}}}{2}} \right) - \left( {9{{\left( 1 \right)}^{1/3}} - \frac{{3{{\left( 1 \right)}^{2/3}}}}{2}} \right) \cr
& {\text{simplifying}} \cr
& = \left( {9\left( 2 \right) - \frac{{3\left( 4 \right)}}{2}} \right) - \left( {9 - \frac{3}{2}} \right) \cr
& = 12 - \frac{{15}}{2} \cr
& = \frac{9}{2} \cr} $$
