Answer
$$\frac{{106}}{{405}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{{y^2}}}{{\sqrt {4 - 3y} }}dy} \cr
& {\text{Let }}u = 4 - 3y{\text{ }} \Rightarrow {\text{ }}y = \frac{{4 - u}}{3},{\text{ }}dy = - \frac{1}{3}du \cr
& {\text{The new limits of integration are:}} \cr
& y = 1 \Rightarrow u = 1 \cr
& y = 0 \Rightarrow 4 \cr
& {\text{Apply the substitution}} \cr
& \int_0^1 {\frac{{{y^2}}}{{\sqrt {4 - 3y} }}dy} = \int_4^1 {\frac{{{{\left( {\frac{{4 - u}}{3}} \right)}^2}}}{{\sqrt u }}\left( { - \frac{1}{3}du} \right)} \cr
& {\text{ }} = \frac{1}{{27}}\int_1^4 {\frac{{16 - 8u + {u^2}}}{{\sqrt u }}du} \cr
& {\text{ }} = \frac{1}{{27}}\int_1^4 {\left( {16{u^{ - 1/2}} - 8{u^{1/2}} + {u^{3/2}}} \right)du} \cr
& {\text{ }} = \frac{1}{{27}}\left[ {\frac{{16{u^{1/2}}}}{{1/2}} - \frac{{8{u^{3/2}}}}{{3/2}} + \frac{{{u^{5/2}}}}{{5/2}}} \right]_1^4 \cr
& {\text{ }} = \frac{1}{{27}}\left[ {32{u^{1/2}} - \frac{{16}}{3}{u^{3/2}} + \frac{2}{5}{u^{5/2}}} \right]_1^4 \cr
& = \frac{1}{{27}}\left[ {32{{\left( 4 \right)}^{1/2}} - \frac{{16}}{3}{{\left( 4 \right)}^{3/2}} + \frac{2}{5}{{\left( 4 \right)}^{5/2}}} \right] - \frac{1}{{27}}\left[ {32{{\left( 1 \right)}^{1/2}} - \frac{{16}}{3}{{\left( 1 \right)}^{3/2}} + \frac{2}{5}{{\left( 1 \right)}^{5/2}}} \right] \cr
& = \frac{1}{{27}}\left[ {\frac{{512}}{{15}}} \right] - \frac{1}{{27}}\left[ {\frac{{406}}{{15}}} \right] \cr
& = \frac{{106}}{{405}} \cr} $$
