Answer
$$76$$
Work Step by Step
$$\eqalign{
& \int_1^4 {\left( {5y\sqrt y + 3\sqrt y } \right)} dy \cr
& {\text{use the radical property }}\sqrt a = {a^{1/2}} \cr
& = \int_1^4 {\left( {5y \cdot {y^{1/2}} + 3{y^{1/2}}} \right)} dy \cr
& = \int_1^4 {\left( {5{y^{3/2}} + 3{y^{1/2}}} \right)} dy \cr
& {\text{integrate by using }}\int_a^b {{y^n}dy} = \left( {\frac{{{y^{n + 1}}}}{{n + 1}}} \right)_a^b \cr
& = \left( {5\left( {\frac{{{y^{5/2}}}}{{5/2}}} \right) + 3\left( {\frac{{{y^{3/2}}}}{{3/2}}} \right)} \right)_1^4 \cr
& = \left( {2{y^{5/2}} + 2{y^{3/2}}} \right)_1^4 \cr
& = 2\left( {{y^{5/2}} + {y^{3/2}}} \right)_1^4 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = 2\left( {{{\left( 4 \right)}^{5/2}} + {{\left( 4 \right)}^{3/2}}} \right) - 2\left( {{{\left( 1 \right)}^{5/2}} + {{\left( 1 \right)}^{3/2}}} \right) \cr
& {\text{simplifying}} \cr
& = 2\left( {32 + 8} \right) - 2\left( {1 + 1} \right) \cr
& = 80 - 4 \cr
& = 76 \cr} $$
