Answer
$$ - \frac{{55}}{3}$$
Work Step by Step
$$\eqalign{
& \int_1^4 {\left( {\frac{3}{{\sqrt t }} - 5\sqrt t - {t^{ - 3/2}}} \right)} dt \cr
& {\text{radical properties}} \cr
& = \int_1^4 {\left( {\frac{3}{{{t^{1/2}}}} - 5{t^{1/2}} - {t^{ - 3/2}}} \right)} dt \cr
& = \int_1^4 {\left( {3{t^{ - 1/2}} - 5{t^{1/2}} - {t^{ - 3/2}}} \right)} dt \cr
& {\text{find the antiderivative by the power rule}} \cr
& = \left( {\frac{{3{t^{1/2}}}}{{1/2}} - \frac{{5{t^{3/2}}}}{{3/2}} - \frac{{{t^{ - 1/2}}}}{{ - 1/2}}} \right)_1^4 \cr
& = \left( {6{t^{1/2}} - \frac{{10{t^{3/2}}}}{3} + 2{t^{ - 1/2}}} \right)_1^4 \cr
& = \left( {6\sqrt t - \frac{{10t\sqrt t }}{3} + \frac{2}{{\sqrt t }}} \right)_1^4 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \left( {6\sqrt 4 - \frac{{10\left( 4 \right)\sqrt 4 }}{3} + \frac{2}{{\sqrt 4 }}} \right) - \left( {6\sqrt 1 - \frac{{10\sqrt 1 }}{3} + \frac{2}{{\sqrt 1 }}} \right) \cr
& {\text{simplify}} \cr
& = \left( {12 - \frac{{80}}{3} + 1} \right) - \left( {6 - \frac{{10}}{3} + 2} \right) \cr
& = \left( { - \frac{{41}}{3}} \right) - \left( {\frac{{14}}{3}} \right) \cr
& = - \frac{{55}}{3} \cr} $$
