Answer
$\dfrac{722}{15}$
Work Step by Step
$I=\int_{1}^{4} (t) \times (\dfrac{dt}{2 \sqrt t})+(t^2) dt+\sqrt t \times (2t) dt$
$=\int_{1}^{4} (\dfrac{t^{1/2}}{2})+t^2+2 \times t^{3/2} dt$
$=(\dfrac{1}{2})(\dfrac{2}{3}) \times t^{3/2}+\dfrac{t^3}{3}+\dfrac{4}{5}(t^2) \times \sqrt t]_1^{4}$
$=[\dfrac{8}{3}+\dfrac{64}{3}+\dfrac{128}{5}]-[\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{4}{5}]$
Hence, we have $I=\dfrac{70}{3}+\dfrac{124}{5}=\dfrac{722}{15}$
