\(\int_{4}^{9} (t-3)/\sqrt{t} dt=\int_{4}^{9} t^{1/2}-3t^{-1/2} dt \\ =\int_{4}^{9} t^{1/2}dt-\int_{4}^{9}3t^{-1/2} dt \\ =\frac{2}{3}t^{3/2}|^9_4-6t^{1/2}|^9_4 \\= \frac{2}{3}9^{3/2}-6(9)^{1/2}-\frac{2}{3}4^{3/2}+6(4)^{1/2} \\= \frac{2}{3}(3^2)^{3/2}-6(3^2)^{1/2}-\frac{2}{3}(2^2)^{3/2}+6(2^2)^{1/2} \\=\frac{2}{3}3^3-6(3)^1-\frac{2}{3}2^3-6(2)^1 \\=\frac{2}{3}27-6(3)-\frac{2}{3}8-6(2) \\= 18-18-\frac{16}{3}-12 \\=\frac{16}{3}-12 \\=\frac{16-36}{3} \\= \frac{-20}{3}\)

Portanto, podemos concluir que: \(\int_{4}^{9} (t-3)/\sqrt{t} dt= \frac{-20}{3}\)