5.3 The Fundamental Theorem of Calculus/29

${\begin{aligned}\int _{1}^{9}{\frac {x-1}{\sqrt {x}}}dx&=\int _{1}^{9}{\frac {x}{\sqrt {x}}}-{\frac {1}{\sqrt {x}}}=\int _{1}^{9}{\frac {x}{x^{1/2}}}-{\frac {1}{x^{1/2}}}=\int _{1}^{9}x^{1/2}-x^{-1/2}=\int _{1}^{9}{\frac {2x^{3/2}}{3}}-2x^{1/2}{\bigg |}_{1}^{9}\\[2ex]&=({\frac {2(9)^{3/2}}{3}})-2(9)^{1/2})-({\frac {2(1)^{3/2}}{3}})-2(1)^{1/2})=({\frac {54}{3}}-6)-({\frac {2}{3}}-2)={\frac {52}{3}}-4={\frac {40}{3}}\end{aligned}}$