5.5 The Substitution Rule/65

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{1}^{2} x \sqrt{x-1} dx }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} u &= x-1 \\ u+1 &= x \\[2ex] du &= 1 dx \\[2ex] du &= dx \end{align} }

${\begin{aligned}\int _{1}^{2}x{\sqrt {x-1}}\,dx&=\int _{0}^{1}(u+1){\sqrt {u}}\,du=\int _{0}^{1}(u+1)({\sqrt {u}})=\int _{0}^{1}(u^{\frac {3}{2}}+{\sqrt {u}})\,du\\[2ex]&={\frac {2}{5}}(u^{\frac {5}{2}}+{\frac {2}{3}}u^{\frac {3}{2}}){\bigg |}_{0}^{1}={\frac {2}{5}}+{\frac {2}{3}}\\[2ex]&={\frac {16}{15}}\\[2ex]\end{aligned}}$