5.5 The Substitution Rule/65

$\int _{1}^{2}x{\sqrt {x-1}}dx$

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 &= 2 dx \\[2ex] \frac{1}{2} du &= dx \end{align} }

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} \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}| _{0}^{1} =\frac{2}{5} + \frac{2}{3} \\[2ex] &= \frac{16}{15}\\[2ex] \end{align} }