5.5 The Substitution Rule/65: Difference between revisions

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\begin{align}
\begin{align}


u &= x-1\\u+1 &= x
u &= x-1 = u+1 &= x
   \\[2ex]
   \\[2ex]
du &= 2 dx \\[2ex]
du &= 2 dx \\[2ex]

Revision as of 09:30, 7 September 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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={\frac {2}{5}}U^{\frac {5}{2}}+{\frac {2}{3}}U^{\frac {3}{2}}|_{0}^{1}={\frac {2}{5}}+{\frac {2}{3}}={\frac {16}{15}}}


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} }