5.5 The Substitution Rule/65: Difference between revisions

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Line 9: Line 9:
\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 09:29, 7 September 2022

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 = \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 (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}u&=x-1u+1&=x\\[2ex]du&=2dx\\[2ex]{\frac {1}{2}}du&=dx\end{aligned}}}