5.5 The Substitution Rule/65: Difference between revisions
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u &= x-1 | u &= x-1 | ||
\\[2ex] | \\[2ex] | ||
du &= | du &= 2 dx \\[2ex] | ||
2du &= \frac{1}{\sqrt{t}} dx | 2du &= \frac{1}{\sqrt{t}} dx | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 09:28, 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 (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 \\[2ex] du &= 2 dx \\[2ex] 2du &= \frac{1}{\sqrt{t}} dx \end{align} }