5.5 The Substitution Rule/63: Difference between revisions
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\frac{1}{2} | \frac{1}{2} | ||
\int_{a^2}^{2a^2} | \int_{a^2}^{2a^2} | ||
\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot | \sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot \bigg|_{2a^2}^{a^2} | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
\frac{a^{\frac{1}{2}+1}{1/2+1} | |||
Revision as of 20:22, 1 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_{0}^{a} x* \sqrt{x^2+a^2}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^2+ a^2\\[2ex] du &=2xdx \\[2ex] \frac{1}{2}du &= xdx \\[2ex] \end{align} }
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}{\frac {1}{2}}\int _{a^{2}}^{2a^{2}}{\sqrt {u}}\,\cdot du={\frac {1}{2}}\,\cdot {\bigg |}_{2a^{2}}^{a^{2}}\end{aligned}}}
\frac{a^{\frac{1}{2}+1}{1/2+1}