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

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} \frac{1}{2} \int_{a^2}^{2a^2} \sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot \bigg|_{2a^2}^{a^2} \end{align} }

\frac{a^{\frac{1}{2}+1}{1/2+1}