5.5 The Substitution Rule/63: Difference between revisions

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\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 \frac{a^{\frac{1}{2}+1}{1/2+1} \bigg|_{2a^2}^{a^2}
{


\end{align}
\end{align}
</math>
</math>

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 (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 \frac{a^{\frac{1}{2}+1}{1/2+1} \bigg|_{2a^2}^{a^2} \end{align} }