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 (unknown function "\begin{align}"): {\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} }