5.3 The Fundamental Theorem of Calculus/37: Difference between revisions
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<math>int\limits_{1/2}^{ | <math>int\limits_{1/2}^{(\sqrt{3})/2}\frac{6}{(\sqrt{1-x^2})} du </math> | ||
[[5.3 The Fundamental Theorem of Calculus/1|1]] | [[5.3 The Fundamental Theorem of Calculus/1|1]] | ||
Revision as of 18:57, 25 August 2022
FTC # 2 Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. TeX parse error: \limits is allowed only on operators"): {\displaystyle int\limits _{1/2}^{({\sqrt {3}})/2}{\frac {6}{({\sqrt {1-x^{2}}})}}du}
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