5.3 The Fundamental Theorem of Calculus/19: Difference between revisions

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<math\begin{align}\int\limits_{-1}^{2}(x^3-2x) dx</math>&=\frac{x^4}{4}-\frac{2x^2}{2}\bigg|_{-1}^{2}</math>
<math>\begin{align}\int\limits_{-1}^{2}(x^3-2x) dx</math>&=\frac{x^4}{4}-\frac{2x^2}{2}\bigg|_{-1}^{2}</math>

Revision as of 19:20, 25 August 2022

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align}\int\limits_{-1}^{2}(x^3-2x) dx} &=\frac{x^4}{4}-\frac{2x^2}{2}\bigg|_{-1}^{2}</math>