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

Revision as of 19:27, 25 August 2022

${\begin{aligned}\int _{-1}^{2}(x^{3}-2x)dx&={\frac {x^{4}}{4}}-{\frac {2x^{2}}{2}}{\Bigg |}_{-1}^{2}\end{aligned}}$

Revision as of 19:26, 25 August 2022 (view source) Kimberlyr70044@students.laalliance.org (talk \| contribs) No edit summary Tag: Manual revert ← Older edit		Revision as of 19:27, 25 August 2022 (view source) Kimberlyr70044@students.laalliance.org (talk \| contribs) No edit summary Newer edit →
Line 1:		Line 1:
	<math>\begin{align}int_{-1}^{2}(x^3-2x)dx &=\frac{x^4}{4}-\frac{2x^2}{2}\Bigg\|_{-1}^{2}\end{align}</math>		<math>\begin{align}\int_{-1}^{2}(x^3-2x)dx &=\frac{x^4}{4}-\frac{2x^2}{2}\Bigg\|_{-1}^{2}\end{align}</math>