5.3 The Fundamental Theorem of Calculus/33: Difference between revisions

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=<math>\frac{4y^3}{3}+\frac{4y^2}{2}\bigg|_{1}^{2}</math>
=<math>\frac{4y^3}{3}+\frac{4y^2}{2}\bigg|_{1}^{2}</math>
=<math>2+\frac{32}{3}+\frac{16}{2}-\left(1+\frac{4}{3}+\frac{4}{2}\right)</math>
=<math>2+\frac{32}{3}+\frac{16}{2}-\left(1+\frac{4}{3}+\frac{4}{2}\right)</math>
=<math>\frac{49}{3}

Revision as of 19:04, 26 August 2022

= = = =<math>\frac{49}{3}