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

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

Revision as of 21:26, 6 September 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{1}^{2}\left(1+2y^{2}\right)dy=\int _{1}^{2}\left(4y^{2}+4y+1\right)dy=1y+{\frac {4y^{3}}{3}}+{\frac {4y^{2}}{2}}{\bigg |}_{1}^{2}=2+{\frac {32}{3}}+{\frac {16}{2}}-\left(1+{\frac {4}{3}}+{\frac {4}{2}}\right)={\frac {49}{3}}}