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

Revision as of 21:27, 6 September 2022