5.3 The Fundamental Theorem of Calculus/33: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
| Line 1: | Line 1: | ||
<math>\int_{1}^{2}\left(1+2y | <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
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\right)^{2}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}}}