∫ 1 2 ( 1 + 2 y ) 2 d y = ∫ 1 2 ( 4 y 2 + 4 y + 1 ) d y = 1 y + 4 y 3 3 + 4 y 2 2 | 1 2 = 2 + 32 3 + 16 2 − ( 1 + 4 3 + 4 2 ) = 49 3 {\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}}}
