∫ 1 2 ( 1 + 2 y ) 2 d y = ∫ 1 2 ( 1 + 4 y + 4 y 2 ) d y = y + 4 y 2 2 + 4 y 3 3 | 1 2 = [ 2 + 4 ( 2 ) 2 2 + 4 ( 2 ) 3 2 ] − [ 1 + 4 ( 1 ) 2 2 + 4 ( 1 ) 3 2 ] = 49 3 {\displaystyle {\begin{aligned}\int _{1}^{2}\left(1+2y\right)^{2}dy&=\int _{1}^{2}\left(1+4y+4y^{2}\right)dy\\[2ex]&=y+{\frac {4y^{2}}{2}}+{\frac {4y^{3}}{3}}{\bigg |}_{1}^{2}\\[2ex]&=\left[2+{\frac {4(2)^{2}}{2}}+{\frac {4(2)^{3}}{2}}\right]-\left[1+{\frac {4(1)^{2}}{2}}+{\frac {4(1)^{3}}{2}}\right]\\[2ex]&={\frac {49}{3}}\end{aligned}}}