∫ − 2 − 1 ( 4 y 3 + 2 y 3 ) d y = ∫ − 2 − 1 ( 4 y 3 + 2 y − 3 ) d y = [ y 4 − y − 2 ] − 2 − 1 = ( 1 − 1 ) − ( 16 − 1 4 ) = − 63 4 {\displaystyle {\begin{aligned}\int _{-2}^{-1}\left(4y^{3}+{\frac {2}{y^{3}}}\right)dy&=\int _{-2}^{-1}\left(4y^{3}+2y^{-3}\right)dy\\[2ex]&=\left[y^{4}-y^{-2}\right]_{-2}^{-1}\\[2ex]&=(1-1)-\left(16-{\frac {1}{4}}\right)\\[2ex]&=-{\frac {63}{4}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{-2}^{-1}\left(4y^{3}+{\frac {2}{y^{3}}}\right)dy&=\int _{-2}^{-1}\left(4y^{3}+2y^{-3}\right)dy\\[2ex]&=\left[y^{4}-y^{-2}\right]_{-2}^{-1}\\[2ex]&=(1-1)-\left(16-{\frac {1}{4}}\right)\\[2ex]&=-{\frac {63}{4}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/9414bcb4e5607f5be5cdc4d4bc92be25a0964afe)