∫ 1 4 ( 5 − 2 t + 3 t 2 ) d t = 5 t − t 2 + t 3 | 1 4 = [ 5 ( 4 ) − ( 4 ) 2 + ( 4 ) 3 ] − [ 5 ( 1 ) − ( 1 ) 2 + ( 1 ) 3 ] = 63 {\displaystyle \int _{1}^{4}(5-2t+3t^{2})\,dt=5t-t^{2}+t^{3}{\bigg |}_{1}^{4}=\left[5(4)-(4)^{2}+(4)^{3}\right]-\left[5(1)-(1)^{2}+(1)^{3}\right]=63}
![{\displaystyle \int _{1}^{4}(5-2t+3t^{2})\,dt=5t-t^{2}+t^{3}{\bigg |}_{1}^{4}=\left[5(4)-(4)^{2}+(4)^{3}\right]-\left[5(1)-(1)^{2}+(1)^{3}\right]=63}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/e77ce0c763d0153070312afe8a419c2f8fd9f408)