∫ 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}=5\cdot 4-4^{2}+4^{3}-(5\cdot 1-1^{2}+1^{3})=63}
