Tính $\underset{-1}{\overset{5}{\int }}{x}^2\left|x-1\right|dx$

(AI giải đề cho tôi)
Tách theo dấu của $|x-1|$ với mốc $x=1$: $${\int }_{-1}^5{x}^2|x-1|dx={\int }_{-1}^1{x}^2(1-x)dx+{\int }_1^5{x}^2(x-1)dx.$$ Tính từng phần: $${\int }_{-1}^1(x^2-x^3)dx={\left[\frac{x^3}{3}-\frac{x^4}{4}\right]}_{-1}^1=\left(\frac{1}{3}-\frac{1}{4}\right)-\left(-\frac{1}{3}-\frac{1}{4}\right)=\frac{1}{12}+\frac{7}{12}=\frac{2}{3}.$$ $${\int }_1^5(x^3-x^2)dx={\left[\frac{x^4}{4}-\frac{x^3}{3}\right]}_1^5=\left(\frac{625}{4}-\frac{125}{3}\right)-\left(\frac{1}{4}-\frac{1}{3}\right)=\frac{1375}{12}+\frac{1}{12}=\frac{1376}{12}=\frac{344}{3}.$$ Vậy tổng: $$\frac{2}{3}+\frac{344}{3}=\frac{346}{3}\approx 115,33.$$