Ta có: $\int\limits_0^2|x^2-x|dx$$=\int\limits_0^1(x-x^2)dx+\int\limits_1^2(x^2-x)dx$$=\left(\dfrac{x^2}{2}-\dfrac{x^3}{3}\right)\left|\begin{array}{l}1\\0\end{array}\right.+\left(\dfrac{x^3}{3}-\dfrac{x^2}{2}\right)\left|\begin{array}{l}2\\1\end{array}\right.=1$
Đặt: $t=1+\dfrac{1}{x} \Rightarrow dt=-\dfrac{1}{x^2}dx$Đổi cận: $x=\dfrac{1}{2} \Rightarrow t=3$ $x=1 \Rightarrow t=2$Ta có: $\int\limits_0^2|x^2-x|dx$$=\int\limits_0^1(x-x^2)dx+\int\limits_1^2(x^2-x)dx$$=\left(\dfrac{x^2}{2}-\dfrac{x^3}{3}\right)\left|\begin{array}{l}1\\0\end{array}\right.+\left(\dfrac{x^3}{3}-\dfrac{x^2}{2}\right)\left|\begin{array}{l}2\\1\end{array}\right.=1$
Ta có: $\int\limits_0^2|x^2-x|dx$$=\int\limits_0^1(x-x^2)dx+\int\limits_1^2(x^2-x)dx$$=\left(\dfrac{x^2}{2}-\dfrac{x^3}{3}\right)\left|\begin{array}{l}1\\0\end{array}\right.+\left(\dfrac{x^3}{3}-\dfrac{x^2}{2}\right)\left|\begin{array}{l}2\\1\end{array}\right.=1$