Question

Вычислите: интеграл от 3 до 0 х-1 по модулю *dx

Answer 1

$\int\limits_0^3|x-1|dx=\left(\begin{array}{c}u=x-1\\du=dx\end{array}\right)=\int\limits_{-1}^2|u|du=\left|\int\limits_{-1}^0u\;du\right|+\left| \int\limits_0^2u\;du\right|=\\=\left|\left.\frac{u^2}2\right|\limits_{-1}^0\right|+\left|\left.\frac{u^2}2\right|\limits_{0}^2\right|=\left|0-\frac12\right|+\left|2-0\right|=\left|-\frac12\right|+|2|=\frac12+2=2\frac12=2,5$