$\int_0^2[x] d x+\int_0^2|x-1| d x=$ (where $[x]$ denotes the greatest integer function.)

Question

Aicharya · Accepted Answer

$\begin{aligned} & \int_0^2[x] d x+\int_0^2[x-1] d x \ & =\int_0^1 0 . d x+\int_1^2 1 . d x+\int_0^1(1-x) d x+\int_1^2(x-1) d x \ & =[0]_0^1+[x]_1^2+\left[x-\frac{x^2}{2}ight]_0^1+\left[\frac{x^2}{2}-xight]_1^2 \ & =0+1+\frac{1}{2}+\frac{1}{2}=2\end{aligned}$

$\int_0^2[x] d x+\int_0^2|x-1| d x=$ (where $[x]$ denotes the greatest integer function.)

Solution