Question

Let [t] denote the greatest integer less than or equal to $t$. Then the value of $\int\limits_{1}^{2}|2 x-[3 x]| d x$ is___

Answer 1

$3<3 x<6$
Take cases when $3<3 x<4,4<3 x<5$
$5<3 x<6$
Now $\int\limits_{1}^{2}|2 x-[3 x]| d x$
$=\int\limits_{1}^{4 / 3}(3-2 x) d x+\int\limits_{4 / 3}^{5 / 3}(4-2 x) d x+\int\limits_{5 / 3}^{2}(5-2 x) d x$
$=\frac{2}{9}+\frac{3}{9}+\frac{4}{9}=1$