Let [t] denote the greatest integer less than or equal to t.
Question:

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

Solution:

$3<3 x<6$

Take cases when $3<3 x<4,4<3 x<5$,

$5<3 x<6$

Now $\int_{1}^{2}|2 x-[3 x]| d x$

$=\int_{1}^{4 / 3}(3-2 x) d x+\int_{4 / 3}^{5 / 3}(4-2 x) d x+\int_{5 / 3}^{2}(5-2 x) d x$

$=\frac{2}{9}+\frac{3}{9}+\frac{4}{9}=1$

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