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:
$\int_{1}^{2}|2 x-[3 x]| d x=\int_{1}^{2}|3 x-[3 x]-x| d x$
$=\int_{1}^{2}|\{3 x\}-x| d x=\int_{1}^{2}(x-\{3 x\}) d x$
$=\int_{1}^{2} x d x-\int_{1}^{2}\{3 x\} d x=\left[\frac{x^{2}}{2}\right]_{1}^{2}-3 \int_{0}^{1 / 3} 3 x d x$
$=\frac{(4-1)}{2}-9\left[\frac{x^{2}}{2}\right]_{0}^{1 / 3}=\frac{3}{2}-\frac{1}{2}=1$
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