Question

Let \([t]\) denote the greatest integer less than or equal to \(t\). Then the value of
\(\displaystyle\int_{1}^{2}\left|2x-[3x]\right|dx\) is _______.

Hint:

Split \([1,2]\) at points where \([3x]\) jumps: \(x=\frac43,\frac53\).

Answer 1

Correct Answer: 1

For \(x\in[1,2)\), \(3x\in[3,6)\). \([3x]=3\) on \([1,\frac43)\), \(4\) on \([\frac43,\frac53)\), \(5\) on \([\frac53,2)\).
On \([1,\frac43)\): \(2x-3<0\) (zero at 1.5), so \(|2x-3|=3-2x\); \(\int=\left[3x-x^2\right]_1^{4/3}=\left(4-\frac{16}{9}\right)-2=\frac29\).
On \([\frac43,\frac53)\): \(|2x-4|=4-2x\); \(\int=\left[4x-x^2\right]_{4/3}^{5/3}=\frac{35}{9}-\frac{32}{9}=\frac13\).
On \([\frac53,2)\): \(|2x-5|=5-2x\); \(\int=\left[5x-x^2\right]_{5/3}^{2}=6-\frac{50}{9}=\frac49\).
Total \(=\frac29+\frac39+\frac49=1\).
Answer: 1