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 $\displaystyle\int_1^2\big|2x-[3x]\big|\,dx$ is \_\_\_\_\_\_\_\_.

Answer: 1

Accepted Answer

$$\int_1^2|2x-[3x]|dx=\int_1^2|3x-[3x]-x|dx=\int_1^2|\{3x\}-x|dx$$

For $x\in[1,2]$, since $\{3x\}<1\le x$, we get $x-\{3x\}\ge0$, so
$$I=\int_1^2(x-\{3x\})dx=\int_1^2x\,dx-\int_1^2\{3x\}dx$$

For the second integral, substitute $u=3x$ and use periodicity of $\{u\}$ with period 1:
$$\int_1^2\{3x\}dx=\frac13\int_3^6\{u\}du=\frac13\cdot3\int_0^1u\,du=\frac13\cdot3\cdot\frac12=\frac12$$

$$I=\left[\frac{x^2}2\right]_1^2-\frac12=\frac{4-1}2-\frac12=\frac32-\frac12=1$$

Solutions · 1