Question

Let $\left[t\right]$ denote the greatest integer less than or equal to $t$. Then the value of ${\int }_1^2|2x-[3x\left]\right|dx$is.

Answer 1

Evaluating the integral:

${\int }_1^2|2x-[3x\left]\right|dx$ , where $\left[x\right]$ is the greatest integer function less than equal to $x$.

Let $I={\int }_1^2\left|2x-\left[3x\right]\right|dx$

Now substituting $x=\frac{t}{3}$ and $dt=3dx$

Then limit changes when $x=2$ then $t=6$ and when $x=1$ then $t=3$

$$\begin{gathered} \Rightarrow I=\begin{array}{l}\frac{1}{3}{\int }_3^6\left|\frac{2t}{3}-\left[t\right]\right|dt\end{array} \\ \Rightarrow I=\begin{array}{l}\frac{1}{9}{\int }_3^6\left|2t-3\left[t\right]\right|dt\end{array} \end{gathered}$$

Let us divide the given interval into $3$ parts as

$\Rightarrow I=\frac{1}{9}\left[{\int }_3^4|2t-9|dt+{\int }_4^5|2t-12|dt+{\int }_5^6|2t-15|dt\right]$ $\{\left[x\right]=3,when3<x<4andsoon\}$

Now by using the identity $\left|-\left(x-a\right)\right|=\left|\left(x-a\right)\right|$, then we get

$\Rightarrow I=\frac{1}{9}\left[{\int }_3^4\left(9-2t\right)dt+{\int }_4^5\left(12-2t\right)dt+{\int }_5^6\left(15-2t\right)dt\right]$

Let us integrate and apply the limit simplification.

$\begin{array}{rcl}I& =& \frac{1}{9}\left[{\left(9t-2\frac{t^2}{2}\right)}_3^4+{\left(12t-2\frac{t^2}{2}\right)}_4^5+{\left(15t-2\frac{t^2}{2}\right)}_5^6\right]\\ & =& \frac{1}{9}\left[9\times 1+12\times 1+15\times 1-\left(4^2-3^2\right)-\left(5^2-4^2\right)-\left(6^2-5^2\right)\right]\\ & =& \frac{1}{9}\left[36-\left[4^2-3^2+5^2-4^2+\left(6^2-5^2\right)\right]\right]\\ & =& \frac{1}{9}\left[36-36+9\right]\\ & =& 1\end{array}$

Hence, the value of integral is $1$