Question

The value $\underset{0}{\overset{2}{\int }}\left[x^2-x+1\right]dx,$ where $[.]$ denotes the greatest integer function is given by

Answer 1

We have,

$I={\int }_0^2[x^2-x+1]dx$

Where [.] denotes greatest integer function.

Since,

$[x^2-x+1]=1,x\in [0,1)$

$[x^2-x+1]=3,x\in (1,2]$

Therefore,

$I={\int }_0^{1}1dx+{\int }_1^{2}3dx$

$I={\left[x\right]}_0^1+{\left[3x\right]}_1^2$

$I=1-0+3(2-1)$

$I=1+3$

$I=4$

Hence, the value is $4$.