Mathematics - Definite Integration Question with Solution | TestHub

We have, ${\int }_0^n\left[x\right]dx$

$={\int }_0^{1}0dx+{\int }_1^{2}1dx+{\int }_2^{3}2dx+\dots +{\int }_{n-1}^n\left(n-1\right)dx$

$=1\left(2-1\right)+2\left(3-2\right)+3\left(4-3\right)+\dots \dots +\left(n-1\right)\left\{n-\left(n-1\right)\right\}$

$=1+2+3+\dots .+\left(n-1\right)=\frac{n\left(n-1\right)}{2}$

and ${\int }_0^n\left\{x\right\}dx={\int }_0^n\left(x-\left[x\right]\right)dx=\frac{n}{2}$

$\therefore \frac{{\int }_0^n\left[x\right]dx}{{\int }_0^n\left\{x\right\}dx}=n-1$