the value of integrand limit [0 to [x]] {x-[x]}dx (where [.] denotes greatest integer function.

Arun Kumar IIT Delhi
8 years ago
$\int_{0}^{[x]}x-[x]dx \\=>[x]^2/2-\int_{1}^{2}1dx-\int_{2}^{3}2dx-\int_{3}^{4}3dx..............depends\, value \,of\, [x]$

Arun Kumar
IIT Delhi
Praneeth
14 Points
8 years ago
x-[x] is periodic with period 1. therefore, ?0[x]x-[x] dx = [x]?01x-[x] dx =[x]/2
we can get integral of x-[x] within limits 0 and 1 using method of areas as ½.