Grade 12Integral Calculus

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

sachin verma

14 Years agoGrade 12

12 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

Askiitians Faculty

12 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.

Therefore the answer is [x]/2

integral calculus

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integral calculus

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integral calculus

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integral calculus

Last Activity: 4 Years ago