Can you find a general rule for this? - Part (1)

Calculus Level 5

If x , z x,z are real numbers with z > 0 z > 0 , there exist unique integer m m and real number r r with 0 ≤ r < z 0 \le r < z so that x = m z + r x = mz + r . Define the remainder { x mod z } \{x \text{ mod } z\} as the value of r r for the corresponding x , z x,z .

Compute 6 ∫ 0 1 { x 2 mod x } d x 6 \int_0^1 \{x^2 \text{ mod } x\}\,dx


The answer is 2.

Alisa Meier by Alisa Meier , 24, Germany

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1 solution

Arturo Presa
Nov 11, 2015

Actually, if 0 < x < 1 , 0<x<1, then 0 < x 2 < x < 1. 0<x^2<x<1. Therefore, if 0 < x < 1 0<x<1 , x^2=0.x+x^2, where the 0 < x 2 < x . 0<x^2<x. So we obtain that { x 2 mod x } = x 2 \{x^2 \text{ mod } x\}=x^2 for any x x in the interval ( 0 , 1 ) . (0, 1). The values of the function { x 2 mod x } \{x^2 \text{ mod } x\} when x = 0 x=0 or x = 1 x=1 are both 0, but they don't make any difference when finding the values of the corresponding integral.

Therefore, 6 ∫ 0 1 { x 2 mod x } d x = 6 ∫ 0 1 x 2 d x = 2. 6 \int_0^1 \{x^2 \text{ mod } x\}\,dx=6 \int_0^1 x^2 \,dx=2.

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