Can you find a general rule for this? - Part (2)

Calculus Level 4

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 2 { x 2 mod x } d x 6 \int_0^2 \{x^2 \text{ mod } x\}\,dx


The answer is 7.

Alisa Meier by Alisa Meier , 24, Germany

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1 solution

For 0 < x < 1 0 \lt x \lt 1 we have that x 2 < x , x^{2} \lt x, and so on this interval r = x 2 . r = x^{2}.

For 1 < x < 2 1 \lt x \lt 2 we have that 0 < x 2 − x < x , 0 \lt x^{2} - x \lt x, and so on this interval r = x 2 − x . r = x^{2} - x.

We thus find that

∫ 0 2 { x 2 m o d x } d x = ∫ 0 1 x 2 d x + ∫ 1 2 ( x 2 − x ) d x = ( x 3 3 ) 0 2 − ( x 2 2 ) 1 2 = 8 3 − ( 2 − 1 2 ) = 7 6 . \displaystyle\int_{0}^{2} \{x^{2} \mod{x}\} dx = \int_{0}^{1} x^{2} dx + \int_{1}^{2} (x^{2} - x) dx = \left(\dfrac{x^{3}}{3}\right)_{0}^{2} - \left(\dfrac{x^{2}}{2}\right)_{1}^{2} = \dfrac{8}{3} - \left(2 - \dfrac{1}{2}\right) = \dfrac{7}{6}.

The desired answer is thus 6 ∗ 7 6 = 7 . 6*\dfrac{7}{6} = \boxed{7}.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Why not use floor function? It would be easier.

Kartik Sharma - 5 years, 8 months ago

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