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Let $ [\cdot] $ denote the greatest integer function. Then how to evaluate the integral $$I=\int_0^2 [x^2]dx?$$

Fakemistake
  • 2,718

1 Answers1

3

Hint:

$$\int_0^2\lfloor x^2\rfloor dx=\int_0^1 0dx+\int_1^\sqrt21dx+\int_\sqrt2^\sqrt3 2dx+\int_\sqrt3^23dx$$

J. W. Tanner
  • 60,406
  • Integrals are additive on intervals; I chose intervals where $\lfloor x^2\rfloor$ is constant, so the integrals are easy to evaluate – J. W. Tanner Apr 19 '20 at 04:19