$\int_0^{\sqrt{2}}\left[x^2\right] d x$ is

$\int_0^{\sqrt{2}}\left[x^2\right] d x$ is
  1. $2-\sqrt{2}$
  2. $2+\sqrt{2}$
  3. $\sqrt{2}-1$
  4. $\sqrt{2}-2$

Solution

$\int_1^0\left[x^2\right] d x+\int_1^{\sqrt{2}}\left[x^2\right] d x=0+\int_1^{\sqrt{2}} d x=\sqrt{2}-1$

Asked in: JEE Main 2002

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