∫ 0 2 [ x 2 ] d x is, 2 − 2 , 2 + 2 , 2 − 1, 2 − 2

∫ 1 0 [ x 2 ] d x + ∫ 1 2 [ x 2 ] d x = 0 + ∫ 1 2 d x = 2 − 1

∫ 0 2 [ x 2 ] d x is

Question: Answered & Verified by Expert

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

MathematicsDefinite IntegrationJEE MainJEE Main 2002

Options:

Solution:

2397 Upvotes Verified Answer

The correct answer is:
$\sqrt{2}-1$

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

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