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Solve the following.
Evaluate: ∫₀^(√2) [x²] dx
Evaluate: $\int_0^{\sqrt{2}} [x^2] \, dx$
Answer
Answer
Official answer keyWe know greatest integer function is discontinuous when $x$ is an integer.
$\therefore \int_0^{\sqrt{2}} [x^2]\,dx = \int_0^1 0\,dx + \int_1^{\sqrt{2}} 1\,dx = x\Big|_1^{\sqrt{2}} = \sqrt{2} - 1$.
Final answer: $\sqrt{2} - 1$