The value of int_0^2 [x^2] dx is (where [x] d | vedclass

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(A) We need to evaluate the integral $I = \int_0^2 [x^2] dx$. Since the integrand $[x^2]$ changes its value at points where $x^2$ is an integer,we fin

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$5 - \sqrt{2} - \sqrt{3}$

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(A) We need to evaluate the integral $I = \int_0^2 [x^2] dx$. Since the integrand $[x^2]$ changes its value at points where $x^2$ is an integer,we find these points in the interval $[0, 2]$. The values of $x^2$ are $1, 2, 3, 4$ at $x = 1, \sqrt{2}, \sqrt{3}, 2$ respectively. We split the integral into sub-intervals: $I = \int_0^1 [x^2] dx + \int_1^{\sqrt{2}} [x^2] dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^2] dx + \int_{\sqrt{3}}^2 [x^2] dx$. In the interval $[0, 1)$,$0 \le x^2 In the interval $[1, \sqrt{2})$,$1 \le x^2 In the interval $[\sqrt{2}, \sqrt{3})$,$2 \le x^2 In the interval $[\sqrt{3}, 2)$,$3 \le x^2 Thus,$I = \int_0^1 0 dx + \int_1^{\sqrt{2}} 1 dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 dx + \int_{\sqrt{3}}^2 3 dx$. $I = 0 + (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3})$. $I = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3}$. $I = 5 - \sqrt{2} - \sqrt{3}$.

The value of $\int_0^2 [x^2] dx$ is (where $[x]$ denotes the greatest integer function not greater than $x$)

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