int_0^{1.5} {[x^2] , dx},where [.] denotes th | vedclass

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(B) To evaluate the integral $I = \int_0^{1.5} [x^2] \, dx$,we break the interval $[0, 1.5]$ based on the values where $x^2$ is an integer.$x^2 = 1 \i

Vedclass · Accepted Answer

$2 - \sqrt{2}$

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(B) To evaluate the integral $I = \int_0^{1.5} [x^2] \, dx$,we break the interval $[0, 1.5]$ based on the values where $x^2$ is an integer.$x^2 = 1 \implies x = 1$$x^2 = 2 \implies x = \sqrt{2} \approx 1.414$Thus,we split the integral as follows:$I = \int_0^1 [x^2] \, dx + \int_1^{\sqrt{2}} [x^2] \, dx + \int_{\sqrt{2}}^{1.5} [x^2] \, dx$For $0 \le x For $1 \le x For $\sqrt{2} \le x Substituting these values:$I = \int_0^1 0 \, dx + \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{1.5} 2 \, dx$$I = 0 + [x]_1^{\sqrt{2}} + 2[x]_{\sqrt{2}}^{1.5}$$I = (\sqrt{2} - 1) + 2(1.5 - \sqrt{2})$$I = \sqrt{2} - 1 + 3 - 2\sqrt{2}$$I = 2 - \sqrt{2}$

$\int_0^{1.5} {[x^2] \, dx}$,where $[.]$ denotes the greatest integer function,equals

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