int_0^{1.5} [x^2] dx is equal to | vedclass

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(B) We need to evaluate the integral $I = \int_0^{1.5} [x^2] dx$, where $[x^2]$ denotes the greatest integer function.Split the interval $[0, 1.5]$ ba

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$2-\sqrt{2}$

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(B) We need to evaluate the integral $I = \int_0^{1.5} [x^2] dx$, where $[x^2]$ denotes the greatest integer function.Split the interval $[0, 1.5]$ based on the values where $[x^2]$ changes:$I = \int_0^1 [x^2] dx + \int_1^{\sqrt{2}} [x^2] dx + \int_{\sqrt{2}}^{1.5} [x^2] dx$$1$. For $0 \le x $\int_0^1 0 dx = 0$$2$. For $1 \le x $\int_1^{\sqrt{2}} 1 dx = [x]_1^{\sqrt{2}} = \sqrt{2} - 1$$3$. For $\sqrt{2} \le x $\int_{\sqrt{2}}^{1.5} 2 dx = 2[x]_{\sqrt{2}}^{1.5} = 2(1.5 - \sqrt{2}) = 3 - 2\sqrt{2}$Adding these values:$I = 0 + (\sqrt{2} - 1) + (3 - 2\sqrt{2})$$I = 2 - \sqrt{2}$

$\int_0^{1.5} [x^2] dx$ is equal to

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