The value of int_{0}^{2} [x^{2}] dx is equal | vedclass

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(B) To evaluate $\int_{0}^{2} [x^{2}] dx$,we split the interval $[0, 2]$ based on the points where $x^{2}$ is an integer: $x^{2} = 1, 2, 3, 4$. This g

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) To evaluate $\int_{0}^{2} [x^{2}] dx$,we split the interval $[0, 2]$ based on the points where $x^{2}$ is an integer: $x^{2} = 1, 2, 3, 4$. This gives the points $x = 1, \sqrt{2}, \sqrt{3}, 2$. Thus,the integral becomes: $\int_{0}^{1} [x^{2}] dx + \int_{1}^{\sqrt{2}} [x^{2}] dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^{2}] dx + \int_{\sqrt{3}}^{2} [x^{2}] dx$ $= \int_{0}^{1} 0 dx + \int_{1}^{\sqrt{2}} 1 dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 dx + \int_{\sqrt{3}}^{2} 3 dx$ $= 0 + [x]_{1}^{\sqrt{2}} + 2[x]_{\sqrt{2}}^{\sqrt{3}} + 3[x]_{\sqrt{3}}^{2}$ $= (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3})$ $= \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3}$ $= 5 - \sqrt{2} - \sqrt{3}$.

The value of $\int_{0}^{2} [x^{2}] dx$ is equal to,where $[.]$ denotes the Greatest Integer Function $(GIF)$.

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