Let f(x) = {x} denote the fractional part of | vedclass

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(C) Let $I = \int_{0}^{\sqrt{3}} \{x^2\} dx$. Since $\{x^2\} = x^2 - [x^2]$, we split the integral based on the values of $[x^2]$: For $0 \le x < 1$,

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$\sqrt{2} - \sqrt{3} + 1$

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(C) Let $I = \int_{0}^{\sqrt{3}} \{x^2\} dx$. Since $\{x^2\} = x^2 - [x^2]$, we split the integral based on the values of $[x^2]$: For $0 \le x For $1 \le x For $\sqrt{2} \le x Thus, $I = \int_{0}^{1} x^2 dx + \int_{1}^{\sqrt{2}} (x^2 - 1) dx + \int_{\sqrt{2}}^{\sqrt{3}} (x^2 - 2) dx$. $I = \left[ \frac{x^3}{3} \right]_{0}^{1} + \left[ \frac{x^3}{3} - x \right]_{1}^{\sqrt{2}} + \left[ \frac{x^3}{3} - 2x \right]_{\sqrt{2}}^{\sqrt{3}}$. $I = \left( \frac{1}{3} - 0 \right) + \left( (\frac{2\sqrt{2}}{3} - \sqrt{2}) - (\frac{1}{3} - 1) \right) + \left( (\frac{3\sqrt{3}}{3} - 2\sqrt{3}) - (\frac{2\sqrt{2}}{3} - 2\sqrt{2}) \right)$. $I = \frac{1}{3} + (-\frac{\sqrt{2}}{3} + \frac{2}{3}) + (-\sqrt{3} + \frac{4\sqrt{2}}{3})$. $I = \frac{1}{3} + \frac{2}{3} - \sqrt{3} + \frac{3\sqrt{2}}{3} = 1 - \sqrt{3} + \sqrt{2} = \sqrt{2} - \sqrt{3} + 1$.

Let $f(x) = \{x\}$ denote the fractional part of a real number $x$. Then, the value of $\int_{0}^{\sqrt{3}} f(x^2) dx$ is

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