The value of int_{0}^{4} { sqrt{x} } dx,where | vedclass

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(C) We know that $\{ \sqrt{x} \} = \sqrt{x} - [\sqrt{x}]$.So,the integral becomes $\int_{0}^{4} (\sqrt{x} - [\sqrt{x}]) dx = \int_{0}^{4} \sqrt{x} dx

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$7/3$

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(C) We know that $\{ \sqrt{x} \} = \sqrt{x} - [\sqrt{x}]$.So,the integral becomes $\int_{0}^{4} (\sqrt{x} - [\sqrt{x}]) dx = \int_{0}^{4} \sqrt{x} dx - \int_{0}^{4} [\sqrt{x}] dx$.First,calculate $\int_{0}^{4} \sqrt{x} dx = \left[ \frac{2}{3} x^{3/2} ight]_{0}^{4} = \frac{2}{3} (4^{3/2} - 0) = \frac{2}{3} 	imes 8 = \frac{16}{3}$.Next,evaluate $\int_{0}^{4} [\sqrt{x}] dx$. The value of $[\sqrt{x}]$ changes at $x = 1, 4, 9, \dots$:For $0 \le x For $1 \le x Thus,$\int_{0}^{4} [\sqrt{x}] dx = \int_{0}^{1} 0 dx + \int_{1}^{4} 1 dx = 0 + [x]_{1}^{4} = 4 - 1 = 3$.Finally,the value of the integral is $\frac{16}{3} - 3 = \frac{16 - 9}{3} = \frac{7}{3}$.

The value of $\int_{0}^{4} \{ \sqrt{x} \} dx$,where $\{ \}$ denotes the fractional part of $x$,is

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