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(B) The integral is given by $I = \int_0^2 x^2 [x] \, dx$. Since $[x]$ is the greatest integer function,we split the interval $[0, 2]$ into $[0, 1)$ a

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$7$

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(B) The integral is given by $I = \int_0^2 x^2 [x] \, dx$. Since $[x]$ is the greatest integer function,we split the interval $[0, 2]$ into $[0, 1)$ and $[1, 2]$. For $0 \le x For $1 \le x Thus,$I = \int_0^1 x^2(0) \, dx + \int_1^2 x^2(1) \, dx$. $I = 0 + \int_1^2 x^2 \, dx$. $I = \left[ \frac{x^3}{3} \right]_1^2 = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$.

If $[x]$ denotes the greatest integer less than or equal to $x,$ then the value of the integral $\int_0^2 {{x^2}[x]\,dx} $ equals (in $/3$)

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