The area (in sq. units) of the region A = {(x | vedclass

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(A) The region is defined by $(x-1)[x] \leq y \leq 2\sqrt{x}$ for $0 \leq x \leq 2$.First,we define the function $f(x) = (x-1)[x]$:For $0 \leq x < 1$,

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$\frac{8}{3}\sqrt{2} - \frac{1}{2}$

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(A) The region is defined by $(x-1)[x] \leq y \leq 2\sqrt{x}$ for $0 \leq x \leq 2$.First,we define the function $f(x) = (x-1)[x]$:For $0 \leq x For $1 \leq x At $x = 2$,$[x] = 2$,so $f(2) = (2-1)(2) = 2$.The upper boundary is $y = 2\sqrt{x}$.The area $A$ is given by the integral of the upper curve minus the lower curve:$A = \int_{0}^{2} 2\sqrt{x} \, dx - \int_{1}^{2} (x-1) \, dx$.Calculating the first integral:$\int_{0}^{2} 2\sqrt{x} \, dx = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{2} = 2 \cdot \frac{2}{3} \cdot 2^{3/2} = \frac{4}{3} \cdot 2\sqrt{2} = \frac{8\sqrt{2}}{3}$.Calculating the second integral (area of the triangle formed by $y = x-1$ from $x=1$ to $x=2$):$\int_{1}^{2} (x-1) \, dx = \left[ \frac{(x-1)^2}{2} \right]_{1}^{2} = \frac{1^2}{2} - 0 = \frac{1}{2}$.Thus,the total area is $A = \frac{8\sqrt{2}}{3} - \frac{1}{2}$.

The area (in sq. units) of the region $A = \{(x, y) : (x-1)[x] \leq y \leq 2\sqrt{x}, 0 \leq x \leq 2\}$,where $[t]$ denotes the greatest integer function,is

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