The area (in square units) bounded by the cur | vedclass

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(D) Given $y = \max \{x-[x], x+|x|\}$. For $x \in [0, 2]$,$x \geq 0$,so $x+|x| = 2x$ and $x-[x] = \{x\}$.Since $2x \geq \{x\}$ for all $x \in [0, 2]$,

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$2$

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(D) Given $y = \max \{x-[x], x+|x|\}$. For $x \in [0, 2]$,$x \geq 0$,so $x+|x| = 2x$ and $x-[x] = \{x\}$.Since $2x \geq \{x\}$ for all $x \in [0, 2]$,we have $y = 2x$.We need the area bounded by $y = 2x^2$ and $y = 2x$ between $x = 0$ and $x = 2$.The curves intersect where $2x^2 = 2x$,i.e.,$x^2 - x = 0$,so $x = 0$ and $x = 1$.For $x \in [0, 1]$,$2x \geq 2x^2$. For $x \in [1, 2]$,$2x^2 \geq 2x$.The area is given by:$A = \int_0^1 (2x - 2x^2) dx + \int_1^2 (2x^2 - 2x) dx$$A = \left[x^2 - \frac{2x^3}{3}ight]_0^1 + \left[\frac{2x^3}{3} - x^2ight]_1^2$$A = \left(1 - \frac{2}{3}ight) - 0 + \left(\left(\frac{16}{3} - 4ight) - \left(\frac{2}{3} - 1ight)ight)$$A = \frac{1}{3} + \left(\frac{4}{3} - (-\frac{1}{3})ight) = \frac{1}{3} + \frac{5}{3} = \frac{6}{3} = 2 	ext{ sq. units.}$

The area (in square units) bounded by the curves $y=2x^2$ and $y=\max \{x-[x], x+|x|\}$ in between the lines $x=0$ and $x=2$ is

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