The area bounded by the curves y=2x^2,y=max { | vedclass

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(A) Given curves are $y=2x^2$ and $y=\max \{x-[x], x+|x|\}$.Since $x-[x] = \{x\}$,we have $y=\max \{\{x\}, x+|x|\}$.For $x \in [0, 2]$,$x+|x| = 2x$ an

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$2$

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(A) Given curves are $y=2x^2$ and $y=\max \{x-[x], x+|x|\}$.Since $x-[x] = \{x\}$,we have $y=\max \{\{x\}, x+|x|\}$.For $x \in [0, 2]$,$x+|x| = 2x$ and $\{x\} \in [0, 1)$.Since $2x \geq \{x\}$ for all $x \in [0, 2]$,the function simplifies to $y=2x$.We need the area bounded by $y=2x^2$ and $y=2x$ between $x=0$ and $x=2$.The curves intersect where $2x^2 = 2x$,which gives $x^2-x=0$,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 required area is $\int_0^1 (2x - 2x^2) dx + \int_1^2 (2x^2 - 2x) dx$.$= [x^2 - \frac{2x^3}{3}]_0^1 + [\frac{2x^3}{3} - x^2]_1^2$.$= (1 - \frac{2}{3}) + [(\frac{16}{3} - 4) - (\frac{2}{3} - 1)]$.$= \frac{1}{3} + \frac{4}{3} - (-\frac{1}{3}) = \frac{1}{3} + \frac{4}{3} + \frac{1}{3} = \frac{6}{3} = 2$.

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

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