Question

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

• A. 2
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{4}{3}$

Answer 1

Answer: (A)

Given curves $y=2x^2$,
$y=\max \{x-[x],x+|x|\}$
$=\max \{(x),x+|x|\}$
Now, graph of $\{x\}$

and graph of $x+$ mode $|x|=\{\begin{array}{ll}0,& x<0\\ 2x,& x\ge 0\end{array}$

Now, area bounded by $y=2x^2$ and
$y=\max \{\{x\},x+|x|\}=x+|x|$
and lines $x=0,x=2$ is

$\therefore$ Required area
$=\underset{0}{\overset{1}{\int }}\left(2x-2x^2\right)dx+{\int }_1^2\left(2x^2-2x\right)dx$
$={\left[\frac{2x^2}{2}-\frac{2x^3}{3}\right]}_0^1+{\left[\frac{2x^3}{3}-\frac{2x^2}{2}\right]}_1^2$
$=\left(1-\frac{2}{3}\right)+\left(\frac{16}{3}-4-\frac{2}{3}+1\right)$
$=\frac{1}{3}+\frac{14}{3}-3=5-3=2$