Question

The area bounded by the curves $y=2 x^{2}$, $y=\max \{x-[x]+|x|\}$ 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 $y=2 x^{2}$,
$y =\max \{x-[x], x+|x|\} $
$=\max \{(x), x+|x|\}$
Now, graph of $\{x\}$

and graph of $x+$ mode $|x| = \begin{cases} 0, & {x<\,0 } \\[2ex] 2x, &{x \geq\, 0} \\ \end{cases}$

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

$\therefore $ Required area
$=\int\limits_{0}^{1}\left(2 x-2 x^{2}\right) d x+\int_{1}^{2}\left(2 x^{2}-2 x\right) d x $
$=\left[\frac{2 x^{2}}{2}-\frac{2 x^{3}}{3}\right]_{0}^{1}+\left[\frac{2 x^{3}}{3}-\frac{2 x^{2}}{2}\right]_{1}^{2} $
$=\left(1-\frac{2}{3}\right)+\left(\frac{16}{3}-4-\frac{2}{3}+1\right) $
$=\frac{1}{3}+\frac{14}{3}-3=5-3=2$