Question

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

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

Answer 1

Answer: (D)

$ y=\max \cdot\{x-[x], x+|x|\} $
$\Rightarrow y=\begin{cases}2 x, & x \geq 0 \\ \{x\}, & x < 0\end{cases}$
Shaded region is required Area
$=\left|\int\limits_{0}^{1}\left(2 x-2 x^{2}\right)_{d x}\right|+\left|\int\limits_{1}^{2}\left(2 x^{2}-2 x\right) d x\right|$

$=\left|\int\limits_{0}^{1} 2 x \,d x-\int\limits_{0}^{1} 2 x^{2}\, d x\right|+\left|\int\limits_{1}^{2} 2 x^{2} \,d x-\int\limits_{1}^{2} 2 x\, d x\right|$
$=\left|\left[\frac{2 x^{2}}{2}\right]_{0}^{1}-\left[\frac{2 x^{3}}{3}\right]_{0}^{1}\right|+\left|\left[\frac{2 x^{3}}{3}\right]_{1}^{2}-\left[\frac{2 x^{2}}{2}\right]_{1}^{2}\right|$
$ =\left|1-\frac{2}{3}\right|+\left|\frac{16}{3}-\frac{2}{3}-4+1\right|=\frac{1}{3}+\frac{5}{3}=2$ sq. units.