Question

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

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

Answer 1

Answer: (D)

$y=\max \cdot \{x-[x],x+|x|\}$
$\Rightarrow y=\{\begin{array}{ll}2x,& x\ge 0\\ \{x\},& x<0\end{array}$
Shaded region is required Area
$=\left|\underset{0}{\overset{1}{\int }}{\left(2x-2x^2\right)}_{dx}\right|+\left|\underset{1}{\overset{2}{\int }}\left(2x^2-2x\right)dx\right|$

$=\left|\underset{0}{\overset{1}{\int }}2xdx-\underset{0}{\overset{1}{\int }}2x^{2}dx\right|+\left|\underset{1}{\overset{2}{\int }}2x^{2}dx-\underset{1}{\overset{2}{\int }}2xdx\right|$
$=\left|{\left[\frac{2x^2}{2}\right]}_0^1-{\left[\frac{2x^3}{3}\right]}_0^1\right|+\left|{\left[\frac{2x^3}{3}\right]}_1^2-{\left[\frac{2x^2}{2}\right]}_1^2\right|$
$=\left|1-\frac{2}{3}\right|+\left|\frac{16}{3}-\frac{2}{3}-4+1\right|=\frac{1}{3}+\frac{5}{3}=2$ sq. units.