The area of the region ≤ft(x, y): y ≥ x^2. and ≤ft.y ≤|x| is

region → $y \geq x^2$

and  $y \leq x$

so plotting regions on cartesion plane

so intersecting point of

$y=x^2$

and $y=|x|$

$\Rightarrow x^2=|x|$

So x = -1, 0, 1

y= 1, 0, 1

Corresponding as for parabola in first quadrant $y = x^2 ⇒ x = \sqrt{2}$

for line

$y = x ⇒ x = y$

→ area of region × 2 = Total area

area of 1 region = $\int\limits_0^1 \sqrt{y}-y d y$

So total area = $2\int\limits_0^1 \sqrt{y}-y d y$

$=2\left[\frac{y^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{y^{1+1}}{1+1}\right]_0^1$

$=2\left[\frac{2}{3} y^{3 / 2}-\frac{y^2}{2}\right]_0^1= 2\left[\frac{2}{3}-\frac{1}{2}\right] =\frac{4}{3}-1=\frac{4-3}{3}$

$=\frac{1}{3}$