Answer
$\frac{7}{12}$
Work Step by Step
The area between two curves $y=f(x)$ and $y=g(x)$ between $x=a$ and $x=b$ is given by the formula:
$$A=\int_a^b |f(x)-g(x)|dx.\quad\quad\quad(1)$$
Given
$$f(x)=1-2 x^2, \quad g(x)=|x|$$
Since for $x>0$, $|x|= x$, find the intersection points
\begin{aligned}
1-2x^2&= x\\
2x^2+x-1&=0\\
(2x-1)(x+1)&=0
\end{aligned}
Then $x=1/2,\ \ x=-1$, we refuse $x=-1<0$
By symmetry, we find the area between $a=0$, $a=1/2$
Since $ 1-2x^2\geq x $ for $0\leq x\leq 1/2$, then
\begin{aligned}
\text{Area} & =2 \int_0^{1 / 2}\left[\left(1-2 x^2\right)-x\right] d x\\
&=2 \int_0^{1 / 2}\left(-2 x^2-x+1\right) d x \\
& =2\left[-\frac{2}{3} x^3-\frac{1}{2} x^2+x\right]_0^{1 / 2}\\
&=2\left[\left(-\frac{1}{12}-\frac{1}{8}+\frac{1}{2}\right)-0\right] \\
& =2\left(\frac{7}{24}\right)\\
&=\frac{7}{12}
\end{aligned}
