Thank you for reporting, we will resolve it shortly
Q.
The intercepts on $x-axis$ made by tangents to the curve, $y = \displaystyle\int_0^x |t| dt, x \in R$, which are parallel to the line $y = 2x$, are equal to
$\frac{dy}{dx} =\left|x\right|=2 \, \, \therefore x =\pm2$
We cari solve for y to get
$ y_{1}=\int\limits_{0}^{2}\left|t\right|dt =\int\limits_{0}^{2}tdt = \frac{t^{2}}{2}|^{2}_{0} =2 $
and $ y_3 {\int\limits_0^{-2} }\left|t\right| dt =-\int\limits_{0}^{-2}tdt =-2 $
Tangents are $y - 2 = 2(x - 2)$ and $y + 2 = 2(x + 2)$ .
Then the $x$ intercepts are obtained by putting $y = 0.$
We then get $x = \pm 1$