= Solution
Let $\Phi(t)=2t^4-t^2$. Its stationary points on $[0,1]$ are the endpoint $0$ and the interior point $1/2$, with
$$
\Phi(0)=0,quad \Phi''(0)=-2,\qquad
\Phi(1/2)=-\frac18,quad \Phi''(1/2)=4.
$$
The endpoint stationary-phase contribution to $\int_0^1e^{ix\Phi(t)}dt$ is
$$
\frac{\sqrt\pi}{2\sqrt x}e^{-i\pi/4},
$$
and the interior contribution is
$$
\sqrt{\frac\pi{2x}}e^{i(\pi/4-x/8)}.
$$
The nonstationary endpoint contributes only $O(x^{-1})$. Taking imaginary parts gives
$$
\boxed{I(x)\sim\sqrt{\frac\pi{2x}}
\left[\sin\left(\frac\pi4-\frac x8\right)-\frac12\right]}.
$$
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