= Solution
If $g'$ never vanishes on $[a,b]$, <integration by parts> gives $I(\lambda)=[f(t)e^{i\lambda g(t)}/(i\lambda g'(t))]_a^b-(i\lambda)^{-1}\int_a^b(f/g')'e^{i\lambda g(t)}\,dt$, hence $\boxed{I(\lambda)=O(\lambda^{-1})}$ for smooth $f,g$. Endpoint terms usually attain this order; special cancellation can give faster decay.
For the final integral, take the imaginary part of $\int_0^1e^{i\lambda(t^3-t)}\,dt$. Its sole stationary point is $t_0=1/\sqrt3$, with $g(t_0)=-2/(3\sqrt3)$ and $g''(t_0)=2\sqrt3$. The <stationary phase> formula in part (a) therefore gives
$$
\boxed{J(\lambda)=\sqrt{\frac\pi{\lambda\sqrt3}}\sin\left(\frac\pi4-\frac{2\lambda}{3\sqrt3}\right)+O(\lambda^{-1}).}
$$
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