= Solution
On an attracting branch of the <critical manifold>, $w=f(v)$ to leading order. The <chain rule> therefore gives
$$
f'(v)\dot v=\dot w=\gamma v-f(v),\qquad
dt=\frac{f'(v)}{\gamma v-f(v)}\,dv.
$$
The initial fast segment ends at $v=1$, and the first slow phase ends at the right <fold point of a critical manifold>, $v_R$. Hence
$$
\boxed{T_1=\int_1^{v_R}g(v)\,dv,\qquad
g(v)=\frac{-3v^2+2(1+a)v-a}{\gamma v-v(a-v)(v-1)},\qquad
v_1=1,\quad v_2=v_R.}
$$
On this interval $f'(v)<0$ and $\gamma v-f(v)>0$. The <integrand> is negative, but the upper limit is smaller than the lower limit, so $T_1>0$. The numerator tends to zero at the fold while the denominator remains positive; the leading <integral> is finite. Fast segments and the <fold delay> give vanishing corrections as $\varepsilon\to0^+$.
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