On odd integers $t=2n-1$, $f$ alternates: $f(1)=2,f(3)=-2,f(5)=2,f(7)=-2,f(9)=2$. Between consecutive odd integers, $f$ is linear (the formula is the linear interpolant). One can write compactly $f(t)=2(-1{)}^{n+1}(2n-t)$ on $(2n-1,2n+1)$.

Integrating piecewise, on each unit interval $f$ is a triangle wave of area zero over each full period $[2n-1,2n+1]$. The antiderivative $g$ vanishes at $x=1$ and also at $x=3,5,7$ (each full positive-then-negative triangle returns to zero). Computing the explicit forms:

On $(1,3]$: $g(x)=-(x-1)(x-3)$.

On $(3,5]$: $g(x)=(x-3)(x-5)$.

On $(5,7]$: $g(x)=-(x-5)(x-7)$.

On $(7,8]$: $g(x)=(x-7)(x-9)$.

$g(x)=0$ in $(1,8]$ occurs at $x=3,5,7$, so $\alpha =3$.

For $\beta$: near $x=1^{+}$, $g(x)=-(x-1)(x-3)$, so $\frac{g(x)}{x-1}=-(x-3)\to 2$ as $x\to 1^{+}$. Hence $\beta =2$.

$\therefore \alpha +\beta =5.$