I think I'm pretty close to the solution. Since $n$ is even, it holds $x_k=-x_{n-k}$. That implies $$\omega_{n+1}(x)=(x-x_{n/2})\prod_{k=0}^{\frac{n}{2}-1}(x-x_k)\prod_{k=0}^{\frac{n}{2}-1}(x+x_k)=x\prod_{k=0}^{\frac{n}{2}-1}(x^2-x_k^2)$$
So, it should hold $$\int_{-t}^t\omega_{n+1}(x)=0$$
I've tested this with $n$ even and some $t>0$. It seems to be true. But I'm still unable to argue why.
Let's look first at $\int_{-1}^{1} f(t) \ dt$.
We want the formula to evaluate $\int_{-1}^{1} dt, \int_{-1}^{1} t \ dt, \int_{-1}^{1} t^{2} \ dt, \int_{-1}^{1} t^{3} \ dt, \int_{-1}^{1} t^{4} \ dt, \int_{-1}^{1} t^{5} \ dt, \int_{-1}^{1} t^{6} \ dt $ and $\int_{-1}^{1} t^{7} \ dt$ exactly.
That leads to the following system of equations:
$$ 2 = \omega_{0} + \omega_{1}$$
$$ 0 = \omega_{0} x_{0} + \omega_{1} x_{1}+ \omega_{2}+\omega_{3}$$
$$ \frac{2}{3} = \omega_{0} x_{0}^{2} + \omega_{1} x_{1}^{2} + 2 \omega_{2} x_{2} + 2 \omega_{3} x_{3}$$
$$ 0 = \omega_{0} x_{0}^{3} + \omega_{1} x_{1}^{3}+ 3 \omega_{2} x_{2}^{2} + 3 \omega_{3} x_{3}^{2} $$
$$ \frac{2}{5} = \omega_{0} x_{0}^{4} + \omega_{1} x_{1}^{4}+ 4 \omega_{2} x_{2}^{3} + 4 \omega_{3} x_{3}^{3} $$
$$0 = \omega_{0} x_{0}^{5} + \omega_{1} x_{1}^{5} + 5 \omega_{2} x_{2}^{4} + 5 \omega_{3} x_{3}^{4} $$
$$\frac{2}{7} = \omega_{0} x_{0}^{6} + \omega_{1} x_{1}^{6} + 6 \omega_{2} x_{2}^{5} + 6\omega_{3} x_{3}^{5} $$
$$ 0 = \omega_{0} x_{0}^{7} + \omega_{1} x_{1}^{7} + 7 \omega_{2} x_{2}^{6} + 7\omega_{3} x_{3}^{6}$$
Solve the system using a numerical solver.
Then use the the fact that $$\int_{a}^{b} f(x) \ dx = \frac{b-a}{2} \int_{-1}^{1} f \Big(a+(1+t)\frac{b-a}{2} \Big) \ dt$$
Best Answer
For $n=1$, we have that $l_0$ and $l_1$ are linear functions. Moreover, $l_0(a) = l_1(b) = 1$ and $l_0(b) = l_1(a) = 0$. It follows that $$A_i = \int_a^b l_i(x) dx = \frac{1}{2}(b-a)$$ for $i=1,2$.