A 1d example won’t do. 
But here is a simple 2d one.
Take f(x, y) = x * exp(y). We can plot this in 3d or as a countour plot:
f = lambda x, y: x * y.exp()
xx = torch.linspace(-1,1)[None].expand(100,100)
yy = torch.linspace(-1,1)[:, None].expand(100,100)
zz = f(xx, yy)
x0 = torch.tensor([-0.5, 0.5])
y0 = torch.tensor([-0.25, 0.5])
z0 = f(x0, y0)
fig = pyplot.figure()
ax = fig.gca(projection='3d')
ax.scatter3D(x0[0], y0[0], z0[0], color=['k'], s=20)
ax.scatter3D(x0[1], y0[1], z0[1], color=['k'], s=20)
ax.set_transform
ax.plot_surface(xx.numpy(),yy.numpy(),zz.numpy(), cmap=pyplot.cm.coolwarm)
ax.view_init(elev=40)
pyplot.figure()
pyplot.contourf(xx.numpy(),yy.numpy(),zz.numpy(), cmap=pyplot.cm.coolwarm, levels=20)
pyplot.plot(x0, y0)

3d:

Contour plot:

Now I added two points (in the 3d) and a line in the countour plot. We take the top right end of the line as x and the lower bottom as x’.
We can parametrize this line:
x_line = (x0[0] + torch.linspace(0, 1, 1000) * (x0[1] - x0[0])).requires_grad_()
y_line = (y0[0] + torch.linspace(0, 1, 1000) * (y0[1] - y0[0])).requires_grad_()

Note that this looks a like the x’ + α(x-x’) you have as an argument to ∂F/∂x_i in the integral in your equation (1).
And indeed we can compute the gradients for each point on the line:
z_line = f(x_line, y_line)
z_line.sum().backward()

We can approximate the integral over 0…1 by taking the mean over the .grads. Thus we can calculate the integrated gradients:
ig_x = (x0[1] - x0[0]) * x_line.grad.mean()
ig_y = (y0[1] - y0[0]) * y_line.grad.mean()

This gives ig_x as 1.1599 and ig_y as 0.0540.
As a sanity check, we can compare ig_x + ig_y with z0[1]-z0[0] and indeed they seem to differ by 0.0001, which looks good.
Best regards
Thomas