I1 = dx/2 * f(x0) + dx * sum_{i=1}^{n-1} f(xi) + dx/2 * f(xn)
I2 = dx/2 * f(xn) + dx * sum_{i=n+1}^{n+m-1} f(xi) + dx/2 * f(x_{n+m})
->
I1 + I2 = dx/2 * f(x0) + dx * sum_{i=1}^{n+m-1} f(xi) + dx/2 * f(x_{n+m})
Thus since the endpoints in the trapezoidal rule are only counted half, all is fine.
