In the following question I am trying to use Adams bashforth method to show for $r=1$ it is the forward Euler method. Here are my workings, are they correct? if not how do I solve my question.
$$y'=f(x,y), y(x_0) = y_0$$
and we know that $$x_{n+1} = x_n + h$$
$$\int^{x_{n+1}}_{x_n} dy = \int ^{x_{n+1}}_{x_n} f(x,y)dx$$ $$y(x_{n+1})- y(x_n) = \int^{x_{n+1}}_{x_n} f(x,y)dx$$ $$y_{n+1} = y_n +\int^{x_{n+1}}_{x_n} f(x,y)dx$$
if we were to use the $0$th order poly which is constant the first point and the present point then $f(x_n,y_n)$
$$y_{n+1} = y_n + \int^{x_{n+1}}_{x_n} f(x_n,y_n) dx \\ = y_n + f(x_n,y_n)(x_{n+1}-x_n)$$
Then $$y_{n+1} = y_n+ hf_n$$
which is Euler's forward method since $x_{n+1} = x_n + h$ and $f(x_n,y_n) = f_n$