Please verify integral and approximation, boundary theory

[fahraynk]

I used Newtons method and taylor approximations to solve this equation $$f'''+\frac{m+1}{2}ff''+m(1-f^{'2})=0$$
It solves for velocity of air over a flat plate.
The velocity is a constant $u_e$ everywhere except in a boundary layer over the plate, where the velocity is a function of distance from the plate = u(y).
$$f'=f'(\eta)=\frac{u(y)}{u_e}\\\\\eta=y\sqrt{\frac{u_e}{\nu x}}$$
$\nu$ is viscosity.

now that I have f', I need to calculate this integral
$$\int_0^\infty (1-f') dy $$

So my question which I would like help with is, can someone please tell me if the following work is correct :
$$\frac{dy}{d\eta}=\sqrt{\frac{\nu x}{u_e}}\\\\
\int_0^\infty (1-f') dy = \int_0^{\eta_{max}} (1-f') \frac{dy}{d\eta}d\eta = \sqrt{\frac{\nu x}{u_e}}\int_0^{\eta_{max}} (1-f') d\eta$$

$\eta_{max}$ is the location where the boundary layer ends, and $\frac{u}{u_e}=1$ when $\eta \longrightarrow \eta_{max}$, thus the integral becomes 1-1=0 at $\eta=\eta_{max}$

Also, I am going to use this rule to approximate in MATLAB :
$$\int_A^B G(x) dx = (B-A)G(\frac{A+B}{2})\\\\
K= B-A = f'[2:N+1]-f'[1:N]\\\\
J = f'(\frac{A+B}{2}) = \frac{1}{2}(f'[1:N]+f'{2:N+1})\\\\
G(\frac{A+B}{2}) = ones(N)-J\\\\
\int_A^B G(x) dx = sum( K * (ones(N)-J) )$$
For this to work, I would have to input a specific x value.

 

[fahraynk]

I messed up K, and some other stuff, so to write over the integration :
instead of $\int G(x)dx=(B-A)G(\frac{A+B}{2})$ it should read
$$
\int_A^B G(x)dx
=\sum_{i=1}^{N}

{\frac{(B-A)}{N}G(\frac{A+B}{2})} \implies \int_0^{\infty}(1-f')dy =

sum[\frac{\eta_{max}}{N} * (ones(N)- \frac{1}{2}( f'[1:N]+f'[2:N+1] )]
$$
$\eta_{max}/N$ represents B-A, the number of steps in the $\eta$ direction. Ones(N) is a vector length N of 1's. $Ones(N) =[1,1,1...1_{N-1},1_N]$