given y'(t) = f(t), y(a)=0 what is y(b) = f(t) dt?

y(t) y(t_a) + y'(t_a) (t-t_a) + (1/2)y''(t_a) (t-t_a)²

but we know y'(t_a) = f(t_a), so y"(t_a) = f'(t_a)
and we can thus approximate the derivative numerically:

y"(t_a) = f'(t_a) [f(t)-f(t_a)]/(t-t_a)

so y(t) y(t_a) + f(t_a) (t-t_a) + (1/2){[f(t)-f(t_a)]/(t-t_a)}(t-t_a)²

y(t) y(t_a) + f(t_a) (t-t_a) + (1/2)[f(t)-f(t_a)](t-t_a)

y(t) y(t_a) + (1/2)[ f(t) + f(t_a) ](t-t_a)

Trapezoidal Rule algorithm:

y(t_n+1)= y(t_n) + (1/2)[ f(t_n+1) + f(t_n) ](t_n+1- t_n)

if t_n+1 = t_n + h (i.e. evenly spaced intervals)

y(t_n+ h) = y(t_n) + (h/2)[ f(t_n+ h) - f(t_n) ]

by induction this becomes:

y(b) = integral (h/2)[f(b)-f(a)] + h f(a+nh)