how would you approximate this data using the Interp1 function and spline method
For example,
year=[1;2;3;4;5;6;7;8;9;10];
pop=[5;10;15;20;25;30;35;40;50;60];
interp1(year,pop, 2.5,'spline')
whats the difference between the cubic method.
It's just a different interpolation kernel. The cubic method will gives an interpolated curve that is only once-differentiable, but it will follow the shape of the data more closely than a spline interpolation. Example:
y=[zeros(1,10) 1 1.15 1.15 1 zeros(1,10)];
x=1:numel(y);
xq=linspace(1,numel(y),1000);
plot(x,y,'o',xq,interp1(y,xq,'spline'), xq, interp1(y,xq,'cubic'))
legend('', 'Spline','Cubic')
