Initially i thought the previous answer was ok, but when reading your entire post you specified that you needed the FIRST positive area. Also, just setting negative values to zero introduces additional errors because when going from last positive value to first negative, say eg from 1 to -1 , the zero crossing is actually between the two indexes not on the first negative index. Therefor interpolation will be a better approximation.
%here is your example, note that the interpolation gives only a very small
%improvement here since your first negative value is so close to zero.
y=[1:1:10];
[C,lags]=autocorr(y,9);
lags2=1:0.1:lags(end)+1;
CI=interp1(1+lags,C,lags2); %inperpolate 10 x.
idx1=find(CI>0,1,'first'); %from first positive value
idx2=find(CI(idx1:end)<0,1,'first'); %to first negative after first pos.
Area=trapz(CI(idx1:idx2-1))*0.1 % either give the selected value of lags2 or
% do as here, calculate with unit spacing and
% multiply by the actual increment.
Area =
1.7345
Note that you will still have errors related to the fact that you dont know the actual zero crossings. To illustrate zoom in on this plot:
figure;
plot(lags2,CI,'-x');
hold on
plot(lags2(idx1:idx2-1),CI(idx1:idx2-1),'-rx');
grid minor
