How to calculate the width that contains 60% of the total area of peaks?

So starting from the center (where the max peak always lies), I want to calculate automatically the width that contains 60% of the total area under multiple peaks. How can I do that? Attached is the normalized intensities and positions.

Below is the code to an attempt to get the contribution of the main peak only from the total area:

TotalArea=trapz(x2,Int);

lims = (x2>= -3.90625E-5 ) & (x2<= 3.90625E-5); %manually selected

MainPeakArea=trapz(x2(lims),Int(lims));

MainLobePower = (MainPeakArea/TotalArea)*100;

Answer 1

[Chunru]

load position
load Intensity
whos

  Name        Size              Bytes  Class     Attributes

  Int         1x2048            16384  double              
  ans         1x38                 76  char                
  cmdout      1x33                 66  char                
  x2          1x2048            16384  double              

plot(x2, Int); hold on
% find the max
[maxI, idx] = max(Int);
maxX = x2(idx)

maxX = 0

n = length(x2);
Atotal = trapz(x2, Int);
for w=1:length(x2)
    ii = max(idx-w, 1):min(idx+w, n);
    A = trapz(x2(ii), Int(ii));
    
    if A/Atotal>=0.6
        break
    end
end
area(x2(ii), Int(ii));
plot(x2(idx), Int(idx), 'r^');
xlim([-1 1]*1e-3)
title(sprintf('Width: %g - %g', x2(ii(1)), x2(ii(end)) ));

[Lope]

@Chunru, Thank you so much for the help.

Answer 2

[Bruno Luong]

load Intensity.mat
load position.mat

ifun=@(x) interp1(x2,Int,x,'linear','extrap');
I0=integral(ifun,x2(1),x2(end))

I0 = 5.6250e-05

x = fzero(@(x) integral(ifun,-x,x)-0.6*I0,0) % the interval is (-x,x) width is 2*x

x = 3.1468e-05

[Lope]

@Bruno Luong Thank you so much for this.