计算gnuplot中曲线下的面积

您的数据以x单位等距，步长为1。因此，您可以简单地将强度值与宽度（即1）相乘。如果您有不规则的数据，那么这将更加复杂。

代码：

### determination of area below curve
reset session
FILE = "SO/spectrum_spl9.txt"

# fitting function      
f(x)  = g1(x)+g2(x)+g3(x)
g1(x) = p1*exp(-(x-m1)**2/(2*s1**2))
g2(x) = p2*exp(-(x-m2)**2/(2*s2**2))
g3(x) = p3*exp(-(x-m3)**2/(2*s3**2))

# Estimation of each parameter      
p1 = 100000
p2 = 2840
p3 = 28000
m1 = 70
m2 = 150
m3 = 350
s1 = 25
s2 = 100
s3 = 90

set fit quiet nolog
fit [0:480] f(x) FILE via p1, m1, s1, p2, m2, s2, p3, m3, s3

set table $Difference
    plot myIntegral=0 FILE u 1:(myIntegral=myIntegral+f($1)-g3($1),f($1)-g3($1)) w table
unset table

set samples 500    # set samples to plot the functions
plot [0:550] FILE u 1:2 w p lc 'blue' ti FILE noenhanced, \
      f(x) ls 1, \
      g1(x) lc rgb 'black', \
      g2(x) lc rgb 'green', \
      g3(x) lc rgb 'orange', \
      $Difference u 1:2 w filledcurves lc rgb 0xddff0000 ti sprintf("Area: %.3g",myIntegral)
### end of code

结果：