Area under gaussian curve

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masoud avaznejad
masoud avaznejad 2020-12-13
评论： Star Strider 2020-12-14
Hi guys
I want the value of area_under_curve to be exactly 30000 from 8.5 to 17
Is this way true to use integral?
Using matlab R2019b
clc;
clear;
close all;
gauss = @(x,mu,sig,amp,vo)amp*exp(-(((x-mu).^2)/(2*sig.^2)))+vo;
x = linspace(8.5,17,1000000);
mu = 12;
sig = 1.19895;
amp = 10000;
vo = 0;
gauss = gauss(x,mu,sig,amp,vo);
plot(x, gauss/1000, 'g-', 'LineWidth',.1)
gauss = @(x)amp*exp(-(((x-mu).^2)/(2*sig.^2)))+vo;
arear_under_crve = integral(gauss,8.5,17);

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Star Strider
Star Strider 2020-12-13
Define ‘gauss’ as:
gauss = @(x,amp) amp*exp(-(((x-mu).^2)/(2*sig.^2)))+vo;
and:
arear_under_crve = @(amp) integral(@(x)gauss(x,amp),8.5,17);
then:
amp = fsolve(@(amp) arear_under_crve(amp) - 30000, 1)
produces:
amp =
9999.98895298012
.
  2 个评论
masoud avaznejad
masoud avaznejad 2020-12-14
Dear Star Strider
It returns amp for me in the value of 10000
And if I uses Function trapz it would be completly diffrent in asnwer!
which one is more accurate?
Star Strider
Star Strider 2020-12-14
Depending on what you have set for your format, it could round up to 10000. (I used format long g to display it.) It might also give slilghtly different results with different MATLAB versions. I am using R2020b, Update 3.
I would trust the integral and fsolve result. The trapz function is useful if you have vectors you want to integrate, however integral is more accurate considering that you have defined your code in terms of functions (specifically, anonymous functions).
The trapz result would also depend on the resolution of the vectors you created from your function. The argument and function results would have to have very fine resolution (very long vectors with small increments) to equal the integral and fsolve result.

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