You can define your own residuals function, including a penalization parameter, like detailed in the code below, where it is known beforehand that the integral along the interval must be 2.. If you test without the penalization you will see that what your are getting is the conventional curve_fit:
import matplotlib.pyplot as plt import scipy from scipy.optimize import curve_fit, minimize, leastsq from scipy.integrate import quad from scipy import pi, sin x = scipy.linspace(0, pi, 100) y = scipy.sin(x) + (0. + scipy.rand(len(x))*0.4) def func1(x, a0, a1, a2, a3): return a0 + a1*x + a2*x**2 + a3*x**3 # here you include the penalization factor def residuals(p, x, y): integral = quad( func1, 0, pi, args=(p[0], p[1], p[2], p[3]))[0] penalization = abs(2.-integral)*10000 return y - func1(x, p[0], p[1], p[2], p[3]) - penalization popt1, pcov1 = curve_fit( func1, x, y ) popt2, pcov2 = leastsq(func=residuals, x0=(1., 1., 1., 1.), args=(x, y)) y_fit1 = func1(x, *popt1) y_fit2 = func1(x, *popt2) plt.scatter(x, y, marker='.') plt.plot(x, y_fit1, color='g', label='curve_fit') plt.plot(x, y_fit2, color='y', label='constrained') plt.legend(); plt.xlim(-0.1, 3.5); plt.ylim(0, 1.4) print ('Exact integral:', quad(sin , 0, pi)[0]) print ('Approx integral1:', quad(func1, 0, pi, args=(popt1[0], popt1[1], popt1[2], popt1[3]))[0]) print ('Approx integral2:', quad(func1, 0, pi, args=(popt2[0], popt2[1], popt2[2], popt2[3]))[0]) plt.show() #Exact integral: 2.0 #Approx integral1: 2.60068579748 #Approx integral2: 2.00001911981 Other related questions:
