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1

You are approximating a definite integral with cumtrapz it won't give you the same result as the integrated equation unless you add a constant and plot with the given x coordinates: import numpy as np import matplotlib.pyplot as plt import scipy.integrate as it x = np.arange(-10,10, 0.01) # start,stop,step f = x**2 f_int=it.cumtrapz(f,x, initial=0) plt....


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When one says an algorithm is of order $O(n)$, that may mean that the complexity is given by: $c + b*n$. With every new element you add you increase in runtime (effectively). What mathematically minded people often forget is that these statements do not include how large the constants are. That of course carries over to $O(n²)$ and such. I can not answer ...


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