Calculus is the core of advanced mathematics, including differentiation and integration . In the previous articles, we introduced differentiation and its inverse operation - indefinite integral (strictly speaking, indefinite integral belongs to the differential module).
Portal: Differentiation and Derivative Indefinite Integrals
Starting today, we enter the integral module. As always, let's start with an example.
1
Using elementary mathematics, the area of general regular figures, such as circles and regular polygons, can be calculated. If the area enclosed by the curve is required, the idea of advanced mathematics needs to be used. For example, the following example: the area enclosed by the parabola and the x-axis (0<x<1) is obviously unsolvable with elementary mathematics.
We have to use the idea of "exhaustion method" to solve this problem. Divide n points equally on the interval [0,1] , and the distance between every two adjacent points is h:
For two adjacent dividing points, we can get two rectangles - the green part and the yellow part (an example in the figure below):
Then the required area has the following relationship:
By mathematical induction, we have:
This gives us the area of the desired figure. Note that if we take any point ξ between two adjacent division points, we can form the area of the blue + green part :
Similarly we have:
According to this, we can extend to another way of thinking to find the area: the segmentation of the interval is no longer equal, as long as we make the maximum distance between two adjacent division points tend to 0, we can also ensure that its limit is equal to the area of the graph . therefore:
This idea can not only be used to find the area of irregular graphics, but also can be applied to other fields, such as finding the distance of variable speed motion:
2
From this we derive the definition of the definite integral :
By definition, we immediately conclude:
The definite integral is a limit of a function, so there is also an "ε-δ" language:
Here is an example to prove that the Dirichlet function is not integrable.
3
Next, we discuss the necessary and sufficient conditions for Riemann integrability . To this end, we will introduce Darboux's theorem . Before that, we must first introduce the Darboux sum concept and two related lemmas.
The Darboux sum (Darboux sum) concept is introduced below :
According to the definition of definite bound, it is obvious that the following inequalities hold:
Introduce two lemmas.
Lemma 1: If one point is added to the original division, the large sum will not increase, and the small sum will not decrease
Lemma 2: For any two partitions, the Darboux sum of one partition must be greater than or equal to the Darboux sum of the other partition
According to Lemma 2, the Darboux large sum that can be divided arbitrarily is taken as a set, and the Daboux small sum that can be divided arbitrarily is taken as another set. They are all bounded, so the definite bound exists, and the following relationship can be deduced:
Now we can prove Darbe's theorem:
There are three points to be explained here. The first point is the inequality in the penultimate line. There are three parts in total, the blue part is obviously less than ε/2 according to the first line of the proof; the yellow part can know that the difference must be less than or equal to 0 according to Lemma 1 , and the green part can be understood as follows:
The editor of the last inequality expands and makes a detailed derivation, just look at a small interval:
The second point to be explained about Darbe's theorem is the red part of the proof, which just satisfies the function "ε-δ" language, so the limit is proved.
The third point is the most important point. From the process of proving Darbe’s theorem, we can know that the logic that can finally prove the conclusion is:
4
With Darboux's theorem, we can look at the necessary and sufficient conditions for Riemann integrability. There is a theorem:
Described in words: the necessary and sufficient condition for Riemann integrability is that the Darboux maximum sum and the Darboux small sum limit under any division of P are equal
The difference between the maximum value and the minimum value of a function in a certain interval represents the amplitude, then the necessary and sufficient condition for Riemann integrability can have a second expression:
According to this theorem, it is very simple to see that the Dirichlet function is not integrable :
Corollary 1: Continuous functions on closed intervals must be integrable
Corollary 2: Monotone functions on closed intervals must be integrable
Recalling the third point of the proof of Darbe’s theorem, we can also give a third expression of the necessary and sufficient conditions for Riemann’s integrability :
The proof can be completed by using the third point to pay attention to in Darbe's theorem, which will not be expanded here. And from this equivalence condition we can get the third inference.
Corollary 3: A bounded function with only a finite number of discontinuities on a closed interval must be integrable
If there are multiple but finite discontinuities, it can be proved by continuously dividing according to the above method.
Now we can prove that the Riemann function is integrable. Recalling the Riemann function, it is a function with a period of 1:
Portal: Continuous Function
appendix:
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.patches as patches
plt.rcParams['font.sans-serif']=['SimHei']
plt.rcParams['axes.unicode_minus']=False
x = np.linspace(0,1,500)
y = x**2
x_bar = np.linspace(0,1,20)
y_bar = x_bar**2
color = ['white' if i != 11 else 'green' for i in range(20)]
x_label = []
for i in range(20):
if i == 11:
x_label.append('x$_{i-1}$')
elif i == 12:
x_label.append('x$_{i}$')
# x_label.append('ξ$_{i}$')
else:
x_label.append('')
fig,ax = plt.subplots()
plt.vlines(1,0,1)
plt.plot(x, y, c='r',label='x**2')
plt.bar(x_bar, y_bar, color=color, edgecolor='green',width=0.5,align='edge') # 绘制y刻度标签
ax.add_patch(patches.Rectangle((x_bar[11], y_bar[11]),x_bar[12]-x_bar[11],y_bar[12]-y_bar[11],edgecolor='#CD8500',facecolor='#CD8500',fill=True))
# ax.add_patch(patches.Rectangle((x_bar[11], y_bar[11]),x_bar[12]-x_bar[11],(y_bar[12]-y_bar[11])/2,edgecolor='#1E90FF',facecolor='#1E90FF',fill=True))
plt.xlim(0, 1)
plt.ylim(0, 1)
# x_bar = list(x_bar)
# x_bar.insert(11,(x_bar[11]+x_bar[12])/2)
# x_label.insert(11,'ξ$_{i}$')
plt.xticks(x_bar, x_label) # 绘制x刻度标签
plt.yticks([])
plt.title("y=x^2曲线")
plt.show()