What is an integral, and what is its physical meaning
The appearance of the concept of an integral was due tothe necessity of finding an antiderivative function with respect to its derivative, as well as determining the amount of work, the area of complex figures, the distance traveled, with parameters outlined by curves described by nonlinear formulas.
From course
But force can change over the course of the work, and in some kind of a natural dependence. The same situation arises with the calculation of the distance traveled if the speed is not constant.
So, it is clear what an integral is for. Determining it as the sum of products of the values of a function by an infinitesimal increment of the argument completely describes the main meaning of this concept as the area of the figure bounded from above by the function line, and along the edges by the boundaries of the definition.
Jean Gaston Darboux, French mathematician, inthe second half of the XIX century very clearly explained what an integral is. He did this so clearly that on the whole, it is not difficult even for a junior high school student to understand this question.
Suppose there is a function of any complex shape. The axis of ordinates, on which the values of the argument are plotted, is divided into small intervals, ideally they are infinitesimal, but since the concept of infinity is rather abstract, it is enough to imagine simply small segments, the value of which is usually denoted by the Greek letter Δ (delta).
The function was "cut" into small bricks.
To each value of the argument there corresponds a point onaxis of ordinates, on which the corresponding values of the function are plotted. But since the boundaries of the selected section are two, then the values of the function will also be two, larger and smaller.
The sum of products of large values onThe increment Δ is called the large Darboux sum, and is denoted as S. Accordingly, smaller values on the bounded portion multiplied by Δ all together form a small Darboux sum. The section itself resembles a rectangular trapezoid, since the curvature of the function line can be neglected with an infinitesimal increment. The simplest way to find the area of such a geometric figure is to add products of a larger and smaller value to a Δ-increment and divide by two, that is, to define it as the arithmetic mean.
This is the Darboux integral:
s = Σf (x) Δ is a small sum;
S = Σf (x + Δ) Δ is a large sum.
So, what is an integral? The area bounded by the function line and the boundaries of the definition will be:
∫f (x) dx = {(S + s) / 2} + c
That is, the arithmetic average of the large and small Darboux sums is a constant value, which is nullified by differentiation.
Proceeding from the geometric expression of thisconcept, the physical meaning of the integral becomes clear. The area of the figure, delineated by the velocity function, and bounded by the time interval along the abscissa, will be the length of the traversed path.
L = ∫f (x) dx on the interval from t1 to t2,
Where
f (x) is the velocity function, that is, the formula by which it varies with time;
L is the path length;
t1 - time of the beginning of the path;
t2 is the end time of the path.
Precisely according to the same principle, the magnitude of the work is determined, only along the abscissa will the distance be plotted, and on the ordinate the magnitude of the force applied at each particular point.
