Let's consider a curved trapezoid bounded by the Ox axis, the curve y=f(x) and two straight lines: x=a and x=b (Fig. 85). Let's take an arbitrary value of x (just not a and not b). Let's give it an increment h = dx and consider a strip bounded by straight lines AB and CD, the Ox axis and the arc BD belonging to the curve under consideration. We will call this strip an elementary strip. The area of an elementary strip differs from the area of the rectangle ACQB by the curvilinear triangle BQD, and the area of the latter is less than the area of the rectangle BQDM with sides BQ = =h=dx) QD=Ay and area equal to hAy = Ay dx. As side h decreases, side Du also decreases and simultaneously with h tends to zero. Therefore, the area of the BQDM is second-order infinitesimal. The area of an elementary strip is the increment of the area, and the area of the rectangle ACQB, equal to AB-AC ==/(x) dx> is the differential of the area. Consequently, we find the area itself by integrating its differential. Within the figure under consideration, the independent variable l: changes from a to b, so the required area 5 will be equal to 5= \f(x) dx. (I) Example 1. Let's calculate the area bounded by the parabola y - 1 -x*, straight lines X =--Fj-, x = 1 and the O* axis (Fig. 86). at Fig. 87. Fig. 86. 1 Here f(x) = 1 - l?, the limits of integration are a = - and £ = 1, therefore J [*-t]\- -fl -- Г -1-±Л_ 1V1 -l-l-Ii-^ 3) |_ 2 3V 2 / J 3 24 24* Example 2. Let's calculate the area limited by the sinusoid y = sinXy, the Ox axis and the straight line (Fig. 87). Applying formula (I), we obtain A 2 S= J sinxdx= [-cos x]Q =0 -(-1) = lf Example 3. Calculate the area limited by the arc of the sinusoid ^у = sin jc, enclosed between two adjacent intersection points with the Ox axis (for example, between the origin and the point with the abscissa i). Note that from geometric considerations it is clear that this area will be twice the area of the previous example. However, let's do the calculations: I 5= | s\nxdx= [ - cosх)* - - cos i-(-cos 0)= 1 + 1 = 2. o Indeed, our assumption turned out to be correct. Example 4. Calculate the area bounded by the sinusoid and the Ox axis at one period (Fig. 88). Preliminary calculations suggest that the area will be four times larger than in Example 2. However, after making calculations, we obtain “i Г,*i S - \ sin x dx = [ - cos x]0 = = - cos 2l -(-cos 0) = - 1 + 1 = 0. This result requires clarification. To clarify the essence of the matter, we also calculate the area limited by the same sinusoid y = sin l: and the Ox axis in the range from l to 2i. Applying formula (I), we obtain 2l $2l sin xdx=[ - cosх]l = -cos 2i~)-c05i=- 1-1 =-2. Thus, we see that this area turned out to be negative. Comparing it with the area calculated in exercise 3, we find that their absolute values are the same, but the signs are different. If we apply property V (see Chapter XI, § 4), we get 2l I 2l J sin xdx= J sin * dx [ sin x dx = 2 + (- 2) = 0What happened in this example is not an accident. Always the area located below the Ox axis, provided that the independent variable changes from left to right, is obtained when calculated using integrals. In this course we will always consider areas without signs. Therefore, the answer in the example just discussed will be: the required area is 2 + |-2| = 4. Example 5. Let's calculate the area of the BAB shown in Fig. 89. This area is limited by the Ox axis, the parabola y = - xr and the straight line y - = -x+\. Area of a curvilinear trapezoid The required area OAB consists of two parts: OAM and MAV. Since point A is the intersection point of a parabola and a straight line, we will find its coordinates by solving the system of equations 3 2 Y = mx. (we only need to find the abscissa of point A). Solving the system, we find l; = ~. Therefore, the area has to be calculated in parts, first square. OAM and then pl. MAV: .... G 3 2, 3 G xP 3 1/2 U 2. QAM-^x = 
[replacement:
] = 
This means that the improper integral converges and its value is equal to .
Problem 1(about calculating the area of a curved trapezoid).
In the Cartesian rectangular coordinate system xOy, a figure is given (see figure) bounded by the x axis, straight lines x = a, x = b (a by a curvilinear trapezoid. It is required to calculate the area of a curvilinear trapezoid.
Solution. Geometry gives us recipes for calculating the areas of polygons and some parts of a circle (sector, segment). Using geometric considerations, we can only find an approximate value of the required area, reasoning as follows.
Let's split the segment [a; b] (base of a curved trapezoid) into n equal parts; this partition is carried out using points x 1, x 2, ... x k, ... x n-1. Let us draw straight lines through these points parallel to the y-axis. Then the given curvilinear trapezoid will be divided into n parts, into n narrow columns. The area of the entire trapezoid is equal to the sum of the areas of the columns.
Let us consider the k-th column separately, i.e. a curved trapezoid whose base is a segment. Let's replace it with a rectangle with the same base and height equal to f(x k) (see figure). The area of the rectangle is equal to \(f(x_k) \cdot \Delta x_k \), where \(\Delta x_k \) is the length of the segment; It is natural to consider the resulting product as an approximate value of the area of the kth column. 
If we now do the same with all the other columns, we will arrive at the following result: the area S of a given curvilinear trapezoid is approximately equal to the area S n of a stepped figure made up of n rectangles (see figure):
\(S_n = f(x_0)\Delta x_0 + \dots + f(x_k)\Delta x_k + \dots + f(x_(n-1))\Delta x_(n-1) \)
Here, for the sake of uniformity of notation, we assume that a = x 0, b = x n; \(\Delta x_0 \) - length of the segment, \(\Delta x_1 \) - length of the segment, etc.; in this case, as we agreed above, \(\Delta x_0 = \dots = \Delta x_(n-1) \)
So, \(S \approx S_n \), and this approximate equality is more accurate, the larger n.
By definition, it is believed that the required area of a curvilinear trapezoid is equal to the limit of the sequence (S n):
$$ S = \lim_(n \to \infty) S_n $$
Problem 2(about moving a point)
A material point moves in a straight line. The dependence of speed on time is expressed by the formula v = v(t). Find the movement of a point over a period of time [a; b].
Solution. If the movement were uniform, then the problem would be solved very simply: s = vt, i.e. s = v(b-a). For uneven movement, you have to use the same ideas on which the solution to the previous problem was based.
1) Divide the time interval [a; b] into n equal parts.
2) Consider a period of time and assume that during this period of time the speed was constant, the same as at time t k. So we assume that v = v(t k).
3) Let’s find the approximate value of the point’s movement over a period of time; we’ll denote this approximate value as s k
\(s_k = v(t_k) \Delta t_k \)
4) Find the approximate value of displacement s:
\(s \approx S_n \) where
\(S_n = s_0 + \dots + s_(n-1) = v(t_0)\Delta t_0 + \dots + v(t_(n-1)) \Delta t_(n-1) \)
5) The required displacement is equal to the limit of the sequence (S n):
$$ s = \lim_(n \to \infty) S_n $$
Let's summarize. Solutions to various problems were reduced to the same mathematical model. Many problems from various fields of science and technology lead to the same model in the process of solution. This means that this mathematical model must be specially studied.
The concept of a definite integral
Let us give a mathematical description of the model that was built in the three considered problems for the function y = f(x), continuous (but not necessarily non-negative, as was assumed in the considered problems) on the interval [a; b]:
1) split the segment [a; b] into n equal parts;
2) make up the sum $$ S_n = f(x_0)\Delta x_0 + f(x_1)\Delta x_1 + \dots + f(x_(n-1))\Delta x_(n-1) $$
3) calculate $$ \lim_(n \to \infty) S_n $$
In the course of mathematical analysis it was proven that this limit exists in the case of a continuous (or piecewise continuous) function. He is called a certain integral of the function y = f(x) over the segment [a; b] and denoted as follows:
\(\int\limits_a^b f(x) dx \)
The numbers a and b are called the limits of integration (lower and upper, respectively).
Let's return to the tasks discussed above. The definition of area given in Problem 1 can now be rewritten as follows:
\(S = \int\limits_a^b f(x) dx \)
here S is the area of the curved trapezoid shown in the figure above. This is geometric meaning of a definite integral.
The definition of the displacement s of a point moving in a straight line with a speed v = v(t) over the period of time from t = a to t = b, given in Problem 2, can be rewritten as follows:
Newton-Leibniz formula
First, let's answer the question: what is the connection between the definite integral and the antiderivative?
The answer can be found in Problem 2. On the one hand, the displacement s of a point moving in a straight line with a speed v = v(t) over the period of time from t = a to t = b is calculated by the formula
\(S = \int\limits_a^b v(t) dt \)
On the other hand, the coordinate of a moving point is an antiderivative for speed - let's denote it s(t); This means that the displacement s is expressed by the formula s = s(b) - s(a). As a result we get:
\(S = \int\limits_a^b v(t) dt = s(b)-s(a) \)
where s(t) is the antiderivative of v(t).
The following theorem was proven in the course of mathematical analysis.
Theorem. If the function y = f(x) is continuous on the interval [a; b], then the formula is valid
\(S = \int\limits_a^b f(x) dx = F(b)-F(a) \)
where F(x) is the antiderivative of f(x).
The given formula is usually called Newton-Leibniz formula in honor of the English physicist Isaac Newton (1643-1727) and the German philosopher Gottfried Leibniz (1646-1716), who received it independently of each other and almost simultaneously.
In practice, instead of writing F(b) - F(a), they use the notation \(\left. F(x)\right|_a^b \) (it is sometimes called double substitution) and, accordingly, rewrite the Newton-Leibniz formula in this form:
\(S = \int\limits_a^b f(x) dx = \left. F(x)\right|_a^b \)
When calculating a definite integral, first find the antiderivative, and then carry out a double substitution.
Based on the Newton-Leibniz formula, we can obtain two properties of the definite integral.
Property 1. The integral of the sum of functions is equal to the sum of the integrals:
\(\int\limits_a^b (f(x) + g(x))dx = \int\limits_a^b f(x)dx + \int\limits_a^b g(x)dx \)
Property 2. The constant factor can be taken out of the integral sign:
\(\int\limits_a^b kf(x)dx = k \int\limits_a^b f(x)dx \)
Calculating the areas of plane figures using a definite integral
Using the integral, you can calculate the areas not only of curved trapezoids, but also of plane figures of a more complex type, for example, the one shown in the figure. The figure P is limited by straight lines x = a, x = b and graphs of continuous functions y = f(x), y = g(x), and on the segment [a; b] the inequality \(g(x) \leq f(x) \) holds. To calculate the area S of such a figure, we will proceed as follows:
\(S = S_(ABCD) = S_(aDCb) - S_(aABb) = \int\limits_a^b f(x) dx - \int\limits_a^b g(x) dx = \)
\(= \int\limits_a^b (f(x)-g(x))dx \)
So, the area S of a figure bounded by straight lines x = a, x = b and graphs of functions y = f(x), y = g(x), continuous on the segment and such that for any x from the segment [a; b] the inequality \(g(x) \leq f(x) \) is satisfied, calculated by the formula
\(S = \int\limits_a^b (f(x)-g(x))dx \)