Y 2 x 1 find the area. Calculate the area of a figure bounded by lines

Task 1(on the calculation of the area of a curvilinear trapezoid).

In the Cartesian rectangular coordinate system xOy, a figure is given (see figure), bounded by the x axis, straight lines x \u003d a, x \u003d b (a curvilinear trapezoid. It is required to calculate the area of \u200b\u200bthe curvilinear trapezoid.
Decision. Geometry gives us recipes for calculating the areas of polygons and some parts of a circle (sector, segment). Using geometric considerations, we will be able to find only an approximate value of the required area, arguing as follows.

Let's split the segment [a; b] (base of a curvilinear trapezoid) into n equal parts; this partition is feasible with the help of points x 1 , x 2 , ... x k , ... x n-1 . Let us draw 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.

Consider separately the k-th column, i.e. curvilinear trapezoid, the base of which 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 $f(x_k) \cdot \Delta x_k $, where $\Delta x_k $ is the length of the segment; it is natural to consider the compiled product as an approximate value of the area of the kth column.

If we now do the same with all the other columns, then we 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 consider that a \u003d x 0, b \u003d x n; $\Delta x_0 $ - segment length , $\Delta x_1 $ - segment length , etc; while, as we agreed above, $\Delta x_0 = \dots = \Delta x_(n-1) $

So, $S \approx S_n $, and this approximate equality is the more accurate, the larger n.
By definition, it is assumed that the desired area of the curvilinear trapezoid is equal to the limit of the sequence (S n):
$$ S = \lim_(n \to \infty) S_n $$

Task 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 displacement of a point over the time interval [a; b].
Decision. If the motion were uniform, then the problem would be solved very simply: s = vt, i.e. s = v(b-a). For uneven motion, one has to use the same ideas on which the solution of the previous problem was based.
1) Divide the time interval [a; b] into n equal parts.
2) Consider a time interval and assume that during this time interval the speed was constant, such as at time t k . So, we assume that v = v(t k).
3) Find the approximate value of the point displacement over the time interval , this approximate value will be denoted by s k
$s_k = v(t_k) \Delta t_k $
4) Find the approximate value of the 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. The solutions of 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. So, this mathematical model should be specially studied.

The concept of a definite integral

Let us give a mathematical description of the model that was constructed in the three considered problems for the function y = f(x), which is continuous (but not necessarily non-negative, as was assumed in the considered problems) on the segment [a; b]:
1) split the segment [a; b] into n equal parts;
2) 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) compute $$ \lim_(n \to \infty) S_n $$

In the course of mathematical analysis, it was proved that this limit exists in the case of a continuous (or piecewise continuous) function. He's called a definite integral of the function y = f(x) over the segment [a; b] and are denoted like this:
$\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 curvilinear trapezoid shown in the figure above. This is what geometric meaning of the definite integral.

The definition of the displacement s of a point moving in a straight line with a speed v = v(t) over the time interval from t = a to t = b, given in Problem 2, can be rewritten as follows:

Newton - Leibniz formula

To begin with, let's answer the question: what is the relationship between a definite integral and an antiderivative?

The answer can be found in problem 2. On the one hand, the displacement s of a point moving along a straight line with a speed v = v(t) over a time interval from t = a to t = b and is calculated by the formula
$S = \int\limits_a^b v(t) dt $

On the other hand, the coordinate of the moving point is the antiderivative for the speed - let's denote it s(t); hence 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 for v(t).

The following theorem was proved in the course of mathematical analysis.
Theorem. If the function y = f(x) is continuous on the segment [a; b], then the formula
$S = \int\limits_a^b f(x) dx = F(b)-F(a) $
where F(x) is the antiderivative for f(x).

This 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 $

Calculating a definite integral, first find the antiderivative, and then carry out a double substitution.

Based on the Newton-Leibniz formula, one can obtain two properties of a 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 area not only of curvilinear trapezoids, but also of plane figures of a more complex type, such as the one shown in the figure. The figure P is bounded 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 the figure bounded by the straight lines x = a, x = b and the 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, is calculated by the formula
$S = \int\limits_a^b (f(x)-g(x))dx $

Table of indefinite integrals (antiderivatives) of some functions

$$ \int 0 \cdot dx = C $$ $$ \int 1 \cdot dx = x+C $$ $$ \int x^n dx = \frac(x^(n+1))(n+1 ) +C \;\; (n \neq -1) $$ $$ \int \frac(1)(x) dx = \ln |x| +C $$ $$ \int e^x dx = e^x +C $$ $$ \int a^x dx = \frac(a^x)(\ln a) +C \;\; (a>0, \;\; a \neq 1) $$ $$ \int \cos x dx = \sin x +C $$ $$ \int \sin x dx = -\cos x +C $$ $ $ \int \frac(dx)(\cos^2 x) = \text(tg) x +C $$ $$ \int \frac(dx)(\sin^2 x) = -\text(ctg) x +C $$ $$ \int \frac(dx)(\sqrt(1-x^2)) = \text(arcsin) x +C $$ $$ \int \frac(dx)(1+x^2 ) = \text(arctg) x +C $$ $$ \int \text(ch) x dx = \text(sh) x +C $$ $$ \int \text(sh) x dx = \text(ch )x+C $$

In the previous section, devoted to the analysis of the geometric meaning of a definite integral, we obtained a number of formulas for calculating the area of a curvilinear trapezoid:

Yandex.RTB R-A-339285-1

S (G) = ∫ a b f (x) d x for a continuous and non-negative function y = f (x) on the segment [ a ; b] ,

S (G) = - ∫ a b f (x) d x for a continuous and non-positive function y = f (x) on the segment [ a ; b] .

These formulas are applicable for solving relatively simple problems. In fact, we often have to work with more complex shapes. In this regard, we will devote this section to the analysis of algorithms for calculating the area of figures, which are limited by functions in an explicit form, i.e. like y = f(x) or x = g(y) .

Theorem

Let the functions y = f 1 (x) and y = f 2 (x) be defined and continuous on the segment [ a ; b ] , and f 1 (x) ≤ f 2 (x) for any value x from [ a ; b] . Then the formula for calculating the area of \u200b\u200ba figure Gbounded by lines x \u003d a, x \u003d b, y \u003d f 1 (x) and y \u003d f 2 (x) will look like S (G) \u003d ∫ a b f 2 (x) - f 1 (x) d x .

A similar formula will be applicable for the area of \u200b\u200bthe figure bounded by the lines y \u003d c, y \u003d d, x \u003d g 1 (y) and x \u003d g 2 (y): S (G) \u003d ∫ c d (g 2 (y) - g 1 (y) d y .

Proof

We will analyze three cases for which the formula will be valid.

In the first case, taking into account the additivity property of the area, the sum of the areas of the original figure G and the curvilinear trapezoid G 1 is equal to the area of the figure G 2 . It means that

Therefore, S (G) = S (G 2) - S (G 1) = ∫ a b f 2 (x) d x - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) d x .

We can perform the last transition using the third property of the definite integral.

In the second case, the equality is true: S (G) = S (G 2) + S (G 1) = ∫ a b f 2 (x) d x + - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) d x

The graphic illustration will look like:

If both functions are non-positive, we get: S (G) = S (G 2) - S (G 1) = - ∫ a b f 2 (x) d x - - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) d x . The graphic illustration will look like:

Let's move on to the consideration of the general case when y = f 1 (x) and y = f 2 (x) intersect the axis O x .

We will denote the intersection points as x i , i = 1 , 2 , . . . , n - 1 . These points break the segment [ a ; b ] into n parts x i - 1 ; x i , i = 1 , 2 , . . . , n , where α = x 0< x 1 < x 2 < . . . < x n - 1 < x n = b . Фигуру G можно представить объединением фигур G i , i = 1 , 2 , . . . , n . Очевидно, что на своем интервале G i попадает под один из трех рассмотренных ранее случаев, поэтому их площади находятся как S (G i) = ∫ x i - 1 x i (f 2 (x) - f 1 (x)) d x , i = 1 , 2 , . . . , n

Hence,

S (G) = ∑ i = 1 n S (G i) = ∑ i = 1 n ∫ x i x i f 2 (x) - f 1 (x)) d x = = ∫ x 0 x n (f 2 (x) - f ( x)) d x = ∫ a b f 2 (x) - f 1 (x) d x

We can make the last transition using the fifth property of the definite integral.

Let us illustrate the general case on the graph.

The formula S (G) = ∫ a b f 2 (x) - f 1 (x) d x can be considered proven.

And now let's move on to the analysis of examples of calculating the area of \u200b\u200bfigures that are limited by the lines y \u003d f (x) and x \u003d g (y) .

Considering any of the examples, we will begin with the construction of a graph. The image will allow us to represent complex shapes as combinations of simpler shapes. If you are having trouble plotting graphs and figures on them, you can study the section on basic elementary functions, geometric transformation of graphs of functions, as well as plotting while examining a function.

Example 1

It is necessary to determine the area of \u200b\u200bthe figure, which is limited by the parabola y \u003d - x 2 + 6 x - 5 and straight lines y \u003d - 1 3 x - 1 2, x \u003d 1, x \u003d 4.

Decision

Let's plot the lines on the graph in the Cartesian coordinate system.

On the interval [ 1 ; 4] the graph of the parabola y = - x 2 + 6 x - 5 is located above the straight line y = - 1 3 x - 1 2 . In this regard, to obtain an answer, we use the formula obtained earlier, as well as the method for calculating a definite integral using the Newton-Leibniz formula:

S (G) = ∫ 1 4 - x 2 + 6 x - 5 - - 1 3 x - 1 2 d x = = ∫ 1 4 - x 2 + 19 3 x - 9 2 d x = - 1 3 x 3 + 19 6 x 2 - 9 2 x 1 4 = = - 1 3 4 3 + 19 6 4 2 - 9 2 4 - - 1 3 1 3 + 19 6 1 2 - 9 2 1 = = - 64 3 + 152 3 - 18 + 1 3 - 19 6 + 9 2 = 13

Answer: S (G) = 13

Let's look at a more complex example.

Example 2

It is necessary to calculate the area of the figure, which is limited by the lines y = x + 2 , y = x , x = 7 .

Decision

In this case, we have only one straight line parallel to the x-axis. This is x = 7 . This requires us to find the second integration limit ourselves.

Let's build a graph and put on it the lines given in the condition of the problem.

Having a graph in front of our eyes, we can easily determine that the lower limit of integration will be the abscissa of the intersection point of the graph with a straight line y \u003d x and a semi-parabola y \u003d x + 2. To find the abscissa, we use the equalities:

y = x + 2 O DZ: x ≥ - 2 x 2 = x + 2 2 x 2 - x - 2 = 0 D = (- 1) 2 - 4 1 (- 2) = 9 x 1 = 1 + 9 2 = 2 ∈ O D G x 2 = 1 - 9 2 = - 1 ∉ O D G

It turns out that the abscissa of the intersection point is x = 2.

We draw your attention to the fact that in the general example in the drawing, the lines y = x + 2 , y = x intersect at the point (2 ; 2) , so such detailed calculations may seem redundant. We have provided such a detailed solution here only because in more complex cases the solution may not be so obvious. This means that it is better to always calculate the coordinates of the intersection of lines analytically.

On the interval [ 2 ; 7 ] the graph of the function y = x is located above the graph of the function y = x + 2 . Apply the formula to calculate the area:

S (G) = ∫ 2 7 (x - x + 2) d x = x 2 2 - 2 3 (x + 2) 3 2 2 7 = = 7 2 2 - 2 3 (7 + 2) 3 2 - 2 2 2 - 2 3 2 + 2 3 2 = = 49 2 - 18 - 2 + 16 3 = 59 6

Answer: S (G) = 59 6

Example 3

It is necessary to calculate the area of \u200b\u200bthe figure, which is limited by the graphs of the functions y \u003d 1 x and y \u003d - x 2 + 4 x - 2.

Decision

Let's draw lines on the graph.

Let's define the limits of integration. To do this, we determine the coordinates of the points of intersection of the lines by equating the expressions 1 x and - x 2 + 4 x - 2 . Provided that x is not equal to zero, the equality 1 x \u003d - x 2 + 4 x - 2 becomes equivalent to the equation of the third degree - x 3 + 4 x 2 - 2 x - 1 \u003d 0 with integer coefficients. You can refresh the memory of the algorithm for solving such equations by referring to the section “Solution of cubic equations”.

The root of this equation is x = 1: - 1 3 + 4 1 2 - 2 1 - 1 = 0.

Dividing the expression - x 3 + 4 x 2 - 2 x - 1 by the binomial x - 1, we get: - x 3 + 4 x 2 - 2 x - 1 ⇔ - (x - 1) (x 2 - 3 x - 1) = 0

We can find the remaining roots from the equation x 2 - 3 x - 1 = 0:

x 2 - 3 x - 1 = 0 D = (- 3) 2 - 4 1 (- 1) = 13 x 1 = 3 + 13 2 ≈ 3 . 3; x 2 \u003d 3 - 13 2 ≈ - 0. 3

We have found an interval x ∈ 1; 3 + 13 2 , where G is enclosed above the blue line and below the red line. This helps us determine the area of the shape:

S (G) = ∫ 1 3 + 13 2 - x 2 + 4 x - 2 - 1 x d x = - x 3 3 + 2 x 2 - 2 x - ln x 1 3 + 13 2 = = - 3 + 13 2 3 3 + 2 3 + 13 2 2 - 2 3 + 13 2 - ln 3 + 13 2 - - - 1 3 3 + 2 1 2 - 2 1 - ln 1 = 7 + 13 3 - ln 3 + 13 2

Answer: S (G) \u003d 7 + 13 3 - ln 3 + 13 2

Example 4

It is necessary to calculate the area of the figure, which is limited by the curves y \u003d x 3, y \u003d - log 2 x + 1 and the x-axis.

Decision

Let's put all the lines on the graph. We can get the graph of the function y = - log 2 x + 1 from the graph y = log 2 x if we place it symmetrically about the x-axis and move it up one unit. The equation of the x-axis y \u003d 0.

Let's denote the points of intersection of the lines.

As can be seen from the figure, the graphs of the functions y \u003d x 3 and y \u003d 0 intersect at the point (0; 0) . This is because x \u003d 0 is the only real root of the equation x 3 \u003d 0.

x = 2 is the only root of the equation - log 2 x + 1 = 0 , so the graphs of the functions y = - log 2 x + 1 and y = 0 intersect at the point (2 ; 0) .

x = 1 is the only root of the equation x 3 = - log 2 x + 1 . In this regard, the graphs of the functions y \u003d x 3 and y \u003d - log 2 x + 1 intersect at the point (1; 1) . The last statement may not be obvious, but the equation x 3 \u003d - log 2 x + 1 cannot have more than one root, since the function y \u003d x 3 is strictly increasing, and the function y \u003d - log 2 x + 1 is strictly decreasing.

The next step involves several options.

Option number 1

We can represent the figure G as the sum of two curvilinear trapezoids located above the abscissa axis, the first of which is located below the midline on the segment x ∈ 0; 1 , and the second one is below the red line on the segment x ∈ 1 ; 2. This means that the area will be equal to S (G) = ∫ 0 1 x 3 d x + ∫ 1 2 (- log 2 x + 1) d x .

Option number 2

The figure G can be represented as the difference of two figures, the first of which is located above the x-axis and below the blue line on the segment x ∈ 0; 2 , and the second one is between the red and blue lines on the segment x ∈ 1 ; 2. This allows us to find the area like this:

S (G) = ∫ 0 2 x 3 d x - ∫ 1 2 x 3 - (- log 2 x + 1) d x

In this case, to find the area, you will have to use a formula of the form S (G) \u003d ∫ c d (g 2 (y) - g 1 (y)) d y. In fact, the lines that bound the shape can be represented as functions of the y argument.

Let's solve the equations y = x 3 and - log 2 x + 1 with respect to x:

y = x 3 ⇒ x = y 3 y = - log 2 x + 1 ⇒ log 2 x = 1 - y ⇒ x = 2 1 - y

We get the required area:

S (G) = ∫ 0 1 (2 1 - y - y 3) d y = - 2 1 - y ln 2 - y 4 4 0 1 = = - 2 1 - 1 ln 2 - 1 4 4 - - 2 1 - 0 ln 2 - 0 4 4 = - 1 ln 2 - 1 4 + 2 ln 2 = 1 ln 2 - 1 4

Answer: S (G) = 1 ln 2 - 1 4

Example 5

It is necessary to calculate the area of the figure, which is limited by the lines y \u003d x, y \u003d 2 3 x - 3, y \u003d - 1 2 x + 4.

Decision

Draw a line on the chart with a red line, given by the function y = x . Draw the line y = - 1 2 x + 4 in blue, and mark the line y = 2 3 x - 3 in black.

Note the intersection points.

Find the intersection points of the graphs of functions y = x and y = - 1 2 x + 4:

x = - 1 2 x + 4 O DZ: x ≥ 0 x = - 1 2 x + 4 2 ⇒ x = 1 4 x 2 - 4 x + 16 ⇔ x 2 - 20 x + 64 = 0 D = (- 20) 2 - 4 1 64 \u003d 144 x 1 \u003d 20 + 144 2 \u003d 16; x 2 = 20 - 144 2 = 4 i is the solution to the equation x 2 = 4 = 2 , - 1 2 x 2 + 4 = - 1 2 4 + 4 = 2 ⇒ x 2 = 4 is the solution to the equation ⇒ (4 ; 2) point of intersection i y = x and y = - 1 2 x + 4

Find the intersection point of the graphs of functions y = x and y = 2 3 x - 3:

x = 2 3 x - 3 O DZ: x ≥ 0 x = 2 3 x - 3 2 ⇔ x = 4 9 x 2 - 4 x + 9 ⇔ 4 x 2 - 45 x + 81 = 0 D = (- 45 ) 2 - 4 4 81 = 729 x 1 = 45 + 729 8 = 9, x 2 45 - 729 8 = 9 4 Check: x 1 = 9 = 3, 2 3 x 1 - 3 \u003d 2 3 9 - 3 \u003d 3 ⇒ x 1 \u003d 9 is the solution to the equation ⇒ (9; 3) point and intersection y = x and y = 2 3 x - 3 x 2 = 9 4 = 3 2 , 2 3 x 1 - 3 = 2 3 9 4 - 3 = - 3 2 ⇒ x 2 = 9 4 is not a solution to the equation

Find the point of intersection of the lines y = - 1 2 x + 4 and y = 2 3 x - 3:

1 2 x + 4 = 2 3 x - 3 ⇔ - 3 x + 24 = 4 x - 18 ⇔ 7 x = 42 ⇔ x = 6 - 1 2 6 + 4 = 2 3 6 - 3 = 1 ⇒ (6 1) point of intersection y = - 1 2 x + 4 and y = 2 3 x - 3

Method number 1

We represent the area of the desired figure as the sum of the areas of individual figures.

Then the area of the figure is:

S (G) = ∫ 4 6 x - - 1 2 x + 4 d x + ∫ 6 9 x - 2 3 x - 3 d x = = 2 3 x 3 2 + x 2 4 - 4 x 4 6 + 2 3 x 3 2 - x 2 3 + 3 x 6 9 = = 2 3 6 3 2 + 6 2 4 - 4 6 - 2 3 4 3 2 + 4 2 4 - 4 4 + + 2 3 9 3 2 - 9 2 3 + 3 9 - 2 3 6 3 2 - 6 2 3 + 3 6 = = - 25 3 + 4 6 + - 4 6 + 12 = 11 3

Method number 2

The area of the original figure can be represented as the sum of the other two figures.

Then we solve the line equation for x, and only after that we apply the formula for calculating the area of \u200b\u200bthe figure.

y = x ⇒ x = y 2 red line y = 2 3 x - 3 ⇒ x = 3 2 y + 9 2 black line y = - 1 2 x + 4 ⇒ x = - 2 y + 8 s i n i i l i n i i

So the area is:

S (G) = ∫ 1 2 3 2 y + 9 2 - - 2 y + 8 d y + ∫ 2 3 3 2 y + 9 2 - y 2 d y = = ∫ 1 2 7 2 y - 7 2 d y + ∫ 2 3 3 2 y + 9 2 - y 2 d y = = 7 4 y 2 - 7 4 y 1 2 + - y 3 3 + 3 y 2 4 + 9 2 y 2 3 = 7 4 2 2 - 7 4 2 - 7 4 1 2 - 7 4 1 + + - 3 3 3 + 3 3 2 4 + 9 2 3 - - 2 3 3 + 3 2 2 4 + 9 2 2 = = 7 4 + 23 12 = 11 3

As you can see, the values match.

Answer: S (G) = 11 3

Results

To find the area of a figure that is limited by given lines, we need to draw lines on a plane, find their intersection points, and apply the formula for finding the area. In this section, we have reviewed the most common options for tasks.

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

In this article, you will learn how to find the area of a figure bounded by lines using integral calculations. For the first time, we encounter the formulation of such a problem in high school, when the study of certain integrals has just been completed and it is time to start the geometric interpretation of the knowledge gained in practice.

So, what is required to successfully solve the problem of finding the area of \u200b\u200ba figure using integrals:

Ability to correctly draw drawings;
Ability to solve a definite integral using the well-known Newton-Leibniz formula;
The ability to "see" a more profitable solution - i.e. to understand how in this or that case it will be more convenient to carry out the integration? Along the x-axis (OX) or y-axis (OY)?
Well, where without correct calculations?) This includes understanding how to solve that other type of integrals and correct numerical calculations.

Algorithm for solving the problem of calculating the area of a figure bounded by lines:

1. We build a drawing. It is advisable to do this on a piece of paper in a cage, on a large scale. We sign with a pencil above each graph the name of this function. The signature of the graphs is done solely for the convenience of further calculations. Having received the graph of the desired figure, in most cases it will be immediately clear which integration limits will be used. Thus, we solve the problem graphically. However, it happens that the values of the limits are fractional or irrational. Therefore, you can make additional calculations, go to step two.

2. If the integration limits are not explicitly set, then we find the points of intersection of the graphs with each other, and see if our graphical solution coincides with the analytical one.

3. Next, you need to analyze the drawing. Depending on how the graphs of functions are located, there are different approaches to finding the area of \u200b\u200bthe figure. Consider various examples of finding the area of a figure using integrals.

3.1. The most classic and simplest version of the problem is when you need to find the area of a curvilinear trapezoid. What is a curvilinear trapezoid? This is a flat figure bounded by the x-axis (y=0), straight x = a, x = b and any curve continuous on the interval from a before b. At the same time, this figure is non-negative and is located not lower than the x-axis. In this case, the area of the curvilinear trapezoid is numerically equal to the definite integral calculated using the Newton-Leibniz formula:

Example 1 y = x2 - 3x + 3, x = 1, x = 3, y = 0.

What lines define the figure? We have a parabola y = x2 - 3x + 3, which is located above the axis OH, it is non-negative, because all points of this parabola are positive. Next, given straight lines x = 1 and x = 3 that run parallel to the axis OU, are the bounding lines of the figure on the left and right. Well y = 0, she is the x-axis, which limits the figure from below. The resulting figure is shaded, as seen in the figure on the left. In this case, you can immediately begin to solve the problem. Before us is a simple example of a curvilinear trapezoid, which we then solve using the Newton-Leibniz formula.

3.2. In the previous paragraph 3.1, the case was analyzed when the curvilinear trapezoid is located above the x-axis. Now consider the case when the conditions of the problem are the same, except that the function lies under the x-axis. A minus is added to the standard Newton-Leibniz formula. How to solve such a problem, we will consider further.

Example 2 . Calculate the area of a figure bounded by lines y=x2+6x+2, x=-4, x=-1, y=0.

In this example, we have a parabola y=x2+6x+2, which originates from under the axis OH, straight x=-4, x=-1, y=0. Here y = 0 limits the desired figure from above. Direct x = -4 and x = -1 these are the boundaries within which the definite integral will be calculated. The principle of solving the problem of finding the area of \u200b\u200ba figure almost completely coincides with example number 1. The only difference is that the given function is not positive, and everything is also continuous on the interval [-4; -1] . What does not positive mean? As can be seen from the figure, the figure that lies within the given x has exclusively "negative" coordinates, which is what we need to see and remember when solving the problem. We are looking for the area of \u200b\u200bthe figure using the Newton-Leibniz formula, only with a minus sign at the beginning.

The article is not completed.

In fact, in order to find the area of \u200b\u200ba figure, you do not need so much knowledge of the indefinite and definite integral. The task "calculate the area using a definite integral" always involves the construction of a drawing, so your knowledge and drawing skills will be a much more relevant issue. In this regard, it is useful to refresh the memory of the graphs of the main elementary functions, and, at a minimum, be able to build a straight line, and a hyperbola.

A curvilinear trapezoid is a flat figure bounded by an axis, straight lines, and a graph of a continuous function on a segment that does not change sign on this interval. Let this figure be located not less abscissa:

Then the area of a curvilinear trapezoid is numerically equal to a certain integral. Any definite integral (that exists) has a very good geometric meaning.

In terms of geometry, the definite integral is the AREA.

I.e, the definite integral (if it exists) corresponds geometrically to the area of some figure. For example, consider the definite integral . The integrand defines a curve on the plane that is located above the axis (those who wish can complete the drawing), and the definite integral itself is numerically equal to the area of the corresponding curvilinear trapezoid.

Example 1

This is a typical task statement. The first and most important moment of the decision is the construction of a drawing. Moreover, the drawing must be built RIGHT.

In this problem, the solution might look like this.
Let's make a drawing (note that the equation defines the axis):

On the segment, the graph of the function is located over axis, That's why:

Answer:

After the task is completed, it is always useful to look at the drawing and figure out if the answer is real. In this case, "by eye" we count the number of cells in the drawing - well, about 9 will be typed, it seems to be true. It is quite clear that if we had, say, the answer: 20 square units, then, obviously, a mistake was made somewhere - 20 cells clearly do not fit into the figure in question, at most a dozen. If the answer turned out to be negative, then the task was also solved incorrectly.

Example 3

Calculate the area of the figure bounded by lines and coordinate axes.

Decision: Let's make a drawing:

If the curvilinear trapezoid is located under axle(or at least not higher given axis), then its area can be found by the formula:

In this case:

Attention! Don't confuse the two types of tasks:

1) If you are asked to solve just a definite integral without any geometric meaning, then it can be negative.

2) If you are asked to find the area of a figure using a definite integral, then the area is always positive! That is why the minus appears in the formula just considered.

In practice, most often the figure is located in both the upper and lower half-planes, and therefore, from the simplest school problems, we move on to more meaningful examples.

Example 4

Find the area of a flat figure bounded by lines , .

Decision: First you need to complete the drawing. Generally speaking, when constructing a drawing in area problems, we are most interested in the intersection points of lines. Let's find the points of intersection of the parabola and the line. This can be done in two ways. The first way is analytical. We solve the equation:

Hence, the lower limit of integration , the upper limit of integration .

It is best not to use this method if possible..

It is much more profitable and faster to build the lines point by point, while the limits of integration are found out as if “by themselves”. Nevertheless, the analytical method of finding the limits still sometimes has to be used if, for example, the graph is large enough, or the threaded construction did not reveal the limits of integration (they can be fractional or irrational). And we will also consider such an example.

We return to our task: it is more rational to first construct a straight line and only then a parabola. Let's make a drawing:

And now the working formula: If there is some continuous function on the interval greater than or equal some continuous function, then the area of the figure bounded by the graphs of these functions and straight lines, can be found by the formula:

Here it is no longer necessary to think where the figure is located - above the axis or below the axis, and, roughly speaking, it matters which chart is ABOVE(relative to another graph), and which one is BELOW.

In the example under consideration, it is obvious that on the segment the parabola is located above the straight line, and therefore it is necessary to subtract from

The completion of the solution might look like this:

The desired figure is limited by a parabola from above and a straight line from below.
On the segment , according to the corresponding formula:

Answer:

Example 4

Calculate the area of the figure bounded by the lines , , , .

Decision: Let's make a drawing first:

The figure whose area we need to find is shaded in blue.(carefully look at the condition - how the figure is limited!). But in practice, due to inattention, a “glitch” often occurs, that you need to find the area of \u200b\u200bthe figure that is shaded in green!

This example is also useful in that in it the area of \u200b\u200bthe figure is calculated using two definite integrals.

Really:

1) On the segment above the axis there is a straight line graph;

2) On the segment above the axis is a hyperbola graph.

It is quite obvious that the areas can (and should) be added, therefore:

Definite integral. How to calculate the area of a figure

We now turn to the consideration of applications of the integral calculus. In this lesson, we will analyze a typical and most common task. How to use a definite integral to calculate the area of a plane figure. Finally, those who seek meaning in higher mathematics - may they find it. You never know. In real life, you will have to approximate a summer cottage with elementary functions and find its area using a certain integral.

To successfully master the material, you must:

1) Understand the indefinite integral at least at an intermediate level. Thus, dummies should first read the lesson Not.

2) Be able to apply the Newton-Leibniz formula and calculate the definite integral. You can establish warm friendly relations with certain integrals on the page Definite integral. Solution examples.

Actually, everyone is familiar with the problem of finding the area using a definite integral since school, and we will go a little ahead of the school curriculum. This article might not exist at all, but the fact is that the problem occurs in 99 cases out of 100, when a student is tormented by a hated tower with enthusiasm mastering a course in higher mathematics.

The materials of this workshop are presented simply, in detail and with a minimum of theory.

Let's start with a curvilinear trapezoid.

Curvilinear trapezoid called a flat figure bounded by the axis , straight lines , and the graph of a function continuous on a segment that does not change sign on this interval. Let this figure be located not less abscissa:

Then the area of a curvilinear trapezoid is numerically equal to a certain integral. Any definite integral (that exists) has a very good geometric meaning. On the lesson Definite integral. Solution examples I said that a definite integral is a number. And now it's time to state another useful fact. From the point of view of geometry, the definite integral is the AREA.

I.e, the definite integral (if it exists) geometrically corresponds to the area of some figure. For example, consider the definite integral . The integrand defines a curve on the plane that is located above the axis (those who wish can complete the drawing), and the definite integral itself is numerically equal to the area of the corresponding curvilinear trapezoid.

Example 1

This is a typical task statement. The first and most important moment of the decision is the construction of a drawing. Moreover, the drawing must be built RIGHT.

When building a blueprint, I recommend the following order: at first it is better to construct all lines (if any) and only after- parabolas, hyperbolas, graphs of other functions. Function graphs are more profitable to build point by point, with the technique of pointwise construction can be found in the reference material Graphs and properties of elementary functions. There you can also find material that is very useful in relation to our lesson - how to quickly build a parabola.

In this problem, the solution might look like this.
Let's make a drawing (note that the equation defines the axis):

I will not hatch a curvilinear trapezoid, it is obvious what area we are talking about here. The solution continues like this:

On the segment, the graph of the function is located over axis, That's why:

Answer:

Who has difficulty calculating the definite integral and applying the Newton-Leibniz formula , refer to the lecture Definite integral. Solution examples.

After the task is completed, it is always useful to look at the drawing and figure out if the answer is real. In this case, “by eye” we count the number of cells in the drawing - well, about 9 will be typed, it seems to be true. It is quite clear that if we had, say, the answer: 20 square units, then, obviously, a mistake was made somewhere - 20 cells obviously do not fit into the figure in question, at most a dozen. If the answer turned out to be negative, then the task was also solved incorrectly.

Example 2

Calculate the area of the figure bounded by the lines , , and the axis

This is a do-it-yourself example. Full solution and answer at the end of the lesson.

What to do if the curvilinear trapezoid is located under axle?

Example 3

Calculate the area of the figure bounded by lines and coordinate axes.

Decision: Let's make a drawing:

If the curvilinear trapezoid is located under axle(or at least not higher given axis), then its area can be found by the formula:
In this case:

Attention! Don't confuse the two types of tasks:

1) If you are asked to solve just a definite integral without any geometric meaning, then it can be negative.

2) If you are asked to find the area of a figure using a definite integral, then the area is always positive! That is why the minus appears in the formula just considered.

In practice, most often the figure is located in both the upper and lower half-planes, and therefore, from the simplest school problems, we move on to more meaningful examples.

Example 4

Find the area of a flat figure bounded by lines , .

Hence, the lower limit of integration , the upper limit of integration .
It is best not to use this method if possible..

It is much more profitable and faster to build the lines point by point, while the limits of integration are found out as if “by themselves”. The point-by-point construction technique for various charts is discussed in detail in the help Graphs and properties of elementary functions. Nevertheless, the analytical method of finding the limits still sometimes has to be used if, for example, the graph is large enough, or the threaded construction did not reveal the limits of integration (they can be fractional or irrational). And we will also consider such an example.

We return to our task: it is more rational to first construct a straight line and only then a parabola. Let's make a drawing:

I repeat that with pointwise construction, the limits of integration are most often found out “automatically”.

Here it is no longer necessary to think about where the figure is located - above the axis or below the axis, and, roughly speaking, it matters which chart is ABOVE(relative to another graph), and which one is BELOW.

In the example under consideration, it is obvious that on the segment the parabola is located above the straight line, and therefore it is necessary to subtract from

The completion of the solution might look like this:

The desired figure is limited by a parabola from above and a straight line from below.
On the segment , according to the corresponding formula:

Answer:

In fact, the school formula for the area of a curvilinear trapezoid in the lower half-plane (see simple example No. 3) is a special case of the formula . Since the axis is given by the equation , and the graph of the function is located not higher axes, then

And now a couple of examples for an independent solution

Example 5

Example 6

Find the area of the figure enclosed by the lines , .

In the course of solving problems for calculating the area using a certain integral, a funny incident sometimes happens. The drawing was done correctly, the calculations were correct, but due to inattention ... found the area of the wrong figure, that's how your obedient servant screwed up several times. Here is a real life case:

Example 7

Calculate the area of the figure bounded by the lines , , , .

Decision: Let's make a drawing first:

…Eh, the drawing came out crap, but everything seems to be legible.

This example is also useful in that in it the area of \u200b\u200bthe figure is calculated using two definite integrals. Really:

1) On the segment above the axis there is a straight line graph;

2) On the segment above the axis is a hyperbola graph.

It is quite obvious that the areas can (and should) be added, therefore:

Answer:

Let's move on to one more meaningful task.

Example 8

Calculate the area of a figure bounded by lines,
Let's present the equations in a "school" form, and perform a point-by-point drawing:

It can be seen from the drawing that our upper limit is “good”: .
But what is the lower limit? It is clear that this is not an integer, but what? May be ? But where is the guarantee that the drawing is made with perfect accuracy, it may well turn out that. Or root. What if we didn't get the graph right at all?

In such cases, one has to spend additional time and refine the limits of integration analytically.

Let's find the points of intersection of the line and the parabola.
To do this, we solve the equation:

,

Really, .

The further solution is trivial, the main thing is not to get confused in substitutions and signs, the calculations here are not the easiest.

On the segment , according to the corresponding formula:

Answer:

Well, and in conclusion of the lesson, we will consider two tasks more difficult.

Example 9

Calculate the area of the figure bounded by lines , ,

Decision: Draw this figure in the drawing.

Damn, I forgot to sign the schedule, and redoing the picture, sorry, not hotz. Not a drawing, in short, today is the day =)

For point-by-point construction, it is necessary to know the appearance of the sinusoid (and in general it is useful to know graphs of all elementary functions), as well as some sine values, they can be found in trigonometric table. In some cases (as in this case), it is allowed to construct a schematic drawing, on which graphs and integration limits must be displayed in principle correctly.

There are no problems with the integration limits here, they follow directly from the condition: - "x" changes from zero to "pi". We make a further decision:

On the segment, the graph of the function is located above the axis, therefore:

Y 2 x 1 find the area. Calculate the area of ​​a figure bounded by lines

The concept of a definite integral

Newton - Leibniz formula

Calculating the areas of plane figures using a definite integral

Table of indefinite integrals (antiderivatives) of some functions

Results

Definite integral. How to calculate the area of ​​a figure

Y 2 x 1 find the area. Calculate the area of a figure bounded by lines

Definite integral. How to calculate the area of a figure