Consider a curvilinear trapezoid bounded by the Ox axis, a curve y \u003d f (x) and two straight lines: x \u003d a and x \u003d b (Fig. 85). Take an arbitrary value of x (only not a and not b). Let us give it an increment h = dx and consider a strip bounded by straight lines AB and CD, by the Ox axis, and by an arc BD belonging to the curve under consideration. This strip will be called the elementary strip. The area of ​​an elementary strip differs from the area of ​​a rectangle ACQB by a curvilinear triangle BQD, and the area of ​​the latter less area rectangle BQDM with sides BQ = h=dx) QD=Ay and area equal to hAy = Ay dx. As the side h decreases, the side Du also decreases and, simultaneously with h, tends to zero. Therefore, the area of ​​BQDM is infinitesimal of the second order. The area of ​​the elementary strip is the area increment, and the area of ​​the rectangle ACQB, equal to AB-AC==/(x) dx> is the area differential. Therefore, we find the area itself by integrating its differential. Within the limits of 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. Calculate the area bounded by the parabola y - 1 -x *, the straight lines X \u003d - Fj-, x \u003d 1 and the axis O * (Fig. 86). at Fig. 87. Fig. 86. 1 Here f(x) = 1 - l?, the limits of integration a = - and t = 1, therefore 3) |_ 2 3V 2 / J 3 24 24* Example 2. Calculate the area bounded by the sinusoid y = sinXy, the Ox axis and the straight line (Fig. 87). Applying formula (I), we obtain L 2 S= J sinxdx= [-cos x] Q =0 -(-1) = lf 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 more area previous example. However, let's do the calculations: i 5= | s \ nxdx \u003d [ - cosx) * - - cos i- (- cos 0) \u003d 1 + 1 \u003d 2. o Indeed, our assumption turned out to be fair. Example 4. Calculate the area bounded by the sinusoid and the ^ axis Ox on one period (Fig. 88). Preliminary ras-figure judgments suggest that the area will turn out to be four times larger than in pr. 2. However, after doing the calculations, we get “i G, * i S - \ sin x dx \u003d [- cos x] 0 = = - cos 2n - (-cos 0) \u003d - 1 + 1 \u003d 0. This result requires clarification. To clarify the essence of the matter, we also calculate the area bounded by the same sinusoid y \u003d sin l: and the Ox axis ranging from l to 2n. Applying formula (I), we obtain Thus, we see that this area turned out to be negative. Comparing it with the area calculated in Ex. 3, we find that their absolute values ​​are the same, but the signs are different. If we apply property V (see Ch. XI, § 4), then we get by accident. Always the area below the x-axis, provided that the independent variable changes from left to right, is obtained by calculating using integrals negative. In this course, we will always consider unsigned areas. Therefore, the answer in the example just analyzed will be as follows: the required area is equal to 2 + |-2| = 4. Example 5. Let's calculate the area of ​​the BAB shown in Fig. 89. This area is limited by the axis Ox, the parabola y = - xr and the straight line y - = -x + \. Area of ​​a curvilinear trapezoid The sought-for area OAB consists of two parts: OAM and MAB. Since point A is the point of intersection of the parabola and the straight line, we will find its coordinates by solving the system of equations 3 2 Y \u003d 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 pl. OAM, and then pl. MAV: .... G 3 2, 3 G xP 3 1/2 Y 2. QAM-^x ,

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

Consequently,

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 plotting graphs and shapes on them is difficult for you, you can study the section on basic elementary functions, geometric transformation of graphs of functions, as well as plotting during the study of 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.

Solution

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 .

Solution

AT 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 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 given such a detailed solution here only because, in a more difficult 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.

Solution

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.

Solution

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.

Solution

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 bounded 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 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 much more topical issue will be your knowledge and drawing skills. 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.

That is, 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 area corresponding curvilinear trapezoid.

Example 1

This is a typical task statement. First and crucial point solutions - building a drawing. Moreover, the drawing must be built RIGHT.

When building a blueprint, I recommend the following order: 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 pointwise.

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.

Solution: 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 sense, 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 , .

Solution: 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 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 , , , .

Solution: 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:

Task number 3. Make a drawing and calculate the area of ​​\u200b\u200bthe figure bounded by lines

Application of the integral to solving applied problems

Area calculation

The definite integral of a continuous non-negative function f(x) is numerically equal to the area of ​​a curvilinear trapezoid bounded by the curve y \u003d f (x), the O x axis and the straight lines x \u003d a and x \u003d b. Accordingly, the area formula is written as follows:

Consider some examples for calculating areas flat figures.

Task number 1. Calculate the area bounded by the lines y \u003d x 2 +1, y \u003d 0, x \u003d 0, x \u003d 2.

Solution. Let's build a figure, the area of ​​​​which we will have to calculate.

y \u003d x 2 + 1 is a parabola whose branches are directed upwards, and the parabola is shifted upwards by one unit relative to the O y axis (Figure 1).

Figure 1. Graph of the function y = x 2 + 1

Task number 2. Calculate the area bounded by the lines y \u003d x 2 - 1, y \u003d 0 in the range from 0 to 1.

Solution. The graph of this function is the parabola of the branch, which is directed upwards, and the parabola is shifted down by one unit relative to the O y axis (Figure 2).

Figure 2. Graph of the function y \u003d x 2 - 1

Task number 3. Make a drawing and calculate the area of ​​\u200b\u200bthe figure bounded by lines

y = 8 + 2x - x 2 and y = 2x - 4.

Solution. The first of these two lines is a parabola with branches pointing downwards, since the coefficient at x 2 is negative, and the second line is a straight line crossing both coordinate axes.

To construct a parabola, let's find the coordinates of its vertex: y'=2 – 2x; 2 – 2x = 0, x = 1 – vertex abscissa; y(1) = 8 + 2∙1 – 1 2 = 9 is its ordinate, N(1;9) is its vertex.

Now we find the points of intersection of the parabola and the line by solving the system of equations:

Equating the right sides of an equation whose left sides are equal.

We get 8 + 2x - x 2 \u003d 2x - 4 or x 2 - 12 \u003d 0, from where .

So, the points are the points of intersection of the parabola and the straight line (Figure 1).

Figure 3 Graphs of functions y = 8 + 2x – x 2 and y = 2x – 4

Let's build a straight line y = 2x - 4. It passes through the points (0;-4), (2; 0) on the coordinate axes.

To build a parabola, you can also have its intersection points with the 0x axis, that is, the roots of the equation 8 + 2x - x 2 = 0 or x 2 - 2x - 8 = 0. By the Vieta theorem, it is easy to find its roots: x 1 = 2, x 2 = four.

Figure 3 shows a figure (parabolic segment M 1 N M 2) bounded by these lines.

The second part of the problem is to find the area of ​​this figure. Its area can be found using a definite integral using the formula .

With regard to this condition, we obtain the integral:

2 Calculation of the volume of a body of revolution

The volume of the body obtained from the rotation of the curve y \u003d f (x) around the O x axis is calculated by the formula:

When rotating around the O y axis, the formula looks like:

Task number 4. Determine the volume of the body obtained from the rotation of a curvilinear trapezoid bounded by straight lines x \u003d 0 x \u003d 3 and a curve y \u003d around the O x axis.

Solution. Let's build a drawing (Figure 4).

Figure 4. Graph of the function y =

The desired volume is equal to

Task number 5. Calculate the volume of the body obtained from the rotation of a curvilinear trapezoid bounded by a curve y = x 2 and straight lines y = 0 and y = 4 around the axis O y .

Solution. We have:

Review questions

Back forward

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Keywords: integral, curvilinear trapezoid, area of ​​figures bounded by lilies

Equipment: whiteboard, computer, multimedia projector

Lesson type: lesson-lecture

Lesson Objectives:

• educational: shape a culture mental labor, to create a situation of success for each student, to form a positive motivation for learning; develop the ability to speak and listen to others.
• developing: formation of independent thinking of the student on the application of knowledge in different situations ability to analyze and draw conclusions development of logic developing the ability to ask questions correctly and find answers to them. Improving the formation of computational, calculating skills, developing the thinking of students in the course of performing the proposed tasks, developing an algorithmic culture.
• educational: to form concepts about a curvilinear trapezoid, about an integral, to master the skills of calculating the areas of flat figures

Teaching method: explanatory and illustrative.

During the classes

In the previous classes, we learned how to calculate the areas of figures whose boundaries are broken lines. In mathematics, there are methods that allow you to calculate the area of ​​\u200b\u200bfigures bounded by curves. Such figures are called curvilinear trapezoids, and their area is calculated using antiderivatives.

Curvilinear trapezoid ( slide 1)

A curvilinear trapezoid is a figure bounded by the function graph, ( w.m.), straight x = a and x = b and abscissa

Various types of curvilinear trapezoids ( slide 2)

We are considering different kinds curvilinear trapezoids and note that one of the lines is degenerate to a point, the role of the limiting function is played by the line

Area of ​​a curvilinear trapezoid (slide 3)

Fix the left end of the interval a, and right X we will change, i.e., we move the right wall of the curvilinear trapezoid and get a changing figure. The area of ​​a variable curvilinear trapezoid bounded by the function graph is the antiderivative F for function f

And on the segment [ a; b] the area of ​​the curvilinear trapezoid formed by the function f, is equal to the increment of the antiderivative of this function:

Exercise 1:

Find the area of ​​a curvilinear trapezoid bounded by the graph of a function: f(x) = x 2 and direct y=0, x=1, x=2.

Solution: ( according to the slide 3 algorithm)

Draw a graph of the function and lines

Find one of the antiderivatives of the function f(x) = x 2 :

Slide Self-Check

Integral

Consider a curvilinear trapezoid given by the function f on the segment [ a; b]. Let's break this segment into several parts. The area of ​​the entire trapezoid will be divided into the sum of the areas of smaller curvilinear trapezoids. ( slide 5). Each such trapezoid can be approximately considered a rectangle. The sum of the areas of these rectangles gives an approximate idea of ​​the entire area of ​​the curvilinear trapezoid. The smaller we break the segment [ a; b], the more accurately we calculate the area.

We write these considerations in the form of formulas.

Divide the segment [ a; b] into n parts with dots x 0 \u003d a, x1, ..., xn \u003d b. Length k- th denote by xk = xk - xk-1. Let's sum up

Geometrically, this sum is the area of ​​the figure shaded in the figure ( sh.m.)

Sums of the form are called integral sums for the function f. (sch.m.)

Integral sums give an approximate value of the area. Exact value obtained by passing to the limit. Imagine that we refine the partition of the segment [ a; b] so that the lengths of all small segments tend to zero. Then the area of ​​the composed figure will approach the area of ​​the curvilinear trapezoid. We can say that the area of ​​a curvilinear trapezoid is equal to the limit of integral sums, Sk.t. (sch.m.) or integral, i.e.,

Definition:

function integral f(x) from a before b is called the limit of integral sums

= (sch.m.)

Newton-Leibniz formula.

Remember that the limit of integral sums is equal to the area of ​​a curvilinear trapezoid, so we can write:

Sk.t. = (sch.m.)

On the other hand, the area of ​​a curvilinear trapezoid is calculated by the formula

S to. t. (sch.m.)

Comparing these formulas, we get:

= (sch.m.)

This equality is called the Newton-Leibniz formula.

For the convenience of calculations, the formula is written as:

= = (sch.m.)

Tasks: (sch.m.)

1. Calculate the integral using the Newton-Leibniz formula: ( check slide 5)

2. Compile integrals according to the drawing ( check on slide 6)

3. Find the area of ​​​​a figure bounded by lines: y \u003d x 3, y \u003d 0, x \u003d 1, x \u003d 2. ( Slide 7)

Finding the areas of plane figures ( slide 8)

How to find the area of ​​figures that are not curvilinear trapezoids?

Let two functions be given, the graphs of which you see on the slide . (sch.m.) Find the area of ​​the shaded figure . (sch.m.). Is the figure in question a curvilinear trapezoid? And how can you find its area, using the additivity property of the area? Consider two curvilinear trapezoids and subtract the area of ​​the other from the area of ​​one of them ( w.m.)

Let's make an algorithm for finding the area from the animation on the slide:

1. Plot Functions
2. Project the intersection points of the graphs onto the x-axis
3. Shade the figure obtained by crossing the graphs
4. Find curvilinear trapezoids whose intersection or union is the given figure.
5. Calculate the area of ​​each
6. Find difference or sum of areas

Oral task: How to get the area of ​​a shaded figure (tell using animation, slide 8 and 9)

Homework: Work out the abstract, No. 353 (a), No. 364 (a).

Bibliography

1. Algebra and the beginning of analysis: a textbook for grades 9-11 of the evening (shift) school / ed. G.D. Glazer. - M: Enlightenment, 1983.
2. Bashmakov M.I. Algebra and the beginning of analysis: a textbook for grades 10-11 of middle school / Bashmakov M.I. - M: Enlightenment, 1991.
3. Bashmakov M.I. Mathematics: a textbook for institutions beginning. and avg. prof. education / M.I. Bashmakov. - M: Academy, 2010.
4. Kolmogorov A.N. Algebra and the beginning of analysis: a textbook for 10-11 cells. educational institutions / A.N. Kolmogorov. - M: Enlightenment, 2010.
5. Ostrovsky S.L. How to make a presentation for the lesson? / S.L. Ostrovsky. – M.: First of September, 2010.