In geometry, the area of ​​a figure is one of the main numerical characteristics of a flat body. What is an area, how to determine it for various figures, as well as what properties it has - we will consider all these questions in this article.

What is area: definition

The area of ​​a figure is the number of unit squares in that figure; informally speaking, this is the size of the figure. Most often, the area of ​​the figure is referred to as "S". It can be measured using a palette or a planimeter device. Also, the area of ​​a figure can be calculated by knowing its basic dimensions. For example, the area of ​​a triangle can be calculated using three different formulas:

The area of ​​a rectangle is equal to the product of its width and length, and the area of ​​a circle is equal to the product of the square of the radius and π = 3.14.

Shape area properties

• the area is equal to that of equal figures;
• area is always non-negative;
• the unit of measure for area is the area of ​​a square with a side equal to 1 unit of length;
• if the figure is divided into two parts, then the total area of ​​the figure is equal to the sum of the areas of its constituent parts;
• figures that are equal in area are called equal in size;
• if one figure belongs to another figure, then the area of ​​the first cannot exceed the area of ​​the second.

Area formula is necessary to determine the area of ​​a figure, which is a real-valued function defined on a certain class of figures in the Euclidean plane and satisfying 4 conditions:

1. Positiveness - The area cannot be less than zero;
2. Normalization - a square with a side of one has an area of ​​1;
3. Congruence - congruent shapes have equal area;
4. Additivity - the area of ​​the union of 2 figures without common internal points is equal to the sum of the areas of these figures.

How to find the area of ​​a shape?

Knowing and being able to calculate the areas of various shapes is necessary not only for solving simple geometric problems. You cannot do without this knowledge when drawing up or checking estimates for the repair of premises, calculating the amount of necessary consumables. So let's figure out how to find the areas of different shapes.

The part of a plane enclosed within a closed contour is called the area of ​​this plane. The area is expressed by the number of square units enclosed in it.

To calculate the area of ​​basic geometric shapes, you must use the correct formula.

Area of ​​a triangle

Legend:

1. If h, a are known, then the area of ​​the desired triangle is determined as the product of the lengths of the side and the height of the triangle dropped to this side, divided in half: S = (a h) / 2
2. If a, b, c are known, then the required area is calculated according to Heron's formula: the square root taken from the product of half of the perimeter of the triangle and three differences of half of the perimeter and each side of the triangle: S = √ (p (p - a) (p - b) (p - c)).
3. If a, b, γ are known, then the area of ​​a triangle is determined as half of the product of 2 sides, multiplied by the value of the sine of the angle between these sides: S = (a b sin γ) / 2
4. If a, b, c, R are known, then the required area is determined as the division of the product of the lengths of all sides of the triangle by the four radii of the circumscribed circle: S = (a b c) / 4R
5. If p, r are known, then the required area of ​​the triangle is determined by multiplying half of the perimeter by the radius of the inscribed circle: S = p r

Square area

Legend:

1. If the side is known, then the area of ​​this figure is determined as the square of the length of its side: S = a 2
2. If d is known, then the area of ​​a square is determined as half the square of the length of its diagonal: S = d 2/2

Rectangle area

Legend:

• S - determined area,
• a, b - the lengths of the sides of the rectangle.

1. If a, b are known, then the area of ​​a given rectangle is determined by the product of the lengths of its two sides: S = a b
2. If the lengths of the sides are unknown, then the area of ​​the rectangle must be divided into triangles. In this case, the area of ​​a rectangle is defined as the sum of the areas of its constituent triangles.

Parallelogram area

Legend:

• S is the required area,
• a, b - side lengths,
• h is the length of the height of this parallelogram,
• d1, d2 - lengths of two diagonals,
• α is the angle between the sides,
• γ is the angle between the diagonals.

1. If a, h are known, then the required area is determined by multiplying the lengths of the side and the height lowered to this side: S = a h
2. If a, b, α are known, then the area of ​​the parallelogram is determined by multiplying the lengths of the sides of the parallelogram and the value of the sine of the angle between these sides: S = a b sin α
3. If d 1, d 2, γ are known, then the area of ​​the parallelogram is determined as half of the product of the lengths of the diagonals and the value of the sine of the angle between these diagonals: S = (d 1 d 2 sinγ) / 2

Rhombus area

Legend:

• S is the required area,
• a - side length,
• h - height length,
• α - smaller angle between two sides,
• d1, d2 - the lengths of the two diagonals.

1. If a, h are known, then the area of ​​a rhombus is determined by multiplying the length of the side by the length of the height that is lowered to this side: S = a h
2. If a, α are known, then the area of ​​a rhombus is determined by multiplying the square of the side length by the sine of the angle between the sides: S = a 2 sin α
3. If d 1 and d 2 are known, then the required area is determined as half of the product of the lengths of the diagonals of the rhombus: S = (d 1 d 2) / 2

Trapezium area

Legend:

1. If a, b, c, d are known, then the required area is determined by the formula: S = (a + b) / 2 * √.
2. With known a, b, h, the required area is determined as the product of half the sum of the bases and the height of the trapezoid: S = (a + b) / 2 h

Area of ​​a convex quadrilateral

Legend:

1. If d 1, d 2, α are known, then the area of ​​a convex quadrilateral is determined as half of the product of the diagonals of the quadrilateral multiplied by the sine of the angle between these diagonals: S = (d 1 d 2 sin α) / 2
2. For known p, r, the area of ​​a convex quadrilateral is defined as the product of the half-perimeter of the quadrangle by the radius of a circle inscribed in this quadrilateral: S = p r
3. If a, b, c, d, θ are known, then the area of ​​a convex quadrilateral is determined as the square root of the products of the difference of the half-perimeter and the length of each side minus the product of the lengths of all sides and the square of the cosine of half the sum of two opposite angles: S 2 = (p - a ) (p - b) (p - c) (p - d) - abcd cos 2 ((α + β) / 2)

Area of ​​a circle

Legend:

If r is known, then the required area is determined as the product of π by the radius squared: S = π r 2

If d is known, then the area of ​​a circle is defined as the product of the number π by the square of the diameter, divided by four: S = (π d 2) / 4

Complex figure area

A complex one can be broken down into simple geometric shapes. The area of ​​a complex figure is defined as the sum or difference of the constituent areas. Consider, for example, a ring.

Designation:

• S is the area of ​​the ring,
• R, r are the radii of the outer and inner circles, respectively,
• D, d - diameters of the outer and inner circles, respectively.

In order to find the area of ​​the ring, it is necessary to subtract the area from the area of ​​the larger circle smaller circle. S = S1-S2 = πR 2 -πr 2 = π (R 2 -r 2).

Thus, if R and r are known, then the area of ​​the ring is determined as the difference between the squares of the radii of the outer and inner circles, multiplied by the number pi: S = π (R 2 -r 2).

If D and d are known, then the area of ​​the ring is determined as a quarter of the difference between the squares of the diameters of the outer and inner circles, multiplied by the number pi: S = (1/4) (D 2 -d 2) π.

Area of ​​the filled shape

Suppose that inside one square (A) there is another (B) (smaller), and we need to find the filled cavity between shapes "A" and "B". Let's just say, the "frame" of a small square. For this:

1. Find the area of ​​figure "A" (calculated by the formula for finding the area of ​​a square).
2. Similarly, we find the area of ​​figure "B".
3. Subtract area "B" from area "A". And thus we get the area of ​​the filled figure.

Now you know how to find the areas of different shapes.

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:

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 that are limited by functions in an explicit form, i.e. as y = f (x) or x = g (y).

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 of x from [a; b]. Then the formula for calculating the area of ​​the figure G bounded by the lines x = a, x = b, y = f 1 (x) and y = f 2 (x) will have the form S (G) = ∫ abf 2 (x) - f 1 (x) dx.

A similar formula will be applicable for the area of ​​the figure bounded by the lines y = c, y = d, x = g 1 (y) and x = g 2 (y): S (G) = ∫ cd (g 2 (y) - g 1 (y) dy.

Proof

Let us consider three cases for which the formula will be valid.

In the first case, taking into account the property of area additivity, 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) = ∫ abf 2 (x) dx - ∫ abf 1 (x) dx = ∫ ab (f 2 (x) - f 1 (x)) dx.

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

In the second case, the following equality holds: S (G) = S (G 2) + S (G 1) = ∫ abf 2 (x) dx + - ∫ abf 1 (x) dx = ∫ ab (f 2 (x) - f 1 (x)) dx

The graphic illustration will look like:

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

Let us turn to the consideration of the general case when y = f 1 (x) and y = f 2 (x) intersect the O x axis.

The intersection points will be denoted as x i, i = 1, 2,. ... ... , n - 1. These points split 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 ∫ xixif 2 (x) - f 1 (x)) dx = = ∫ x 0 xn (f 2 (x) - f ( x)) dx = ∫ abf 2 (x) - f 1 (x) dx

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 an analysis of examples of calculating the area of ​​figures that are bounded by the lines y = f (x) and x = g (y).

We will begin consideration of any of the examples by building a graph. The image will allow us to represent complex shapes as combinations of simpler shapes. If plotting graphs and shapes on them causes you difficulty, you can study the section on basic atomic functions, geometric transformation of graphs of functions, and plotting while exploring a function.

Example 1

It is necessary to determine the area of ​​the figure, which is bounded by the parabola y = - x 2 + 6 x - 5 and straight lines y = - 1 3 x - 1 2, x = 1, x = 4.

Solution

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

On the segment [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 according to the Newton-Leibniz formula:

S (G) = ∫ 1 4 - x 2 + 6 x - 5 - - 1 3 x - 1 2 dx = = ∫ 1 4 - x 2 + 19 3 x - 9 2 dx = - 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 bounded by the lines y = x + 2, y = x, x = 7.

Solution

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

Let's build a graph and draw on it the lines given in the problem statement.

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

y = x + 2 О Д З: 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 ∈ О Д З x 2 = 1 - 9 2 = - 1 ∉ О Д З

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 the coordinates of the intersection of lines are always best calculated analytically.

On the interval [2; 7] the graph of the function y = x is located above the graph of the function y = x + 2. Let's apply the formula for calculating the area:

S (G) = ∫ 2 7 (x - x + 2) dx = 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 ​​the figure, which is limited by the graphs of the functions y = 1 x and y = - x 2 + 4 x - 2.

Solution

Let's draw lines on the chart.

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 zero, the equality 1 x = - x 2 + 4 x - 2 becomes equivalent to the equation of the third degree - x 3 + 4 x 2 - 2 x - 1 = 0 with integer coefficients. You can refresh your memory of the algorithm for solving such equations by referring to the section "Solving 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 = 3 - 13 2 ≈ - 0. 3

We found the interval x ∈ 1; 3 + 13 2, in which the figure G is enclosed above the blue 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 xdx = - 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) = 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 = x 3, y = - log 2 x + 1 and the abscissa axis.

Solution

Let's draw all the lines on the chart. We can get the graph of the function y = - log 2 x + 1 from the graph y = log 2 x, if we arrange it symmetrically about the abscissa axis and lift it up one unit. The abscissa equation is y = 0.

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

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

x = 2 is the only root of the equation - log 2 x + 1 = 0, therefore 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 = x 3 and y = - log 2 x + 1 intersect at the point (1; 1). The last statement may not be obvious, but the equation x 3 = - log 2 x + 1 cannot have more than one root, since the function y = x 3 is strictly increasing, and the function y = - log 2 x + 1 is strictly decreasing.

Further solution assumes 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 centerline on the segment x ∈ 0; 1, and the second is below the red line on the segment x ∈ 1; 2. This means that the area will be 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 abscissa axis and below the blue line on the segment x ∈ 0; 2, and the second is between the red and blue lines on the segment x ∈ 1; 2. This allows us to find the area as follows:

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) = ∫ 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.

Solve the equations y = x 3 and - log 2 x + 1 for 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) dy = - 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 bounded by the lines y = x, y = 2 3 x - 3, y = - 1 2 x + 4.

Solution

With the red line, draw on the chart the line specified by the function y = x. Draw the line y = - 1 2 x + 4 in blue, and draw the line y = 2 3 x - 3 in black.

Let's mark the points of intersection.

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

x = - 1 2 x + 4 О Д З: 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 = 144 x 1 = 20 + 144 2 = 16; x 2 = 20 - 144 2 = 4 Check: x 1 = 16 = 4, - 1 2 x 1 + 4 = - 1 2 16 + 4 = - 4 ⇒ x 1 = 16 not I have a solution x 2 = 4 = 2, - 1 2 x 2 + 4 = - 1 2 4 + 4 = 2 ⇒ x 2 = 4 i s e r t e r s ⇒ (4; 2) point of intersection i y = x and y = - 1 2 x + 4

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

x = 2 3 x - 3 О Д З: 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 = 2 3 9 - 3 = 3 ⇒ x 1 = 9 I have a solution ⇒ (9; 3) point 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 no solution

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

Method number 1

Let's imagine the area of ​​the required figure as the sum of the areas of the individual figures.

Then the area of ​​the figure is equal to:

S (G) = ∫ 4 6 x - - 1 2 x + 4 dx + ∫ 6 9 x - 2 3 x - 3 dx = = 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 shape can be thought of as the sum of the other two shapes.

Then we will solve the equation of the line with respect to x, and only after that we will apply the formula for calculating the area of ​​the 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

Thus, the area is equal to:

S (G) = ∫ 1 2 3 2 y + 9 2 - - 2 y + 8 dy + ∫ 2 3 3 2 y + 9 2 - y 2 dy = = ∫ 1 2 7 2 y - 7 2 dy + ∫ 2 3 3 2 y + 9 2 - y 2 dy = = 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 ​​are the same.

Answer: S (G) = 11 3

Outcomes

To find the area of ​​a figure, which is bounded by the given lines, we need to build lines on a plane, find their intersection points, apply the formula to find the area. In this section, we examined the most common task options.

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

We now turn to consideration of applications of integral calculus. In this lesson we will analyze a typical and most common task. - how to calculate the area of ​​a flat figure using a definite integral... Finally, those who are looking for meaning in higher mathematics - may they find it. You never know. We'll have to bring the suburban area closer in life with elementary functions and find its area using a definite integral.

To successfully master the material, you must:

1) Understand the indefinite integral at least at the middle level. Thus, dummies should first familiarize themselves with the lesson Not.

2) Be able to apply the Newton-Leibniz formula and calculate a definite integral. You can build warm friendships with definite integrals on the page Definite integral. Examples of solutions.

In fact, in order to find the area of ​​a figure, one does not need so much knowledge of the indefinite and definite integral. The task "calculate area using a definite integral" always involves building a drawing, therefore, your knowledge and drawing skills will be a much more pressing issue. In this regard, it is useful to refresh the memory of the graphs of the basic elementary functions, and, at least, to be able to build a straight line, a parabola and a hyperbola. This can be done (many people need it) with the help of methodological material and an article on geometric transformations of graphs.

Actually, everyone is familiar with the problem of finding the area using a definite integral since school, and we will not go far 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 suffers from the hated tower with enthusiasm to master the course of higher mathematics.

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

Let's start with a curved trapezoid.

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

Then the area of ​​a curvilinear trapezoid is numerically equal to the definite integral... Any definite integral (that exists) has a very good geometric meaning. At the lesson Definite integral. Examples of solutions 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.

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

Example 1

This is a typical formulation of the assignment. The first and most important point of the solution is the construction of the drawing... Moreover, the drawing must be built RIGHT.

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

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

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

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

Answer:

Who has difficulty calculating a definite integral and applying the Newton-Leibniz formula , refer to the lecture Definite integral. Examples of solutions.

After the task is completed, it is always helpful to look at the blueprint and estimate 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 looks like the truth. It is quite clear that if we got, say, the answer: 20 square units, then, obviously, a mistake has been made somewhere - the figure under consideration clearly does not fit 20 cells, at most ten. If the answer is negative, then the task was also solved incorrectly.

Example 2

Calculate the area of ​​a shape bounded by lines, and an axis

This is an example for a do-it-yourself solution. Complete solution and answer at the end of the tutorial.

What to do if the curved trapezoid is located under the axis?

Example 3

Calculate the area of ​​a shape bounded by lines and coordinate axes.

Solution: Let's execute the drawing:

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

Attention! The two types of tasks should not be confused:

1) If you are asked to solve just a definite integral without any geometric meaning, then it may 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 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 problems on an area, we are most interested in the points of intersection of lines. Find the intersection points 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 better not to use this method, if possible..

It is much more profitable and faster to construct the lines point by point, while the limits of integration become clear, as it were, "by themselves." The technique of point-by-point plotting 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 has to be applied sometimes if, for example, the graph is large enough, or the precise construction did not reveal the limits of integration (they can be fractional or irrational). And we will also consider such an example.

Returning to our problem: it is more rational to first construct a straight line and only then a parabola. Let's execute the drawing:

I repeat that in the case of a pointwise construction, the limits of integration are most often found out by an “automaton”.

And now the working formula: If on a segment some continuous function greater than or equal of 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 you no longer need to think about where the figure is located - above the axis or below the axis, and, roughly speaking, it is important which schedule 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 required figure is bounded by a parabola at the top and a straight line at the bottom.
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 axis, then

And now a couple of examples for self-solution

Example 5

Example 6

Find the area of ​​the figure bounded by lines,.

In the course of solving problems for calculating the area using a definite integral, a funny incident sometimes happens. The drawing is done correctly, the calculations are correct, but through inattention ... the area of ​​the wrong figure is found, this is how your humble servant screwed up several times. Here's a real life case:

Example 7

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

Solution: First, let's execute the drawing:

... Eh, a lousy drawing came out, but everything seems to be legible.

The figure whose area we need to find is shaded in blue(carefully look at the condition - what the figure is limited to!). But in practice, due to inattention, a "glitch" often arises, that you need to find the area of ​​the figure, which is shaded in green!

This example is also useful in that it calculates the area of ​​a figure using two definite integrals. Really:

1) A line graph is located on the segment above the axis;

2) The hyperbola graph is located on the segment above the axis.

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 shape bounded by lines,
Let's represent the equations in a "school" form, and we will execute 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 which one? May be ? But where is the guarantee that the drawing is made with perfect accuracy, it may well be that. Or root. What if we plotted the graph incorrectly at all?

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

Find the intersection points 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 are not the easiest here.

On the segment , according to the corresponding formula:

Answer:

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

Example 9

Calculate the area of ​​a figure bounded by lines,

Solution: Let's depict this figure in the drawing.

Damn, I forgot to sign the schedule, but to redo the picture, sorry, not hotts. Not drawing, in short, today is the day =)

For point-by-point construction, you need 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 a number of cases (as in this one), it is allowed to construct a schematic drawing, on which the graphs and limits of integration should be displayed correctly in principle.

There are no problems with the limits of integration, 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: