Task No. 3. Make a drawing and calculate the area of the figure, limited by lines
Application of the integral to the solution of 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 = f(x), the O x axis and the straight lines x = a and x = b. In accordance with this, the area formula is written as follows:
Let's look at some examples of calculating the areas of plane figures.
Task No. 1. Calculate the area bounded by the lines y = x 2 +1, y = 0, x = 0, x = 2.
Solution. Let's construct a figure whose area we will have to calculate.
y = x 2 + 1 is a parabola whose branches are directed upward, and the parabola is shifted upward by one unit relative to the O y axis (Figure 1).
Figure 1. Graph of the function y = x 2 + 1
Task No. 2. Calculate the area bounded by the lines y = x 2 – 1, y = 0 in the range from 0 to 1.
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Solution. The graph of this function is a parabola of branches that are directed upward, and the parabola is shifted relative to the O y axis downward by one unit (Figure 2).
Figure 2. Graph of the function y = x 2 – 1
Task No. 3. Make a drawing and calculate the area of the figure bounded by the lines
y = 8 + 2x – x 2 and y = 2x – 4.
Solution. The first of these two lines is a parabola with its branches directed downward, since the coefficient of x 2 is negative, and the second line is a straight line intersecting both coordinate axes.
To construct a parabola, we find the coordinates of its vertex: y’=2 – 2x; 2 – 2x = 0, x = 1 – abscissa of the vertex; y(1) = 8 + 2∙1 – 1 2 = 9 is its ordinate, N(1;9) is the vertex.
Now let’s find the intersection points of the parabola and the straight 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 = 2x – 4 or x 2 – 12 = 0, whence 
.
So, the points are the intersection points of a parabola and a straight line (Figure 1).
Figure 3 Graphs of functions y = 8 + 2x – x 2 and y = 2x – 4
Let's construct a straight line y = 2x – 4. It passes through the points (0;-4), (2;0) on the coordinate axes.
To construct a parabola, you can also use 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. Using Vieta’s theorem, it is easy to find its roots: x 1 = 2, x 2 = 4.
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 according to the formula 
.
In relation to this condition, we obtain the integral: 
2 Calculation of the volume of a body of rotation
The volume of the body obtained from the rotation of the curve y = f(x) around the O x axis is calculated by the formula:
When rotating around the O y axis, the formula looks like:
Task No. 4. Determine the volume of the body obtained from the rotation of a curved trapezoid bounded by straight lines x = 0 x = 3 and curve y = around the O x axis.
Solution. Let's draw a picture (Figure 4).
Figure 4. Graph of the function y =
The required volume is 
Task No. 5. Calculate the volume of the body obtained from the rotation of a curved trapezoid bounded by the curve y = x 2 and straight lines y = 0 and y = 4 around the O y axis.
Solution. We have:
Review questions
In fact, in order to find the area of a figure, you don’t need that much knowledge of the indefinite and definite integral. The task “calculate the area using a definite integral” always involves constructing a drawing, so your knowledge and drawing skills will be a much more pressing issue. In this regard, it is useful to refresh your memory of the graphs of basic elementary functions, and, at a minimum, be able to construct a straight line and a hyperbola.
A curved trapezoid is a flat figure bounded by an 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 x-axis:
Then the area of a curvilinear trapezoid is numerically equal to a definite integral. Any definite integral (that exists) has a very good geometric meaning.
From the point of view of geometry, the definite integral is AREA.
That is, a certain integral (if it exists) geometrically corresponds to the area of a certain figure. For example, consider the definite integral. The integrand defines a curve on the plane located above the axis (those who wish can make a drawing), and the definite integral itself is numerically equal to area corresponding curved trapezoid.
Example 1
This is a typical assignment statement. First and the most important moment solutions - drawing drawing. Moreover, the drawing must be constructed RIGHT.
When constructing a drawing, I recommend the following order: at first it is better to construct all straight lines (if they exist) and only Then- parabolas, hyperbolas, graphs of other functions. It is more profitable to build graphs of functions point by point.
In this problem, the solution might look like this.
Let's draw the drawing (note that the equation defines the axis):
On the segment, the graph of the function is located above the axis, That's why:
Answer:
After the task is completed, it is always useful to look at the drawing and figure out whether the answer is real. In this case, “by eye” we count the number of cells in the drawing - well, there will be about 9, it seems to be true. It is absolutely clear that if we got, say, the answer: 20 square units, then it is obvious that a mistake was made somewhere - 20 cells obviously do not fit into the figure in question, at most a dozen. If the answer is negative, then the task was also solved incorrectly.
Example 3
Calculate the area of the figure bounded by the lines and coordinate axes.
Solution: Let's make a drawing:
If a curved trapezoid is located under the axle(or at least not higher given axis), then its area can be found using the formula:
In this case:
Attention! The two types of tasks should not be confused:
1) If you are asked to solve simply 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 the minus appears in the formula just discussed.
In practice, most often the figure is located in both the upper and lower half-plane, and therefore, from the simplest school problems we move on to more meaningful examples.
Example 4
Find the area of a plane figure bounded by the lines , .
Solution: First you need to complete the drawing. Generally speaking, when constructing a drawing in area problems, we are most interested in the points of intersection of lines. Let's find the intersection points of the parabola and the straight line. This can be done in two ways. The first method is analytical. We solve the equation:
This means that the lower limit of integration is , the upper limit of integration is .
If possible, it is better not to use this method..
It is much more profitable and faster to construct lines point by point, and the limits of integration become clear “by themselves.” Nevertheless, the analytical method of finding limits still sometimes has to be used if, for example, the graph is large enough, or the detailed construction did not reveal the limits of integration (they can be fractional or irrational). And we will also consider such an example.
Let's return to our task: it is more rational to first construct a straight line and only then a parabola. Let's make the drawing:
And now the working formula: If there is some continuous function on the segment greater than or equal to some continuous function, then the area of the figure limited by the graphs of these functions and the lines , , can be found using 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 matters which graph is HIGHER(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 completed solution might look like this:
The desired figure is limited by a parabola above and a straight line below.
On the segment, according to the corresponding formula:
Answer:
Example 4
Calculate the area of the figure bounded by the lines , , , .
Solution: First, let's make a drawing:
The figure whose area we need to find is shaded blue(look carefully 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 a figure that is shaded in green!
This example is also useful in that it calculates the area of a figure using two definite integrals.
Really:
1) On the segment above the axis there is a graph of a straight line;
2) On the segment above the axis there is a graph of a hyperbola.
It is quite obvious that the areas can (and should) be added, therefore:
In the previous section, devoted to the analysis of the geometric meaning of a definite integral, we received 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 interval [ a ; b ] ,
S (G) = - ∫ a b f (x) d x for a continuous and non-positive function y = f (x) on the interval [ a ; b ] .
These formulas are applicable to solving relatively simple problems. In reality, we will often have to work with more complex figures. In this regard, we will devote this section to an analysis of algorithms for calculating the area of figures that are limited by functions in explicit form, i.e. like y = f(x) or x = g(y).
TheoremLet the functions y = f 1 (x) and y = f 2 (x) be defined and continuous on the interval [ a ; b ] , and f 1 (x) ≤ f 2 (x) for any value 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 look like S (G) = ∫ a b f 2 (x) - f 1 (x) d x .
A similar formula will be applicable for the area of a figure bounded by the lines y = c, y = d, x = g 1 (y) and x = g 2 (y): S (G) = ∫ c d (g 2 (y) - g 1 (y) d y .
Proof
Let's look at three cases for which the formula will be valid.
In the first case, taking into account the property of additivity of 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)) dx.
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 consider general case, when y = f 1 (x) and y = f 2 (x) intersect the O x axis.
We denote the intersection points 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 ∫ 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.
Now let's move on to analyzing examples of calculating the area of figures that are limited by the lines y = f (x) and x = g (y).
We will begin our consideration of any of the examples by constructing a graph. The image will allow us to represent complex shapes as unions of simpler shapes. If constructing graphs and figures on them causes you difficulties, you can study the section on basic elementary functions, geometric transformation of function graphs, as well as the construction of graphs during the study of a function.
Example 1
It is necessary to determine the area of the figure, which is limited 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 the 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 the answer we use the formula obtained earlier, as well as the method of calculating the 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
In this case, we have only one straight line located parallel to the x-axis. This is x = 7. This requires us to find the second limit of integration ourselves.
Let's build a graph and plot 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 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 DZ x 2 = 1 - 9 2 = - 1 ∉ O DZ
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 unnecessary. We brought this here detailed solution only because there are more difficult cases the solution may not be so obvious. This means that it is always better to 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. Let's 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 the figure, which is limited by the graphs of the functions y = 1 x and y = - x 2 + 4 x - 2.
Solution
Let's plot the 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 zero, the equality 1 x = - x 2 + 4 x - 2 becomes equivalent to the third degree equation - x 3 + 4 x 2 - 2 x - 1 = 0 with integer coefficients. To refresh your memory of the algorithm for solving such equations, we can refer 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 contained above the blue and below the red line. This helps us determine the area of the figure:
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) = 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 plot 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 position it symmetrically about the x-axis and move it up one unit. The equation of the x-axis is y = 0.
Let us 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 happens 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, 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 = 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.
The further solution involves several options.
Option #1
We can represent figure G as the sum of two curvilinear trapezoids, located above the x-axis, the first of which is located below the midline 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 equal to S (G) = ∫ 0 1 x 3 d x + ∫ 1 2 (- log 2 x + 1) d x .
Option No. 2
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 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 figure can be represented as functions of the argument y.
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 = x, y = 2 3 x - 3, y = - 1 2 x + 4.
Solution
We will draw a line on the graph with a red line, given by the function y = x. We draw the line y = - 1 2 x + 4 in blue, and the line y = 2 3 x - 3 in black.
Let's mark the intersection points.
Let's find the intersection points of the graphs of the 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 = 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 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
Let's find the intersection point of the graphs of the 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 = 2 3 9 - 3 = 3 ⇒ x 1 = 9 is the solution to the equation ⇒ (9 ; 3) point a s 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 There is no solution to the equation
Let's 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 No. 1
Let us imagine 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 No. 2
The area of the original figure can be represented as the sum of two other figures.
Then we solve the equation of the line relative to x, and only after that we 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 s i n i a l i n e
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 are the same.
Answer: S (G) = 11 3
Results
To find the area of a figure that is limited by given lines, we need to construct lines on a plane, find their intersection points, and apply the formula to find the area. In this section, we examined the most common variants of tasks.
If you notice an error in the text, please highlight it and press Ctrl+Enter
A)
Solution.
The first and most important point of the decision is the construction of the drawing.
Let's make the drawing:
The equation y=0 sets the “x” axis;
- x=-2 And x=1 - straight, parallel to the axis OU;
- y=x 2 +2 - a parabola, the branches of which are directed upward, with the vertex at the point (0;2).
Comment. To construct a parabola, it is enough to find the points of its intersection with the coordinate axes, i.e. putting x=0 find the intersection with the axis OU and deciding accordingly quadratic equation, find the intersection with the axis Oh .
The vertex of a parabola can be found using the formulas: 
You can also build lines point by point.
On the interval [-2;1] the graph of the function y=x 2 +2 located above the axis Ox , That's why:
Answer: S =9 sq. units
After the task is completed, it is always useful to look at the drawing and figure out whether the answer is real. In this case, “by eye” we count the number of cells in the drawing - well, there will be about 9, it seems to be true. It is absolutely clear that if we got, say, the answer: 20 square units, then it is obvious that a mistake was made somewhere - 20 cells obviously do not fit into the figure in question, at most a dozen. If the answer is negative, then the task was also solved incorrectly.
What to do if the curved trapezoid is located under the axle Oh?
b) Calculate the area of a figure bounded by lines y=-e x , x=1 and coordinate axes.
Solution.
Let's make a drawing.
If a curved trapezoid completely located under the axis Oh , then its area can be found using the formula:
Answer: S=(e-1) sq. units" 1.72 sq. units
Attention! The two types of tasks should not be confused:
1) If you are asked to solve simply 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 the minus appears in the formula just discussed.
In practice, most often the figure is located in both the upper and lower half-plane.
With) Find the area of a plane figure bounded by lines y=2x-x 2, y=-x.
Solution.
First you need to complete the drawing. Generally speaking, when constructing a drawing in area problems, we are most interested in the points of intersection of lines. Let's find the intersection points of the parabola 
and straight 
This can be done in two ways. The first method is analytical.
We solve the equation:
This means that the lower limit of integration a=0 , upper limit of integration b=3 .
|
We build the given lines: 1. Parabola - vertex at point (1;1); axis intersection Oh - points (0;0) and (0;2). 2. Straight line - bisector of the 2nd and 4th coordinate angles. And now Attention! If on the segment [ a;b] some continuous function f(x) greater than or equal to some continuous function g(x), then the area of the corresponding figure can be found using the formula: And it doesn’t matter where the figure is located - above the axis or below the axis, but what matters is which graph is HIGHER (relative to another graph), and which 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 |
.You can construct lines point by point, and the limits of integration become clear “by themselves.” Nevertheless, the analytical method of finding limits still sometimes has to be used if, for example, the graph is large enough, or the detailed construction did not reveal the limits of integration (they can be fractional or irrational).
The desired figure is limited by a parabola above and a straight line below.
On the segment 
, according to the corresponding formula:
Answer: S =4.5 sq. units
Problem 1(about calculating the area of a curved trapezoid).
In the Cartesian rectangular coordinate system xOy, a figure is given (see figure) bounded by the x axis, straight lines x = a, x = b (a by a curvilinear trapezoid. It is required to calculate the area of a curvilinear trapezoid.
Solution. Geometry gives us recipes for calculating the areas of polygons and some parts of a circle (sector, segment). Using geometric considerations, we can only find an approximate value of the required area, reasoning as follows.
Let's split the segment [a; b] (base of a curved trapezoid) into n equal parts; this partition is carried out using points x 1, x 2, ... x k, ... x n-1. Let us draw straight lines through these points parallel to the y-axis. Then the given curvilinear trapezoid will be divided into n parts, into n narrow columns. The area of the entire trapezoid is equal to the sum of the areas of the columns.
Let us consider the k-th column separately, i.e. a curved trapezoid whose base is a segment. Let's replace it with a rectangle with the same base and height equal to f(x k) (see figure). The area of the rectangle is equal to \(f(x_k) \cdot \Delta x_k \), where \(\Delta x_k \) is the length of the segment; It is natural to consider the resulting product as an approximate value of the area of the kth column. 
If we now do the same with all the other columns, we will come to the following result: the area S of a given curvilinear trapezoid is approximately equal to the area S n of a stepped figure made up of n rectangles (see figure):
\(S_n = f(x_0)\Delta x_0 + \dots + f(x_k)\Delta x_k + \dots + f(x_(n-1))\Delta x_(n-1) \)
Here, for the sake of uniformity of notation, we assume that a = x 0, b = x n; \(\Delta x_0 \) - length of the segment, \(\Delta x_1 \) - length of the segment, etc.; in this case, as we agreed above, \(\Delta x_0 = \dots = \Delta x_(n-1) \)
So, \(S \approx S_n \), and this approximate equality is more accurate, the larger n.
By definition, it is believed that the required area of a curvilinear trapezoid is equal to the limit of the sequence (S n):
$$ S = \lim_(n \to \infty) S_n $$
Problem 2(about moving a point)
A material point moves in a straight line. The dependence of speed on time is expressed by the formula v = v(t). Find the movement of a point over a period of time [a; b].
Solution. If the movement were uniform, then the problem would be solved very simply: s = vt, i.e. s = v(b-a). For uneven movement, you have to use the same ideas on which the solution to the previous problem was based.
1) Divide the time interval [a; b] into n equal parts.
2) Consider a period of time and assume that during this period of time the speed was constant, the same as at time t k. So we assume that v = v(t k).
3) Let’s find the approximate value of the point’s movement over a period of time; we’ll denote this approximate value as s k
\(s_k = v(t_k) \Delta t_k \)
4) Find the approximate value of displacement s:
\(s \approx S_n \) where
\(S_n = s_0 + \dots + s_(n-1) = v(t_0)\Delta t_0 + \dots + v(t_(n-1)) \Delta t_(n-1) \)
5) The required displacement is equal to the limit of the sequence (S n):
$$ s = \lim_(n \to \infty) S_n $$
Let's summarize. Solutions to various problems were reduced to the same mathematical model. Many problems from various fields of science and technology lead to the same model in the process of solution. So this mathematical model need to be specially studied.
The concept of a definite integral
Let us give a mathematical description of the model that was built in the three considered problems for the function y = f(x), continuous (but not necessarily non-negative, as was assumed in the considered problems) on the interval [a; b]:
1) split the segment [a; b] into n equal parts;
2) make up the sum $$ S_n = f(x_0)\Delta x_0 + f(x_1)\Delta x_1 + \dots + f(x_(n-1))\Delta x_(n-1) $$
3) calculate $$ \lim_(n \to \infty) S_n $$
I know mathematical analysis it has been proven that this limit exists in the case of a continuous (or piecewise continuous) function. He is called a certain integral of the function y = f(x) over the segment [a; b] and denoted as follows:
\(\int\limits_a^b f(x) dx \)
The numbers a and b are called the limits of integration (lower and upper, respectively).
Let's return to the tasks discussed above. The definition of area given in Problem 1 can now be rewritten as follows:
\(S = \int\limits_a^b f(x) dx \)
here S is the area of the curved trapezoid shown in the figure above. This is geometric meaning of a definite integral.
The definition of the displacement s of a point moving in a straight line with a speed v = v(t) over the period of time from t = a to t = b, given in Problem 2, can be rewritten as follows:
Newton-Leibniz formula
First, let's answer the question: what is the connection between the definite integral and the antiderivative?
The answer can be found in Problem 2. On the one hand, the displacement s of a point moving in a straight line with a speed v = v(t) over the period of time from t = a to t = b is calculated by the formula
\(S = \int\limits_a^b v(t) dt \)
On the other hand, the coordinate of a moving point is an antiderivative for speed - let's denote it s(t); this means that the displacement s is expressed by the formula s = s(b) - s(a). As a result we get:
\(S = \int\limits_a^b v(t) dt = s(b)-s(a) \)
where s(t) is the antiderivative of v(t).
The following theorem was proven in the course of mathematical analysis.
Theorem. If the function y = f(x) is continuous on the interval [a; b], then the formula is valid
\(S = \int\limits_a^b f(x) dx = F(b)-F(a) \)
where F(x) is the antiderivative of f(x).
The given formula is usually called Newton-Leibniz formula in honor of the English physicist Isaac Newton (1643-1727) and the German philosopher Gottfried Leibniz (1646-1716), who received it independently of each other and almost simultaneously.
In practice, instead of writing F(b) - F(a), they use the notation \(\left. F(x)\right|_a^b \) (it is sometimes called double substitution) and, accordingly, rewrite the Newton-Leibniz formula in this form:
\(S = \int\limits_a^b f(x) dx = \left. F(x)\right|_a^b \)
When calculating a definite integral, first find the antiderivative, and then carry out a double substitution.
Based on the Newton-Leibniz formula, we can obtain two properties of the definite integral.
Property 1. The integral of the sum of functions is equal to the sum of the integrals:
\(\int\limits_a^b (f(x) + g(x))dx = \int\limits_a^b f(x)dx + \int\limits_a^b g(x)dx \)
Property 2. The constant factor can be taken out of the integral sign:
\(\int\limits_a^b kf(x)dx = k \int\limits_a^b f(x)dx \)
Calculating the areas of plane figures using a definite integral
Using the integral, you can calculate the areas not only of curved trapezoids, but also of plane figures of a more complex type, for example, the one shown in the figure. The figure P is limited by straight lines x = a, x = b and graphs of continuous functions y = f(x), y = g(x), and on the segment [a; b] the inequality \(g(x) \leq f(x) \) holds. To calculate the area S of such a figure, we will proceed as follows:
\(S = S_(ABCD) = S_(aDCb) - S_(aABb) = \int\limits_a^b f(x) dx - \int\limits_a^b g(x) dx = \)
\(= \int\limits_a^b (f(x)-g(x))dx \)
So, the area S of a figure bounded by straight lines x = a, x = b and graphs of functions y = f(x), y = g(x), continuous on the segment and such that for any x from the segment [a; b] the inequality \(g(x) \leq f(x) \) is satisfied, calculated by the formula
\(S = \int\limits_a^b (f(x)-g(x))dx \)