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 of calculating the areas of plane 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 down, 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

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 ​​the rectangle ACQB by a curvilinear triangle BQD, and the area of ​​the latter is less than the area of ​​the rectangle BQDM with sides BQ = =h=dx) QD=Ay and area equal to hAy = Ay dx. As 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 \u003d J sinxdx \u003d [-cos x] Q \u003d 0 - (-1) \u003d lf Example 3. Calculate the area bounded by the arc of the sinusoid ^y \u003d sin jc enclosed between two adjacent intersection points with the Ox axis (for example, between the origin and the point with the abscissa i). Note that from geometric considerations it is clear that this area will be twice the area of ​​the previous example. However, let's do the calculations: i 5= | s \ nxdx \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 = [replacement:

] =

Hence, the improper integral converges and its value is equal to .