Examples of solving problems to the section integral calculus

Example N 1

Find the indefinite integral:

Decision.

Answer:

Example N 2

Find the indefinite integral:

Decision.

Answer:

Example N 3

Find the indefinite integral:

.

Decision.

To integrate this function, it is necessary to make a change of variable .

We got the wrong fraction. Select the integer part in it, dividing the polynomial standing in the numerator by the denominator polynomial by a corner.

Answer:

Example N 4

Find the indefinite integral:

Decision.

We got an irregular fraction; we single out the integer part in it, dividing the polynomial in the numerator into a denominator polynomial by a corner:

Answer:

Example N 5

Find the indefinite integral:

Decision.

Answer:

Example N 6

Find the indefinite integral:

Decision.

Answer:

Example N 7

Find the area of ​​the figure bounded by the lines:

Decision.

Fig. sixteen

Find the intersection points of the functions y ₁ and y ₂ :

Since 4 - x ² ≤ 2 - x on the segment [-1; 2], the area S of this figure is calculated as follows:

The answer is 4.5 .

Example N 8

Find the length of the arc curve .

Decision.

Answer:

Example N 9

Find the length of the arc curve .

Decision.

The arc is explicitly specified. , the limits of integration are given 2 ≤ x ≤ 3, we make the integral

Answer:

Example N 10

Find the length of the astroid arc

.

Decision.

The curve is set parametrically, therefore, its length L is calculated as follows:

The answer is: 4.5

Example N 11

Find the volume of the body obtained by rotating around the OX axis of the figure, which is limited by parabolas y = 3 - x ² , y = x ² + 1.

Decision.

We construct a figure on the plane of the OXU , bounded by parabolas y = 3 - x ² , y = x ² + 1 (Fig. 17).

Fig. 17

Find the intersection points of the curves:

.

Then

Answer:

Example N 12

Change the order of integration:

Decision.

We build the integration domain D , which consists of two areas - D ₁ and D ₂ (Fig. 18):

Area D _{1 is} bounded by a semicircle and straight lines y = 0, x = 2.

The domain D _{2 is} bounded by the straight lines y = x - 4, y = 0, x = 2.

Fig. 18

Region D is correct in the direction of the axis of the shelter . The largest value of y in the domain D is the number y = 0, and the smallest y = -2. From the equation Express x through y :

Function sets the left boundary of the domain D. From the equation y = x - 4 we express x through y : x = y + 4. The function x = y + 4 sets the right boundary of the domain D. Therefore, the domain D is given by the inequalities:

Now we change the order of integration in the double integral:

Answer:

Example N 13

Find the area of ​​the figure bounded by lines _{, , , .}

Decision.

Let's build a shape D bounded by given lines (fig. 19):

- circle centered at a point radius ;

- a circle with a center at the point (1; 0), radius 1;

- OX axis;

- straight with the slope

Fig. nineteen

Expressing circles in polar coordinates:

we conclude that the domain D can be given by the inequalities:

Find the area S of the area D according to the formula of the area of ​​the figures in polar coordinates:

Answer:

Example N 14

Calculate the volume of the body bounded by the surfaces _{, , ,} .

Decision.

The body is bounded by parabolic cylinders, which form parallel to the axis OZ : _, ; coordinate plane ; and the plane z = 4 - y . Let us depict the body (Fig. 20) and its projection (Fig. 21) in the DCS plane in the figures .

Fig. 20	Fig. 21

The basis of a given body is the domain D , given by the inequalities: .

Find the volume V of the body using a double integral:

First we find the inner integral:

Then we find the external integral:

Answer: