Математика. Материалы курса

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We begin to calculate double integrals, and to get a little used to this concept, we start with the simplest situation where the integration domain is a rectangle.

Of course, we need methods of integration that we have used for the definite integral, integral table, Newton-Leibniz formula. Let's start

Calculate the double integral, the integrand (x + 1) * ydxdy, dxdy is the element of area, and it is said under the sign of the double integral that integration is carried out over area D.

What is D?

This is a set of points on the plane (x, y) each coordinate satisfies the inequation (see. Video).

Let's build this area.

We are on plane Oху, x ∈ [0; 2], x=0 is a straight line, coinciding with te axis Oy, x = 2 is another vertical line, y = 0 is axis Ox, and y = 1 is a horizontal line (see the video.).

We got a rectangle enclosed between horizontal and vertical lines, let us sign the axis and coordinates.

The integration domain is a shaded area (see the video.), this is area D.

How to carry out integration?

This situation is special.

Why is it special?

The integrand is a function, a product of two functions, each of which depends only on one variable.

What can we do?

All these symbols: two integral signs, the product of two functions and area element can be regarded as a product, we will separate dxdy by two integrals.

Exterior integral is left on dx, function x + 1 will go there, and the inner integral is ydy.

Look (see. Video), just all symbols that we had in the original record, and are repeated here.

If we have a rectangle integration domain, then we write that x varies from 0 to 2, and y varies from 0 to 1 (see. Video).

What situation arises?

Each of these integrals depends only on one variable, comprises only one variable, and the variable varies from one number to another.

In this case we are dealing with two definite integrals, all you need to do is to calculate two definite integrals.

Look at the first integral. It is calculated apparently simply (see the video.).

It remains to find these values: ½ * 4 = 2, we substitute 2,for x  plus 2 get (2 + 2), and here, substituting 1, we get 1/2.

The answer is obtained (see. Video).

Please note that the double integral becomes the product of two definite integrals only if the integrand is a product of functions of one variable, and each of the variables changes from a constant to a constant, integration domain is a rectangle.

The second problem.

Please note that the recording of the integral is a bit different.

The integration domain is not particularly intricate.

The type of domain is often written immediately under the integrand.

Here the sign of the direct product connects two segments, which means that it is  a direct product of segments, which means that here again we are dealing with a rectangle: x ∈ [0, 1], y ∈ [1, 2].

Each variable changes from one number to another - it is a rectangle.

Build  the rectangle again: direct product of segments is a rectangle, x ∈ [0, 1], y ∈ [1, 2].

I propose to solve this problem in two ways: using, firstly, a rule, or rather a double integral property - the property of linearity.

Let me write the first method (see video.) denote domain D, not to rewrite, in the second integral take out two, it has the same properties as the definite integral.

It turned out, of course, in a generalized form.

Then we get the same situation as it was in the first example, we have the product of two specific in each of these integrals.

See (see. Video).

It remains to perform calculations.

Calculating the first integral (see. Video).

The result is -2.5. This is the first method.

Now let us see how we’ll calculate the second integral by another way.

We remember the answer - 2,5.

The second way: we will not split into two integrals, and will calculate as it is.

Here we leave expression x-2y under the inner integral, x ∈ [0, 1], y ∈ [1, 2].

Calculation begins with the inner integral.

Let's do it separately, and then go back to what we have.

The integral is from 1 to 2, (x-2y) dy.

What should we keep in mind?

There is variable y under the sign of the differential, and x is a constant, tis is some number, which values are not important to us at this stage.

Hence, the integral of the constant is xy, to avoid confusion, remember that this is ∈ [1, 2], you can even sign which variable (see the video.), and here is minus integral, then the variable coincides with the variable of integration, we obtain - y², here, you can write from 1 to 2.

Calculate: (See video.) x-3.

We return to the integral.

We see the inner integral of x-3, and this function becomes integrand.

Begin to calculate the integral of x: this is ½ * x², let's just write the limits of integration from 0 to 1, subtract 3 * x from 0 to 1.

Substitute: (see the video.) 1 / 2-3 = -2.5. We get the same answer (see. Video) that received by the first way.

Go to the next example.

There is an innovation in designation of a set of the integration domain.

Look, it is a segment square - Cartesian square.

This means that both variables satisfy an inequality, are between 0 and π / 2, i.e. x ∈ [0, π / 2], and y is in the same segment.

We will not build a square with such sides

Let's just go over to integration

Here, the first method is not applicable at all, because the properties of the integral can not be applied as there are no properties that allow us to split this integral into two integrals.

Therefore, we will immediately proceed to the repeated integral.

Here we’ll  change a little the order of integration (see the video.), And immediately draw your attention: not necessarily the outer integral - the integral by dx, and the inner one is by dy.

You can change the order of integration, the calculation remains the same, here we have solved differently.

Let us just look at the integral of sin.

We calculate the inner integral by variable x, y is a constant.

The integral of sin = -cos.

Let us write: -cos (x + y), where the limits of integration x changed, let us also write down to avoid confusion, from 0 to π / 2.

Look, this calculation refers to the inner integral, that is the inner integral, the integral from 0 to π / 2, sin (x + y), its calculation can be found here.

There is left dy from the outer integral (see. Video).

This method of recording can begin with calculating the inner integral somewhere separately, and then substituting it, or, if the calculation does not seem to be too cumbersome it can be calculated immediately here.

Let's substitute π / 2in place of x, and the integration limits are 0 and π / 2.

Look, we substitute π / 2 instead of x, I’ll write minus here before the integral at the top point cos (y + π / 2).

At the lowest point, substitute 0 instead of x, we obtain cos (y) dy, y varies from 0 to π / 2.

Minus, calculate the integral cos, the integral of cos is sin, the constant does not interfere, we can always carry it under the sign of the differential.

Remember integration methods: -sin (y + π / 2), the integration limits from y 0 to π/2, and here minus and again minus give plus.

Tabular integral of cos is sin (y), y varies from 0 to π / 2.

Let us calculate: - (sin (π) - sin (π / 2)) + sin (π / 2) - sin 0. sin (π) and sin 0 is 0, and sin (π / 2) = 1.

We got: 1 + 1 = 2.

The problem is solved.

The integration is carried out over a rectangular area which is set by this description here (see. Video).

This is another way of recording the domain D.

You can specify the segments product - direct or Cartesian product.

You can write a system of two inequalities, or describe as a set of points in the plane.

Let's see: leave the outer integral on dx, to calculate the inner one, this is apparently integral of the power function, we write g (x-y)²dy, x varies from 1 to 2, y varies from 3 to 4 (see the video.).

We write out the inner integral separately. (See. Video)

What do we do now to be able to apply the formula for integration?

We must put (x-y) under the sign of the differential

Look, the sign is reversed (see. Video).

The first integral is equal to the difference of fractions 1 / (x-4) - 1 / (x-3).

Now we move on to the original integral instead of the inner integral we write an expression that we have just received (see the video.).

We will not repeat how we get the first integral, this is ln | x-4 | from 1 to 2, and the second integral is -ln | x-3 | from 1 to 2.

Calculate: ln 2 - ln 3 - (ln 1 - ln 2) = 2 * ln 2 - ln 3.

This is the final answer.