10. Properties of integrated functions and double integrals

1. If area (P) is divided into two parts (P’) and (P’’) then if f(x,y) is integrated in (P)

Proof. Let’s divide areas (P’) and (P’’) into parts, at the same time area (P) also will be

divided into parts (P1),P(2),…,(Pn). If we denote by symbol i’ parts contained in (P’)

and by symbol i’’ parts contained in (P’’) then .

If f(x,y) is integrated so that left sum approaches to zero with then both sum

at right will also approach to zero; it means f(x,y) is integrated in (P’) and (P’’) too.

And vice versa, if sums at right approach to zero then sum at left will be approach to 0.

2. If integrated in (P) func f(x,y) will be multiplied by k then got function will be

3. If in area (P) functions f(x,y) and g(x,y) are integrated then :

4. If then

5. If f(x,y) is integrated then is integrated too and

Functionwill be integrated because variationof this func in any area (P)

doesn’t exceed variation of function f. Thenand approaching

to zero of the second sum result in approaching to zero the first one. And necessary

inequality have obtained by passage to the limit in

6. If f(x,y) integrated in (P) then(1.1)

It is obtained by passage to the limit in obvious inequality .

and let’s denote middle ratio as μ so we have another record of (1) :

(5.2) which is called mean value theorem. By theorem of

Bolzano-Cauchy about that continues bounded function must take on all values between

m and M and denote by such point thatthen formula (1.2) must

be rewrote as (1.3) . This expression is the most generally used

form of mean value theorem.

11. Double integral calculation.

Reduction of double integral to repeated one in a case of rectangular area.

Let area (P) will be rectangle [a,b,c,d].]

Th 1. If for function f(x,y) defined in rectangle [a,b,c,d] double integral (2.1)

exists and for any permanent value x in [a,b] integral (2.2)

exists then repeated integral (2.3) also exists and

(2.4)

Proof. Let’s divide segments [a,b] and [c,d] into parts by inserting points i n

So rectangle (P) will be divided into partial rectangles .

Let’s denote by andupper and low bounds of function f(x,y) in rectangle

(P_i,k) so that for all points in this rectangle . Then let’s fix

arbitrary value of in segment [x_i,x_i+1] integrate in relation to y from y_k to y_k+1 :

, where ; integral in relation

to y exist because existence of integral (2.2) is supposed for all segment [c,d]. Now let’s

add such inequalities from k=0 to m-1.We have got

. Next let’s multiply all parts

of these inequalities by and sum up in relation to I from 0 to n-1:

We have the integral sum in middle for function f(x,y). And extremes are Darboux sums for double integral (2.1). If all andwill be approach to zero at the same time the n double integral’s (2.1) existence result to approaching s and S to this integral. So

, i.e. double integral (2.1) represents itself integral

on function f(x) - , q.e.d.

Changing variables x and y it may be proved that (2.4*)