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Def. 2.Let a function be given on a segment [a, b] (fig. 2).

1. We divide the segment into n parts (subintervals)

by points (division points)





2. We take arbitrary point in every subinterval find the value
Fig. 2 of the function at this point and multiply this value by the length of the subinterval.

3. Adding all these products we get a sum (integral sum)

. ( 8 )

4. If there exists the limit of the integral sum (8) as , this limit is called the definite integralof the function over the segment [a, b] and is denoted

. ( 9 )

We read the left side of (9) as “definite integral from a to b of ”.

have the same names as for indefinite integral; a is called lower limit of integration, b upper limit of integration.

Def. 3.A function is called integrable on [over] the segment [a, b], if its definite integral (9) exists.

Theorem 1 (existence theorem).If a function is continuous one on the segment [a, b] then it is integrable over this segment.

Geometric sense ofadefinite integral. If a function is nonnegative, , then by (2), (3) its definite integral is the area of a curvilinear trapezium (1), fig. 1,

( 10 )

Economical sense ofadefinite integral. If a function is a labour producti-vity of some factory then its produced quantity U during a time interval [0, T] by vir-tue of (4), (5) is represented by a definite integral,

. ( 11 )

Physical sense ofadefinite integral. If a function is the velocity of a material point then, on the base of (6), (7), the length path L traveled by the point during a time interval from t = 0 to t = T is given by a definite integral

( 12 )

Ex. 1. Prove that

( 13 )

■The integrand , and so the integral sum (8) equals the length of the segment , that is


therefore its limit, which is the integral (13), equals .■

Note 1. Definite integral doesn’t depend on a variable of integration. It means that

( 14 )

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