Конспект лекций по математическому анализу. Шерстнев А.Н. - 204 стр.

Z +14. eSLI FUNKCIQ  f (x)  0 (x  0) NE WOZRASTAET, TO INTEGRAL
                   P
                   1
     f (t) dt I RQD f (j ) SHODQTSQ ILI RASHODQTSQ ODNOWREMENNO.
 0                j =1
      x130. aBSOL@TNO SHODQ]IESQ INTEGRALY
      1. iNTEGRAL
                                Zb
(1)                                  f (t) dt;
                                 a
IME@]IJ OSOBENNOSTX W PRAWOM  Z b KONCE, NAZYWAETSQ ABSOL@TNO SHODQ]IM-
SQ , ESLI SHODITSQ INTEGRAL a jf (t)j dt.
    2. eSLI INTEGRAL SHODITSQ ABSOL@TNO, TO ON SHODITSQ.
          Zb
  pUSTX a jf (t)j dt < +1. w SILU KRITERIQ kO[I 127.3
                                                  Zy
            8" > 0 9c < b 8x; y (c < x < y < b) ( x jf (t)j dt < "):
                                  Zy           Zy
nO TOGDA DLQ UKAZANNYH x; y : x f (t) dt  x jf (t)j dt < ". sNOWA W SILU
KRITERIQ kO[I \TO OZNA^AET, ^TO INTEGRAL (1) SHODITSQ. >
    kAK MY UWIDIM NIVE, IZ SHODIMOSTI INTEGRALA (1) EGO ABSOL@TNAQ
SHODIMOSTX NE SLEDUET. pO\TOMU POLEZNO RASPOLAGATX PRIZNAKAMI SHODI-
MOSTI BOLEE TONKIMI, ^EM PRIZNAKI DLQ INTEGRALOW OT ZNAKOPOSTOQNNYH
FUNKCIJ. pRIWEDEM DWA POLEZNYH NA PRAKTIKE PRIZNAKA, KOTORYE W BOLEE
OB]EJ FORME BUDUT DOKAZANY NIVE (SM. 135.4).
                                              Zb
    3. pUSTX b 2 R [ f+1g, INTEGRAL J =           f (t)g(t) dt IMEET EDINST-
                                                a
WENNU@ OSOBENNOSTX W TO^KE b 2 R [ f+1g, PRI^EM f NEPRERYWNA, A g
NEPRERYWNO DIFFERENCIRUEMA NA [a; b). pUSTX, KROME TOGO, WYPOLNENY
USLOWIQ (PRIZNAK dIRIHLE Zx )
    1D) FUNKCIQ F (x) = f (t) dt (a  x < b) OGRANI^ENA,
                          a
    2D) g(t) UBYWAET I tlim
                         !b,
                             g(t) = 0;
 LIBO WYPOLNENY USLOWIQ (PRIZNAK aBELQ)
                   Zb
    1A) INTEGRAL f (t) dt SHODITSQ,
                    a

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