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' K =" & < 5 f : [0, 1] → [0, 2] ( ! 4 5 f : [0, 1] → [0, 2] 4 !5 μ (f(K)) = 1, K − = & /$ = 5 K x 3 = 1 2 · K (x), x ∈ [0, 1] 4 5 K 2 3 + x 3 = 1 2 + 1 2 · K (x), x ∈ [0, 1]. /% [a, b] " /& * f [a, b] |f| /' |f| f ! < f (x) = ⎧ ⎨ ⎩ −1, x ∈ Q, 1, x ∈ R\Q. /. * f [a, b] |f| f / / * f n : [a, b] → R, n ∈ N x ∈ [a, b] ∞ n =1 f n (x) f " ; f [a, b] / * f n : [a, b] → R " ! " f : [a, b] → R ! + lim n →∞ V b a [f − f n ] = 0 ; f / * f n : [a, b] → R " ! f : [a, b] → R ! + lim n →∞ V b a [f − f n ] = 0 ; f / * F : [a, b] → R " c 1 , c 2 , . . . [a, b] $ [a,b] f (x)dF (x) = k =1 f (c k )[F (c k ) − F (c k − 0)] (10.5) / * F : [a, b] → R $ [a,b] f (x) dF (x) = $ [a,b] f (x)F (x) dμ (10.6) / $ * F (x) = const f : [a, b] → R ! " ! $ [a, b] f (x) dF (x) = 0 / % < $ A 1 · dF (x) = μ F (A). / & / "2 ! _ ( f : [a, b] → R g : [a, b] → R ! $ [a, b] f (x) dg(x) = f(b)g(b) − f(a)g(a) − $ [a, b] g (x) df(x). (10.7) / ' * x = ϕ(t), α ≤ t ≤ β ! ϕ (α) = a, ϕ(β) = b f : [a, b] → R F : [a, b] → R , $ [a, b] f (x) dF (x) = $ [α, β] f (ϕ(t)) dF (ϕ(t)) ( / "2 / . * f : [a, b] → R g : [a, b] → R ! Z "2 + / "2 + ( (L − S) $ [a, b] f (x) dg(x) = (R − S) $ b a f (x) dg(x) /$/ + [0,1] K(x)dF (x) / "2 ' K(x)− =" & F (x) = 2x + 1. /$ + [0,1] K(x)dF (x) / "2 ' K(x)− =" & F (x) = [3x] + 2x. /$ < f : [0, 1] → R [0, 1] g ! [0, 1] ! " f (x) = ⎧ ⎨ ⎩ 0, x = 0 x 2 sin 1 x 2 , 0 < x ≤ 1 . /$ f : [a, b] → R F : [a, b] → R ! ( $ [a,b] f (x) dF (x) = f(c)[F (b) − F (a)], ∃ c ∈ [a, b]. /$ < f (x) F (x) / "2 $ [a,b] f (x) dF (x) #5 f (x) = x, F (x) = cos x, x ∈ [0, π]. %5 f (x) = sin x, F (x) = |x|, x ∈ [−π, π]. `5 f (x) = x 2 + 3, F (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ x + 2, x ∈ [−2, −1], 2, x ∈ (−1, 0), x 2 + 1, x ∈ [0, 2]. b5 f (x) = x, F (x) = [x], x ∈ [0, n], n ∈ N. a5 f (x) = x 2 , F (x) = [x], x ∈ [0, n], n ∈ N. 75 f (x) = x + 2, F (x) = exp x · sign(cos x), x ∈ [−π, π]. 85 f (x) = x − 1, F (x) = cos x, x ∈ [0, π]. W5 f (x) = x 2 , F (x) = K (x), x ∈ [0, 1]. $5 f (x) = 1 + 2x, F (x) = K (x), x ∈ [0, 1]. #>5 f (x) = [3x], F (x) = K (x), x ∈ [0, 1]. /$$ < - f (x) = ∞ n =0 b n cos(a n πx ), b ∈ (0, 1), a ∈ N va toq son. ' f : R → R * ab > 1 + 3 2 π f : R → R ! /$% < - 5 f (x) = x 2 + 2, x ∈ [−1, 2], 5 f (x) = exp x 2 , x ∈ [−1, 2], !5 f (x) = | cos x|, x ∈ [−π, π], 5 f (x) = sin x, x ∈ [−π, π]. - 0 # - " * < - f + (x) = | f (x)| + f(x) 2 , f − (x) = f (x) − |f(x)| 2 , < ! + #5 f + ! f ! %5 f − ! f ! `5 f + f − ! f ! *5 # % '5 # ` d5 % ` \5 ` < ! + #5 |f| ! f ! %5 f 2 ! f ! `5 f + f − ! f ! *5 # % '5 # ` d5 % ` \5 ` f (x) = 2x, x ∈ E = [0, 5] ! E (f < 6) & & *5 [0, 2] '5 [0, 3) d5 [0, 5) \5 [0, 2) f (x) = [2x], x ∈ E = [0, 5] ! E (f = 4) & & *5 [0, 2] '5 [2, 5 2 ) d5 [2, 5) \5 [2, 3) $ f (x) = ln x 2 − 2x + 1 , x ∈ E = (0, ∞) ! {x : f(x) < 0} & & *5 (0, 2) '5 (0, 1) ∪ (1, 2) d5 (0, ∞) \5 (0, 3) % f (x) = 2 x −1, x ∈ [0, 5] ! {x : 3 < f(x) < 7} & & *5 [0, 3] '5 (2, 3) d5 [0, 3) \5 [2, 3) & x ∈ [0, π] : sin x ≤ 2 −1 & ! & *5 π 3 '5 2π 3 d5 π 6 \5 π 4 ' {x ∈ [0, π] : sin x ≤ cos x} & ! & *5 π 4 '5 2π 3 d5 3π 4 \5 π 3 . A ⊂ [0, 1]− ! & D− \ f 1 (x) = ⎧ ⎨ ⎩ 0, x ∈ A 1, x ∈ [0, 1]\A , f 2 (x) = f 1 (x) + D(x), f 3 (x) = f 1 (x) · D(x). . ! & *5 f 1 '5 f 1 , f 2 d5 f 1 , f 3 \5 f 2 , f 3 / A ⊂ [0, 1] − ! & R− Z f 1 (x) = 1 − χ A (x), f 2 (x) = f 1 (x) − R(x), f 3 (x) = f 2 (x) · D(x). . ! & *5 f 1 '5 f 2 , f 3 d5 f 3 \5 f 2 [0, 1] &- f 1 (x) = R(x), f 2 (x) = D(x), f 3 (x) = x. *5 f 1 '5 f 2 , f 3 d5 f 3 \5 f 1 , f 2 f 1 (x) = 1 + R(x), f 2 (x) = 1 + D(x), f 3 (x) = 1 + R(x) + D(x) ! f (x) ≡ 1 & *5 f 1 , f 2 '5 f 1 , f 2 , f 3 d5 f 2 , f 3 \5 f 1 , f 3 f 1n (x) = sin 2n x, f 2n (x) = cos n x, f 3n (x) = nx 1 + n 2 x 2 , f 4n (x) = 1 + x n ! " [0, 1] & f (x) ≡ 0 c *5 f 1n '5 f 2n, f 3n d5 f 1n , f 3n , f 4n \5 f 1n , f 2n f 1n (x) = sin n x, f 2n (x) = cos n x, f 3n (x) = nx 1 + n 2 x 2 , f 4n (x) = x n ! " [0, 1] & f (x) ≡ 0 c *5 f 1n '5 f 1n, f 3n d5 f 1n , f 3n , f 4n \5 f 1n , f 2n $ E = [0, 4] & f & *5 f (x) = x '5 f (x) = [x] d5 f (x) = e x \5 f (x) = sin x % E = [0, 3] & f 1 (x) = x, f 2 (x) = 1, f 3 (x) = D(x) *5 f 1 , f 2 , f 3 '5 f 1 , f 3 d5 f 1 , f 2 \5 f 2 , f 3 & f (x) = 1 + sign x ! A 1 = {x ∈ [−2, 3] : f(x) = 0} & & *5 [−2, 0] '5 [−2, 0) d5 [−2, −1] \5 [−2, 3] ' f (x) = 5 − [2x] ! A 1 = {x ∈ [−2, 3] : f(x) = 1} & & *5 [2, 3] '5 [2, 3) d5 [2, 2, 5) \5 [2, 2, 6] . & *5 * μ (A) = 0 + A f (x)dμ = 0 '5 + A [f(x) + g(x)]dμ = + A f (x)dμ + + A g (x)dμ d5 + A k · f(x)dμ = k + A f (x)dμ \5 * f ( Download 1.57 Mb. Do'stlaringiz bilan baham: |
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