得分评卷人 四、综合题（本大题共10分） 证明：因为()=f(x,)dz在{a,一致收敛，所以由一致收敛的定 义e>0,Ao>a,A>Ao,u∈a,l,有f红，四d<系所以对o∈ [a,取o+△uela,闭，有f(,uo)d<专与f红，o+△u)dr< CAfx,o+△u)dz-∫Af(z,o)dr<5. 于是，>0，(3A0>0,A>A0,36>0,△4<d,有 o+a-ol=fe,o+at-广w =fe+t+/ f,o+△u-fe,od ≤eo+a-广fe+广fe,o+ah 即(u)在区间a,月连续。 数学分析四试题第8页（共8页）✚➞ ➭ò❁ ♦✦♥Ü❑ ( ✢➀❑✁ 10 ➞ ) ❡➻ê f(x, u) ✸➠➁ D = {(x, u) ∈ R2 | x ∈ [a, +∞), u ∈ [α, β]} ë❨, ➹→➮ ➞ φ(u) = R +∞ a f(x, u)dx ✸➠♠ [α, β] ➌➋➶ñ, ❑②➨ φ(u) ✸ [α, β] ë❨. ②➨: Ï➃ φ(u) = R +∞ a f(x, u)dx ✸ [α, β] ➌➋➶ñ, ↕➧❞➌➋➶ñ✛➼ ➶ ∀ > 0, ∃A0 > a, A > A0, ∀u ∈ [α, β], ❦    R +∞ A f(x, u)dx    <  3 . ↕➧é u0 ∈ [α, β], ✒ u0 + ∆u ∈ [α, β], ❦    R +∞ A f(x, u0)dx    <  3 ❺    R +∞ A f(x, u0 + ∆u)dx    <  3 . ❞ f(x, u) ✸ D ✛ë❨✺⑧✗ p(u) = R A a f(x, u)dx ✸ [α, β] ë❨, ↕➧ p(u)  ✸ u0 ë❨, ❂ ∀ > 0, ∃δ > 0, |∆u| < δ, ❦ |p(u0 + ∆u) − p(u)| =   R A a f(x, u0 + ∆u)dx − R A a f(x, u0)dx    <  3 . ✉➫, ∀ > 0,(∃A0 > 0, ∀A > A0), ∃δ > 0, |∆u| < δ, ❦ |φ(u0 + ∆u) − φ(u0)| =     Z +∞ a f(x, u0 + ∆u)dx − Z +∞ a f(x, u0)dx     =     Z A a f(x, u0 + ∆u)dx + Z +∞ A f(x, u0 + ∆u)dx − Z A a f(x, u0)dx − Z +∞ A f(x, u0)dx     ≤     Z A a f(x, u0 + ∆u)dx − Z A a f(x, u0)dx     +     Z +∞ A f(x, u0 + ∆u)dx     +     Z +∞ A f(x, u0)dx     <  3 +  3 +  3 = , ❂ φ(u) ✸➠♠ [α, β] ë❨. ê➷➞Û(III)➪❑ ✶ 8 ➄↔✁ 8 ➄↕