《数学分析》课程教学课件（讲稿，打印版）第5章 单变量函数的积分学 5.1.6 定积分的计算 5.1.7 用积分定义函数 5.1.8 Taylor展开中余项的积分表示

5.1.6 5.1.7 5.1.8 换元法 分部积分法 5.1.6定积分的计算 1°定积分的换元法 定理1（定积分的换元法）设函数f(x)在区间[a,b]上连续，而函数 x=p(t)在[a,β]上连续可导,a≤p(t)≤b,且p(a)=a, φ(B)=b.则有 下面的换元公式 1 foarde 1 fothe(trat. 证明由定理中的条件可知，上式两端的积分都存在，且函数f(x)和 f(φ(t)φ'(t) 分别在区间 [a,b]及 [α，3]上有原函数.设 F(x) 是 f(x)(在 [a,b]上)的一个原函数，则根据复合函数的求导法则可知，F(φ(t)可导，且 (F(4(t)'=F'(4(t)4'(t)=f(4(t)4'(t). 因此 F(φ(t)是 f(φ(t)φ'(t) 在 [α，3]上的一个原函数. 由 Newton-Leibniz 公 ‖返回全屏关闭退出 1/24

5.1.6 5.1.7 5.1.8 †{ ©ÜÈ©{ 5.1.6 ½È©OŽ 1 ◦ ½È©†{ ½n 1 (½È©†{) ¼ê f(x) 3«m [a, b] þëY, ¼ê x = ϕ(t) 3 [α, β] þëYŒ, a 6 ϕ(t) 6 b, … ϕ(α) = a, ϕ(β) = b. Kk e¡†úª Z b a f(x)dx = Z β α f(ϕ(t))ϕ 0 (t)dt. y² d½n¥^‡Œ, þªüàÈ©Ñ3, …¼ê f(x) Ú f(ϕ(t))ϕ0 (t) ©O3«m [a, b] 9 [α, β] þk¼ê.  F(x) ´ f(x) (3 [a, b] þ) ‡¼ê, KŠâEܼê¦{KŒ, F(ϕ(t)) Œ, … (F(ϕ(t)))0 = F 0 (ϕ(t))ϕ 0 (t) = f(ϕ(t))ϕ 0 (t). Ïd F(ϕ(t)) ´ f(ϕ(t))ϕ0 (t) 3 [α, β] þ‡¼ê. d Newton–Leibniz ú 1/24 kJ Ik J I ˆ£ ¶ '4 òÑ

5.1.6 5.1.7 5.1.8 †{ ©ÜÈ©{ ª, ·‚k Z b a f(x)dx = F(b) − F(a), ±9 Z β α f[ϕ(t)]ϕ 0 (t)dt = F[ϕ(β)] − F[ϕ(α)] = F(b) − F(a). ùÒy² ¤`ª. y. 51 l½ny²5w, vk7‡‡¦ ϕ0 (t) 3 [α, β] þëY, ‡§Œ ÈҌ± . 52 †Ø½È©†{', ùpvk‡¦ x = ϕ(t) î‚üN. ù´ ϏØI‡Ø½È©@ªAò#Cþ†£5È©Cþ. 2/24 kJ Ik J I ˆ£ ¶ '4 òÑ

5.1.6 5.1.7 5.1.8 †{ ©ÜÈ©{ ~ 1  f(x) ´4«m [−a, a] þëYÛ¼ê, ¦ Z a −a f(x) dx. ) Z a −a f(x) dx = Z 0 −a f(x) dx + Z a 0 f(x) dx = − Z 0 a f(−t)dt + Z a 0 f(x) dx = Z 0 a f(t) dt + Z a 0 f(x) dx = − Z a 0 f(t) dt + Z a 0 f(x) dx = 0. 3/24 kJ Ik J I ˆ£ ¶ '4 òÑ

5.1.6 5.1.7 5.1.8 †{ ©ÜÈ©{ ~ 2 ¦ Z a 0 p a2 − x2 dx (a > 0). ) - x = a sin t (0 6 t 6 π/2). K x = 0 ž, t = 0;  x = a ž, t = π 2 . ¤± (d½È©†{K) Z a 0 p a2 − x2 dx = Z π 2 0 a 2 cos2 t dt = a 2 2 Z π 2 0 [1 + cos 2t] dt = a 2 2  t + 1 2 sin 2t     π 2 0 = π 4 a 2 . 5: ù‡~f`²Œ» a ¡Èo©ƒu π 4 a 2 . Ïd¡ È´ πa2 . 4/24 kJ Ik J I ˆ£ ¶ '4 òÑ

5.1.6 5.1.7 5.1.8 †{ ©ÜÈ©{ ~ 3 OŽÈ© I = R 1 0 ln(1+x) 1+x2 dx. ) ŠC† x = tan ϕ, K dϕ = 1 1+x2dx, … x = 0 ž, ϕ = 0;  x = 1 ž, ϕ = π 4 . u´ I = Z π 4 0 ln  cos ϕ + sin ϕ cos ϕ  dϕ = Z π 4 0  ln √ 2  1 √ 2 cos ϕ + 1 √ 2 sin ϕ  − ln(cos ϕ)  dϕ = Z π 4 0 n ln √ 2 + ln  sin  ϕ + π 4  − ln(cos ϕ) o dϕ = π 8 ln 2 + Z π 4 0 ln  sin  ϕ + π 4  dϕ − Z π 4 0 ln(cos ϕ)dϕ. Ϗ Z π 4 0 ln  sin  ϕ + π 4  dϕ ϕ= π 4 −t ====== − Z 0 π 4 ln  sin(π 2 − t)  dt = Z π 4 0 ln(cos t)dt. ¤± I = π 8 ln 2. 5/24 kJ Ik J I ˆ£ ¶ '4 òÑ

5.1.6 5.1.7 5.1.8 †{ ©ÜÈ©{ ~ 4  f(x) ´ [a, b] þŒÈà¼ê, ¦y (b − a)f  a + b 2  6 Z b a f(x) dx. y² Ϗ f(x) ´à¼ê, ¤±é x ∈ [a, b] k f(x) + f(a + b − x) > 2f  a + b 2  . ü>È©Œ Z b a f(x) dx + Z b a f(a + b − x) dx > 2(b − a)f  a + b 2  . Š† t = a + b − x, Œ Z b a f(a + b − x) dx = − Z a b f(t) dt = Z b a f(t) dt. dd=¤y. 6/24 kJ Ik J I ˆ£ ¶ '4 òÑ

5.1.6 5.1.7 5.1.8 †{ ©ÜÈ©{ ~ 5  f(x) 3 R þk½Â, 3?¿k«mþŒÈ, …÷v f(x + y) = f(x) + f(y). ¦ f(x). )  ¼ê F(x) = Z x 0 f(t) dt. é?¿ x, y k F(x + y) = Z x+y 0 f(t) dt = Z x 0 f(t) dt + Z x+y x f(t) dt = F(x) + Z y 0 f(t + x) dt = F(x) + Z y 0 ￾ f(t) + f(x)
dt = F(x) + F(y) + yf(x). 7/24 kJ Ik J I ˆ£ ¶ '4 òÑ

5.1.6 5.1.7 5.1.8 †{ ©ÜÈ©{ =é?¿ x, y k F(x + y) = F(x) + F(y) + yf(x). † x, y  F(x + y) = F(x) + F(y) + xf(y). 'þ¡ª,  yf(x) = xf(y).  y = 1,  f(x) = ax, Ù¥ a = f(1) ´?¿~ê. 8/24 kJ Ik J I ˆ£ ¶ '4 òÑ

5.1.6 5.1.7 5.1.8 †{ ©ÜÈ©{ 2 ◦ ½È©©ÜÈ©{ ½n 2 (½È©©ÜÈ©{) ¼ê u(x) Ú v(x) 3«m [a, b] þkëY ¼ê u 0 (x) † v 0 (x). K Z b a u(x)v 0 (x) dx = u(x)v(x)    b a − Z b a u 0 (x)v(x) dx. y² d‡©¥¦{K (u(x)v(x))0 = u 0 (x)v(x) + u(x)v 0 (x), 9®^‡Œ, þªü>Ñ´ëY, ÏdŒÈ. éþªü>?1È©, ¿^ Newton–Leibniz úª, Ñ u(x)v(x)    b a = Z b a u 0 (x)v(x) dx + Z b a u(x)v 0 (x) dx. =¤‡y²ª. 9/24 kJ Ik J I ˆ£ ¶ '4 òÑ

5.1.6 5.1.7 5.1.8 †{ ©ÜÈ©{ ~ 6 OŽ Z 1 0 ln(1 + x) dx. ) Šâ©ÜÈ©{, Z 1 0 ln(x + 1) dx = x ln(x + 1)    1 0 − Z 1 0 x · 1 x + 1 dx = ln 2 − Z 1 0  1 − 1 x + 1 dx = ln 2 −  1 − ln(x + 1)    1 0  = 2 ln 2 − 1 = ln 4 e . 10/24 kJ Ik J I ˆ£ ¶ '4 òÑ