# 不定积分

  1. 定义
    • F′(x)=f(x)F'(x) = f(x) \quad F(x)F(x) 是 f(x)f(x) 的一个原函数
  2. 原函数存在定理
    • 连续一定有原函数 [F(x)+C]′=f(x)[F(x) + C]' = f(x)
  3. 几何含义
    • 一组平行的曲线簇
  4. 性质
    • ∫[f(x)±g(x)]dx=∫f(x)dx±∫g(x)dx\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx
    • ∫kf(x)dx=k∫f(x)dx\int kf(x) dx = k \int f(x) dx

# 积分表

  1. ∫0dx=C\int 0 dx = C
  2. ∫kdx=kx+C\int k dx = kx + C
  3. ∫xμdx=xμ+1μ+1+C\int x^\mu dx = \dfrac{x^{\mu + 1}}{\mu + 1} + C
  4. ∫1xdx=ln⁡∣x∣+C\int \dfrac{1}{x} dx = \ln|x| + C
  5. ∫11+x2=arctan⁡(x)+C\int \dfrac{1}{1 + x^2} = \arctan(x) + C
  6. ∫11−x2dx=arcsin⁡(x)+C\int \dfrac{1}{\sqrt{1 - x^2}} dx = \arcsin(x) + C
  7. ∫cos⁡(x)dx=sin⁡(x)+C\int \cos(x) dx = \sin(x) + C
  8. ∫sin⁡(x)dx=−cos⁡(x)+C\int \sin(x) dx = -\cos(x) + C
  9. ∫1cos⁡2(x)dx=∫sec⁡2(x)dx=tan⁡(x)+C\int \dfrac{1}{\cos^2(x)} dx =\int \sec^2(x) dx = \tan(x) +C
  10. ∫1sin2(x)dx=∫csc⁡2(x)dx=−cot⁡(x)+C\int \dfrac{1}{sin^2(x)} dx = \int \csc^2(x) dx = -\cot(x) + C
  11. ∫sec⁡(x)tan⁡(x)dx=sec⁡(x)+C\int \sec(x)\tan(x) dx = \sec(x) + C
  12. ∫csc⁡(x)cot⁡(x)dx=−csc⁡(x)+C\int \csc(x)\cot(x) dx = -\csc(x) + C
  13. ∫exdx=ex+C\int e^x dx = e^x + C
  14. ∫axdx=axln⁡(a)+C\int a^x dx = \dfrac{a^x}{\ln(a)} + C
  15. ∫csc⁡2(x)=−cot⁡(x)+C\int \csc^2(x) = -\cot(x) + C
  16. ∫sec⁡2(x)=tan⁡2(x)+C\int \sec^2(x) = \tan^2(x) + C
  17. ∫1a2−x2dx=arcsin⁡(xa)+C\int \dfrac{1}{\sqrt{a^2 - x^2}} dx = \arcsin(\dfrac{x}{a}) + C
  18. ∫1x2+a2dx=1aarctan⁡(xa)+C\int \dfrac{1}{x^2 + a^2} dx = \dfrac{1}{a}\arctan(\dfrac{x}{a}) + C
  19. ∫1x2−a2dx=12aln⁡∣x−ax+a∣+C\int \dfrac{1}{x^2 - a^2} dx = \dfrac{1}{2a}\ln|\dfrac{x - a}{x + a}| + C
  20. ∫1x2+a2dx=ln⁡(x+x2+a2)+C\int \dfrac{1}{\sqrt{x^2 + a^2}} dx = \ln(x + \sqrt{x^2 + a^2}) + C
  21. ∫1x2−a2dx=ln⁡(x+x2−a2)+C\int \dfrac{1}{\sqrt{x^2 - a^2}} dx = \ln(x + \sqrt{x^2 - a^2}) + C
  22. ∫tan⁡(x)dx=−ln⁡∣cos⁡(x)∣+C\int \tan(x) dx = -\ln|\cos(x)| + C
  23. ∫cot⁡(x)dx=ln⁡∣sin(x)∣+C\int \cot(x) dx = \ln|sin(x)| + C
  24. ∫sec⁡(x)=ln⁡∣sec⁡(x)+tan⁡(x)∣+C\int \sec(x) = \ln|\sec(x) + \tan(x)| + C

# 换元积分法

  • 核心思想 (凑), 把 d 前面的某一部分,求原函数拿到 d 的里面去
    • ∫1dx=x+C\int 1 dx = x + C
    • ∫1du=u+C\int 1 du = u + C
    • ∫1dF(u)=F(u)+C\int 1 dF(u) = F(u) + C
    • ∫1d(x2−3)=x2−3+C\int 1 d(x^2 - 3) = x^2 - 3 + C
  • x=ψ(t)x = \psi(t), 先积分出来,然后再换回 xx
    • 第二类换元思路 (常用是前三项)
    1. a2−x2x=asin⁡(t)→a2−a2sin⁡2(t)=a2cos⁡2(t){\sqrt{a^2 - x^2}} \qquad {x = a\sin(t)} \rightarrow {\sqrt{a^2 - a^2\sin^2(t)}} = {\sqrt{a^2\cos^2(t)}}
    2. x2−a2x=asec⁡(t)→a2sec⁡2(t)−a2=a2tan⁡2(t){\sqrt{x^2 - a^2}} \qquad {x = a\sec(t)} \rightarrow {\sqrt{a^2\sec^2(t) - a^2}} = {\sqrt{a^2\tan^2(t)}}
    3. x2+a2x=atan⁡(t)→a2tan⁡2(t)+a2=a2sec⁡2(t){\sqrt{x^2 + a^2}} \qquad {x = a\tan(t)} \rightarrow {\sqrt{a^2\tan^2(t) + a^2}} = {\sqrt{a^2\sec^2(t)}}
    4. a2−x2x=acos⁡(t)→a2−a2cos⁡2(t)=a2sin⁡2(t){\sqrt{a^2 - x^2}} \qquad {x = a\cos(t)} \rightarrow {\sqrt{a^2 - a^2\cos^2(t)}} = {\sqrt{a^2\sin^2(t)}}
    5. x2−a2x=acsc⁡(t)→a2cos⁡2(t)sin⁡2(t)=a2cot⁡2(t){\sqrt{x^2 - a^2}} \qquad {x = a\csc(t)} \rightarrow {\sqrt{a^2\dfrac{\cos^2(t)}{\sin^2(t)}}} = {\sqrt{a^2\cot^2(t)}}
    6. x2+a2x=acot⁡(t)→a2cot⁡2(t)+a2=a2csc⁡2(t){\sqrt{x^2 + a^2}} \qquad {x = a\cot(t)} \rightarrow {\sqrt{a^2\cot^2(t) + a^2}} = {\sqrt{a^2\csc^2(t)}}

# 分部积分法

  • ∫xexdx=∫xdex=xex−∫exdx=xex−ex+C\int x e^x dx = \int x de^x = xe^x - \int e^x dx = xe^x - e^x + C
  • ∫ln⁡(x)dx=xln⁡(x)−x+C\int \ln(x) dx = x\ln(x) - x + C

# 往里拿的优先级

  1. exe^x
  2. sin⁡(x),cos⁡(x)\sin(x) , \cos(x)
  3. xnx^n

# 有理函数的积分

  • 待定系数法

# 定积分

# 概念与性质

  • 设函数f(x)f(x) 在区间 [a,b] 上有定义且有界,然后分割,求和,取极限。
  • PS: 只与被积函数f(x)f(x) 被积区域[a,b][a,b] 有关,与积分变量无关
  • 定理 1: 连续,可积
  • 定理 2: 有界、有限个间断点,可积
  • 几何意义:

# 定积分性质

  • ∫aaf(x)dx=0\int^a_a f(x) dx= 0
  • 性质一
    • ∫abf(x)dx=−∫baf(x)dx\int^b_a f(x) dx= -\int^a_b f(x) dx
  • 性质二
    • ∫ab(αf(x)+βg(x))dx=α∫abf(x)dx+β∫abg(x)dx\int^b_a (\alpha f(x) + \beta g(x)) dx = \alpha \int^b_a f(x) dx + \beta \int^b_a g(x) dx
  • 性质三
    • ∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx\int^b_a f(x) dx = \int^c_a f(x) dx + \int^b_c f(x) dx
  • 性质四
    • f(x)≡1∫ab1dx=b−a∫abkdx=b−af(x) \equiv 1 \quad \int^b_a 1 dx = b - a \quad \int^b_a k dx = b - a
  • 性质五
    • f(x)≥0∫abf(x)dx≥0f(x) \geq 0 \qquad \int^b_a f(x) dx \geq 0
    • f(x)≤0∫abf(x)dx≤0f(x) \leq 0 \qquad \int^b_a f(x) dx \leq 0
  • 推理一
    • f(x)≥g(x)∫abf(x)dx≥∫abg(x)dxf(x) \geq g(x) \qquad \int^b_a f(x) dx \geq \int^b_a g(x) dx
    • g(x)−f(x)≥0∫ab[g(x)−f(x)]dx≥0g(x) - f(x) \geq 0 \qquad \int^b_a [g(x) - f(x)] dx \geq 0
更新于 阅读次数 次

请我喝[茶]~( ̄▽ ̄)~*

洪宽 微信支付

微信支付

洪宽 支付宝

支付宝

洪宽 贝宝

贝宝