MIT_单变量微积分_18

1.微积分第二基本定理

均值定理：
$$\begin{gathered} \Delta F=F(b)-F(a),\Delta x=b-a \\ \Delta F={\int }_a^{b}f(x)dx(FTC1) \\ \frac{\Delta F}{\Delta x}=\frac{1}{b-a}{\int }_a^{b}f(x)dx,相等于将f平分。 \\ \Delta F=Avg(F^{\prime })\cdot \Delta x\le (max{F}^{\prime })\Delta x \\ (min{F}^{\prime })\Delta x\le \Delta F=F^{\prime }(c)\Delta x(MVT)\le (max{F}^{\prime })\Delta x \end{gathered}$$

Ex:$F^{\prime }(x)=\frac{1}{x+1},F(0)=1$，由均值定理推断可知：$A<F(4)<B$,$A和B$分别等于多少？
$$\begin{gathered} F(4)-F(0)=F^{\prime }(C)\ast (4-0)=\frac{1}{c+1}\cdot 4 \\ \frac{1}{1+4}=\frac{4}{5}\le \frac{1}{c+1}\cdot 4\le 4=\frac{1}{1+0}\cdot 4 \\ \frac{9}{5}\le F(4)\le 5 \end{gathered}$$

• 解释-FTC1:
  $F(4)-F(0)={\int }_0^4\frac{1}{1+x}dx<{\int }_0^{4}dx=4$
  $F(4)-F(0)={\int }_0^4\frac{1}{1+x}dx>{\int }_0^4\frac{1}{5}dx=\frac{4}{5}$
  几何意义：$下黎曼和<黎曼和<上黎曼和$
  
  FTC2:已知函数$f$是连续的，$G(x)={\int }_a^{x}f(t)dt(a\le t\le x),G^{\prime }(x)=f(x),G(x)可解出方程$
  $$\{\begin{array}{cc}{y}^{\prime }=f& \\ y(a)=0& \end{array}$$
  Ex:$\frac{d}{dx}{\int }_1^x\frac{dt}{t^2}=\frac{1}{x^2}$
  
  $\Delta G\approx \Delta xf(x)$
  ${\lim }_{\Delta x\to 0}\frac{\Delta G}{\Delta x}=f(x)$

FTC1 证明：
$$\begin{gathered} F^{\prime }=f,假设f连续 \\ G(x)={\int }_a^{x}f(t)dt \\ 由FTC2\Rightarrow G^{\prime }(x)=f(x) \\ {F}^{\prime }(x)=G^{\prime }(x)\Rightarrow F(x)=G(x)+C \\ F(b)-F(a)=(G(b)+C)-(G(a)+C) \\ =G(b)-G(a) \\ ={\int }_a^{b}f(x)dx-0 \end{gathered}$$

Ex:$L^{\prime }(x)=\frac{1}{x},L(1)=0,L(x)={\int }_1^x\frac{dt}{t}$

Ex:
新函数：$y^{\prime }=e^{-x^2},y(0)=0,F(x)={\int }_0^x{e}^{-t^2}dt$
如图所示：时钟函数。$y=e^{-x^2}$