变限积分函数

微积分基本公式

1. 变限积分函数

定义：

\[设f(x)在[a,b]上可积，则F(x)=\int_a^xf(t)dt,x\in[a,b]称为变上限积分函数 \]

\[证明：当x\in(a,b)时\\ \Delta F=F(x+\Delta x)-F(x)\\ =\int_a^{x}f(t)dt-\int_a^xf(t)dt+\int_x^{x+\Delta x}f(t)dt\\ =\int_x^{x+\Delta x}f(t)dt\\ \because f(x)可积\\ \therefore \exist M>0,有|f(x)|\leq M\\ 又|\Delta F|=|\int_x^{x+\Delta x}f(t)dt|\leq \int_x^{x+\Delta x}|f(t)|dt\\ \leq M\cdot |\Delta x|\\ 由夹逼准则有\lim_{\Delta x \rightarrow 0}\Delta F=0\\ 当x=a时\\ \Delta F=F(a+\Delta x)-F(a)\qquad (\Delta x>0)\\ 同理有\lim_{\Delta x\rightarrow 0^ +}\Delta F=0 \]

\[(2)若f(x)在区间[a,b]上连续，则F(x)=\int_a^xf(t)dt在[a,b]上可导，且F'(x)=f(x) \]

\[证明：当x\in(a,b)时，\\ \Delta F=F(x+\Delta x)-F(x)\\ =\int_x^{x+\Delta x}f(t)dt\\ 则\lim_{\Delta x\rightarrow0}\frac{\Delta F}{\Delta x}=\lim_{\Delta x\rightarrow0}\frac{\int_{x}^{x+\Delta x}f(t)dt}{\Delta x}\\ =\lim_{\Delta x\rightarrow0}\frac{f(\xi)\cdot \Delta x}{\Delta x}\qquad (\xi介于x与x+\Delta x之间)\\ =\lim_{\xi\rightarrow x}f(\xi)=f(x)\\ \therefore F'(x)=f(x)\\ 当x=a时，\Delta x>0\\ \Delta F=F(a+\Delta x)-F(a)=\int_a^{a+\Delta x}f(t)dt\\ F'_+(a)=\lim_{\Delta x\rightarrow 0^+}\frac{\Delta F}{\Delta x}\lim_{\Delta x\rightarrow 0^+}\frac{\int_a^{a+\Delta x}f(t)dt}{\Delta x}=\lim_{\Delta x\rightarrow 0^+}\frac {f(\xi)\cdot \Delta x}{\Delta x}\qquad (\xi 介于a与a+\Delta x)之间\\ =\lim_{\xi\rightarrow a^+}f(\xi)=f(a)\\ 同理当x=b时\qquad \Delta x<0\\ F'(b)=f(b)\\ 综上F'(x)=f(x)\qquad x\in[a,b]\\ F(x)=\int_a^xf(t)dt \qquad x\in[a,b] \]

【例】计算下列各导数

\[(1)\frac d{dx}\int_{x^2}^{x^3}\frac {dt}{\sqrt{1+t^4}}\\ 解：=\frac 1{\sqrt{1+(x^3)^4}}\cdot3x^2-\frac1{\sqrt{1+(x^2)^4}}\cdot2x \]

\[(2)\frac d{dx}\int_{\sin x}^{\cos x}\cos (\pi t^2)dt\\ =-\cos (\pi\cdot \cos^2x)\sin x-\cos (\pi \sin^2x)\cos x \]

\[(3)\frac d{dx}\int_{x^2}^0 x\cos t^2dt\\ =-\Big[\int_o^{x^2} x\cos t^2dt\Big]'\\ =-\Big[x\int_0^{x^2} \cos t^2dt\Big]'\\ =-\Big[\int_0^{x^2} \cos t^2dt+x\cos x^4\cdot2x\Big]\\ =-\Big(\int_0^{x^2}\cos t^2dt+2x^2\cos x^4\Big) \]

\[(4)\frac d{dx}\int_0^x \sin(x-t)^2dt\\ 解： \int_0^x\sin(x-t)^2dt=^{令u=x-t}-\int_x^0 \sin u^2\cdot du\\ \therefore \frac d{dx}\int_0^x\sin(x-t)^2dt=\frac d{dx}\int_0^x \sin u^2du\\ =\sin x^2 \]

\[【例】设f(x)在[0,+\infty)内连续且f(x)>0,证明F(x)=\frac{\int_0^x tf(t)dt}{\int_0^x f(t)dt}在(0,+\infty)内为单调增加函数 \]

\[证明：当x>0时\\ F'(x)=\frac{xf(x)\int_0^x f(t)dt-\int_0^x tf(t)dt\cdot f(x)}{\Big(\int_0^xf(t)dt\Big)^2}\\ =\frac{f(x)}{\Big(\int_0^xf(t)dt\Big)^2}\Big[x\int_0^x f(t)dt-\int_0^xtf(t)dt\Big]\\ 令g(x)=x\int_0^xf(t)dt-\int_0^xtf(t)dt\\ g'(x)=\int_0^xf(t)dt>0\qquad (x>0)\\ g(0)=0\qquad g(x)>0\\ 故有f'(x)>0 \qquad x\in(0,\infty)\\ 从而f(x)在(0,\infty)单调递增 \]

\[【例】求\lim_{x\rightarrow 0}\frac{\int_0^x\Big(3\sin t +t^2+t^2\cos \frac1t\Big)dt}{(1+\cos x)\int_0^x\arctan \big[\ln(1+t)\big]dt} \]

\[=\frac12\lim_{x\rightarrow 0}\frac{\int_0^x(3\sin t+t^2\cos \frac1t)dt}{\int_0^x\arctan \big[\ln(1+t)\big]dt}\\ =\frac12\lim_{x\rightarrow0}\frac{3\sin x+x^2\cos \frac1x}{\arctan \big[\ln(1+x)\big]}\\ =\frac12\lim_{x\rightarrow0}(\frac{3\sin x}{x}+\frac{x^2\cos \frac1x}{x})\\ =\frac32 \]

\[【例】求\lim_{x\rightarrow0}\frac{-\int_{\cos x}^1e^{-t^2}dt}{x^2}\\ =\lim_{x\rightarrow0}\frac{e^{-(\cos x)^2}}2=\frac{e^{-1}}2 \]