定积分的计算方法

单变量函数的积分

张瑞
中国科学技术大学数学科学学院

定积分的计算方法

Newton-Leibniz公式

[习题] $f(x)$在$[a,b]$上可积，且有原函数$F(x)$， 则

\[\int_a^b f(x)dx=F(x)\big|_a^b=F(b)-F(a) \]

$f(x)$在$[a,b]$上可积，在$(a,b)$内有原函数$F(x)$，且$F(a+), F(b-)$存在，则

\[\displaystyle\int_a^b f(x)dx=F(b-)-F(a+) \]

例 1. [习题] $\displaystyle\int_0^{2\pi}\dfrac{\sec^2x}{2+\tan^2x}dx$

解. 由Newton-Leibniz公式

\[=\int_0^{2\pi}\dfrac{1}{2+\tan^2x}d(\tan x) =\left.\dfrac1{\sqrt 2}\arctan\dfrac{\tan x}{\sqrt 2}\right|_0^{2\pi} \]

例 2. $\displaystyle\int_{-2}^2|x^2+2|x|-3|dx$

例 3. $\displaystyle\int_{-\pi}^{\pi}\dfrac{1-r^2}{1-2r\cos x+r^2}dx, r\in(0,1)$

14. $\displaystyle\int_0^{2\pi}\dfrac{\sec^2x}{2+\tan^2x}dx$

注意$F(x)=\dfrac1{\sqrt 2}\arctan\dfrac{\tan x}{\sqrt 2}$在$[0,2\pi]$上不是原函数

\[\begin{aligned} F(\dfrac{\pi}2-)=\dfrac{\pi}2\dfrac1{\sqrt 2} , &F(\dfrac{\pi}2+)=\dfrac{-\pi}2\dfrac1{\sqrt 2} \\ F(\dfrac{3\pi}2-)=\dfrac{\pi}2\dfrac1{\sqrt 2} , &F(\dfrac{3\pi}2+)=\dfrac{-\pi}2\dfrac1{\sqrt 2} \\ \end{aligned} \]

所以$\dfrac{\pi}2,\dfrac{3\pi}2$是$F(x)$的间断点，不连续，显然不可导。

\[f(x)=\dfrac{\sec^2x}{2+\tan^2x}=\dfrac1{2\cos^2x+\sin^2x} \]

在$[0,2\pi]$上连续

\[\begin{aligned} \int_0^{2\pi}f(x)dx =&\left(\int_0^{\frac{\pi}2}+\int_{\frac{\pi}2}^{\frac{3\pi}2}+\int_{\frac{3\pi}2}^{2\pi}\right)f(x)dx \\ =&F(\dfrac{\pi}2-)-F(0)+F(\dfrac{3\pi}2-)-F(\dfrac{\pi}2+) \\ &+F(2\pi)-F(\dfrac{3\pi}2+) \\ =&\sqrt{2}\pi \end{aligned} \]

15. $\displaystyle\int_{-\pi}^{\pi}\dfrac{1-r^2}{1-2r\cos x+r^2}dx, r\in(0,1)$

先求原函数。令$t=\tan\dfrac{x}2$，则

\[\begin{aligned} F(x) =&\int\dfrac{1-r^2}{1-2r\cos x+r^2}dx \\ =&2(1-r^2)\int\dfrac1{(1-r)^2+(1+r)^2t^2}dt \\ =&2\arctan\left(\dfrac{1-r}{1+r}t\right)+c \\ =&2\arctan\left(\dfrac{1-r}{1+r}\tan\dfrac{x}2\right)+c \end{aligned} \]

注意到$F(-\pi+)=-\pi, F(\pi-)=\pi$

换元

定理 1. (换元)
设$f(x)$在$[a,b]$上连续，$x=\phi(t)$满足

  1. $\phi(t)$在$[\alpha,\beta]$上有连续导数
  2. $\phi(\alpha)=a, \phi(\beta)=b$，且$a\leq\phi(t)\leq b, t\in[\alpha,\beta]$

则有

\[\int_a^b f(x)dx=\int_{\alpha}^{\beta}f(\phi(t))\phi'(t)dt \]
  1. 不用换回$t$到$x$，而是直接在$t$上积分
  2. 积分限也发生变化

证： $f(x)$在$[a,b]$上连续，则$f(x)$在$[a,b]$上可积，且有原函数$F(x)$，即

\[\int_a^bf(x)=F(b)-F(a) \]

由$\phi'(t)$连续，则$f(\phi(t))\phi'(t)$在$[\alpha,\beta]$上连续，可积，有原函数。

由

\[\dfrac{d}{dt}F(\phi(t))=\dfrac{dF}{dx}\phi'(t)=f(x)\phi'(t)=f(\phi(t))\phi'(t) \]

可知，$F(\phi(t))$是$f(\phi(t))\phi'(t)$在$[\alpha,\beta]$上的一个原函数

\[\begin{aligned} \int_{\alpha}^{\beta}f(\phi(t))\phi'(t)dt =&\left.F(\phi(t))\right|_{\alpha}^{\beta} \\ =&F(\phi(\beta))-F(\phi(\alpha)) \\ =&F(b)-F(a) \end{aligned} \]
\[\begin{aligned} \int\dfrac{\ln x}{x}dx=&\int\ln xd(\ln x) (\ln x=t)\\ =&\int t dt=\dfrac{t^2}{2} =\dfrac{(\ln x)^2}{2} \end{aligned} \]
\[\begin{aligned} \int_1^2\dfrac{\ln x}{x}dx=&\int_1^2\ln xd(\ln x) (\ln x=t) \\ =&\int_0^{\ln 2}t dt=\left.\dfrac{t^2}{2}\right|_0^{\ln 2} =\dfrac{(\ln 2)^2}2 \end{aligned} \]

换元法的条件中，$\phi[\alpha,\beta]\subset[a,b]$这个条件可以放宽。

  • 若$f(x)$在包含$[a,b]$的更大的区间$[A,B]$上连续，则$\phi[\alpha,\beta]\subset[A,B]$即可。

例 4. $\displaystyle\int_{-1}^1 x^2dx$

解. (看看问题出在哪里？) 取$x^2=t$，有

\[\int_{-1}^1 x^2dx=\int_1^1 t\dfrac12\dfrac1{\sqrt{t}}dt=0 \]

$\displaystyle\int_{-1}^1 x^2dx$中，令$t=x^2$。注意：

  • $x>0$时，有$x=\sqrt t$
  • $x<0$时，有$x=-\sqrt t$

因此，

\[\begin{aligned} \int_{-1}^1 x^2dx =&\int_{-1}^0 x^2 dx+\int_0^1 x^2 dx \\ =&\int_{1}^0 t \frac{-1}{2\sqrt t}dt+\int_0^1 t\frac1{2\sqrt t}dt \\ =&\int_0^1\sqrt tdt =\frac23t^{\frac32}\big|_0^1 \end{aligned} \]

例 5. 比较积分的大小

$\displaystyle\int_0^\pi e^{-x^2}\cos^2xdx$，$\displaystyle\int_\pi^{2\pi} e^{-x^2}\cos^2xdx$

例 6. 判定积分的符号 $\displaystyle\int_0^{2\pi}\dfrac{\sin x}{x}dx$

例 7. 判定积分的符号 $\displaystyle\int_{-2}^2x^3e^xdx$

例 8. [习题] $f(x)$以$T$为周期的连续函数，则

\[\int_{\alpha}^{\alpha+T}f(x)dx=\int_0^Tf(x)dx \]

1. $\displaystyle\int_0^\pi e^{-x^2}\cos^2xdx$ ， $\displaystyle\int_\pi^{2\pi} e^{-x^2}\cos^2xdx$

\[\begin{aligned} \int_{\pi}^{2\pi}e^{-x^2}dx =&\int_{0}^{\pi}e^{-(x+\pi)^2}d(x+\pi) \\ =&\int_0^{\pi}e^{-(x+\pi)^2}\cos^2xdx \\ <&\int_0^{\pi}e^{-x^2}\cos^2xdx \end{aligned} \]

2. $\displaystyle\int_0^{2\pi}\dfrac{\sin x}{x}$

\[\begin{aligned} =&\int_0^{\pi}\dfrac{\sin x}{x}dx+\int_{\pi}^{2\pi}\dfrac{\sin x}{x}dx \\ =&\int_0^{\pi}\dfrac{\sin x}{x}dx+\int_0^{\pi}\dfrac{\sin(t+\pi)}{t+\pi}d(t+\pi) \\ =&\int_0^{\pi}\dfrac{\sin x}{x}dx+\int_0^{\pi}\dfrac{-\sin(t)}{t+\pi}d(t) \\ =&\int_0^{\pi}\left(\dfrac{\sin(x)}x-\dfrac{\sin(x)}{x+\pi}\right)dx \\ =&\int_0^{\pi}\dfrac{\pi\sin(x)}{x(x+\pi)}dx > 0\\ \end{aligned} \]

3. $\displaystyle\int_{-2}^2x^3e^xdx$

\[\begin{aligned} =&\int_0^2x^3e^xdx+\int_{-2}^0x^3e^xdx \\ =&\int_0^2x^3e^xdx+\int_2^{0}(-t)^3e^{-t}d(-t) \\ =&\int_0^2x^3e^xdx+\int_0^{2}(-t)^3e^{-t}d(t) \\ =&\int_0^2x^3(e^x-e^{-x})dx >0 \\ \end{aligned} \]

4.

\[\begin{aligned} \int_{\alpha}^{\alpha+T}f(x)dx =&\left(\int_{\alpha}^0+\int_0^T+\int_T^{T+\alpha}\right)f(x)dx \\ =&\int_0^T f(x)dx+\int_\alpha^0f(x)dx+\int_T^{T+\alpha}f(x)dx \end{aligned} \]

$f(x)$以$T$为周期，令$x=t+T$，则

\[\begin{aligned} \int_T^{T+\alpha}f(x)dx =&\int_0^\alpha f(t+T)d(t+T) =\int_0^\alpha f(t)dt \end{aligned} \]

当$f(x)$在$[0,a]$上连续，则

\[\int_0^af(x)dx=\int_a^0f(a-t)d(a-t)=\int_0^af(a-t)dt \]

特别地，当$f(x)$连续，则有

\[\int_0^{\frac{\pi}2}f(\sin x)dx =\int_0^{\frac{\pi}2}f\left(\sin \left(\dfrac{\pi} 2-x\right)\right)dx =\int_0^{\frac{\pi}2}f(\cos x)dx \]

例 9. [例4.5.4] $m$为正整数，$\displaystyle \int_0^{\frac{\pi}2} \sin^mxdx=\int_0^{\frac{\pi}2} \cos^mxdx$

对称函数的积分

定理 2.
$f(x)$在$[-l,l]$上连续，则

  1. 若$f(x)$为奇函数，则$\displaystyle\int_{-l}^l f(x)dx=0$
  2. 若$f(x)$为偶函数，则$\displaystyle\int_{-l}^l f(x)dx=2\int_0^l f(x)dx$

1. $f(x)$为奇函数，$-f(x)=f(-x)$，则

\[\begin{aligned} \int_{-l}^lf(x)dx=\int_0^lf(x)dx+\int_{-l}^0f(x)dx \\ =\int_0^lf(x)dx+\int_l^0f(-t)d(-t) \\ =\int_0^lf(x)dx+\int_0^lf(-t)dt \\ =\int_0^lf(x)dx-\int_0^lf(t)dt = 0 \end{aligned} \]

2. $f(x)$为偶函数，$f(-x)=f(x)$，则

\[\begin{aligned} \int_{-l}^lf(x)dx=\int_0^lf(x)dx+\int_{0}^lf(-t)dt \\ =2\int_0^lf(x)dx \end{aligned} \]

例 10. $\displaystyle\int_{-1}^1 (x+\sqrt{1-x^2})^2dx$

例 11. $\displaystyle\int_{-\frac\pi2}^{\frac\pi2} \dfrac{(1-x)^2\cos^3x}{1+x^2}dx$

例 12. $\displaystyle\int_{-2}^{2} \ln(x+\sqrt{1+x^2})\ln(1+x^2)dx$

3. $\displaystyle\int_{-1}^1 (x+\sqrt{1-x^2})^2dx$

\[\begin{aligned} (x+\sqrt{1-x^2})^2=x^2+1-x^2+2x\sqrt{1-x^2} \\ =1+2x\sqrt{1-x^2} \end{aligned} \]

由$x\sqrt{1-x^2}$为奇函数，则

\[\int_{-1}^1x\sqrt{1-x^2}dx=0 \]

所以，有

\[=\int_{-1}^1 1 dx=2 \]

4.

\[\begin{aligned} \int_{-\frac\pi2}^{\frac\pi2} \dfrac{(1-x)^2\cos^3x}{1+x^2}dx =\int_{-\frac\pi2}^{\frac\pi2} \dfrac{(1+x^2-2x)\cos^3x}{1+x^2}dx \\ =\int_{-\frac\pi2}^{\frac\pi2} \cos^3xdx+\int_{-\frac\pi2}^{\frac\pi2} \dfrac{(-2x)\cos^3x}{1+x^2}dx \end{aligned} \]

$\dfrac{(-2x)\cos^3x}{1+x^2}$为奇函数，$\cos^3x$为偶函数

\[=2\int_0^{\frac{\pi}2}\cos^3xdx =2\int_0^{\frac{\pi}2}\cos^2xd(\sin x) =\dfrac43 \]

5. $\displaystyle\int_{-2}^{2} \ln(x+\sqrt{1+x^2})\ln(1+x^2)dx$

\[\begin{aligned} \ln(-x+\sqrt{1+(-x)^2})=\ln(-x+\sqrt{1+x^2}) \\ =\ln\dfrac1{x+\sqrt{1+x^2}}=-\ln(x+\sqrt{1+x^2}) \end{aligned} \]

所以$\ln(x+\sqrt{1+x^2})$为奇函数，$\ln(1+x^2)$为偶函数。

\[\int_{-2}^{2} \ln(x+\sqrt{1+x^2})\ln(1+x^2)dx=0 \]
  • $f(x)$在$[a,b]$上连续，若$f(a+b-x)=-f(x)$（称$f(x)$关于$\dfrac{a+b}2$奇对称），则$\displaystyle\int_a^b f(x)=0$
  • $f(x)$在$[a,b]$上连续，若$f(a+b-x)=f(x)$（称$f(x)$关于$\dfrac{a+b}2$偶对称），则$\displaystyle\int_a^b f(x)=2\int_a^{\frac{a+b}2} f(x)dx=2\int_{\frac{a+b}2}^b f(x)dx$

定理 3.
$f(x),g(x)$在$[a,b]$上连续，且

  1. $f(a+b-x)=f(x)$，即$f(x)$关于$\dfrac{a+b}2$偶对称；
  2. $g(a+b-x)+g(x)=A$，$A$为常数

则有

\[\begin{aligned} \int_a^b f(x)g(x)dx =&\dfrac{A}2\int_a^bf(x)dx \\ =&A\int_a^{\frac{a+b}2}f(x)dx =A\int_{\frac{a+b}2}^b f(x)dx \end{aligned} \]

证： $h(x)=g(x)-\dfrac{A}2$，则$h(a+b-x)+h(x)=0$，为$\dfrac{a+b}2$奇对称，则

\[\int_a^bf(x)h(x)dx=0 \]

即

\[\begin{aligned} \int_a^bf(x)\left(g(x)-\dfrac{A}2\right)=&0 \\ \int_a^bf(x)g(x)dx =\dfrac{A}{2}\int_a^b f(x)dx =&A\int_{\frac{a+b}2}^bf(x)dx \end{aligned} \]

例 13. $\displaystyle\int_0^{\frac{\pi}4}\ln(1+\tan x)dx$

解. 令$x=\dfrac{\pi}4-t$，可算得

\[\begin{aligned} \int_0^{\frac{\pi}4}\ln(1+\tan x)dx =&\int_{0}^{\frac{\pi}4}\ln\left(1+\dfrac{1-\tan t}{1+\tan t}\right)dt \\ =&\int_{0}^{\frac{\pi}4}\ln\frac{2}{1+\tan t}dt \\ =&\dfrac{\pi}4\ln 2-\int_0^{\frac{\pi}4}\ln(1+\tan t)dt \end{aligned} \]

1. $\displaystyle\int_0^{\frac{\pi}4}\ln(1+\tan x)dx$

$g(x)=\ln(1+\tan x)$，则有

\[\begin{aligned} g\left(\dfrac{\pi}4-x\right) =&\ln\left(1+\tan(\dfrac{\pi}4-x)\right) =\ln\left(1+\dfrac{1-\tan x}{1+\tan x}\right) \\ =&\ln2-\ln(1+\tan x) =\ln2-g(x) \end{aligned} \]

$g\left(\dfrac{\pi}4-x\right)+g(x)=\ln 2$，则有

\[\int_0^{\frac{\pi}4}g(x)dx =\dfrac{\ln 2}2\int_0^{\frac{\pi}4}dx =\dfrac{\pi}8\ln 2 \]

定理 4.
$f(x),g(x)$在$[\frac1a,a]$上连续，且

  1. $f(x)=f\left(\dfrac1x\right)$；
  2. $g(x)+g\left(\dfrac1x\right)=A$，$A$为常数

则有

\[\begin{aligned} \int_{\frac1a}^a\dfrac{f(x)g(x)}xdx =&\dfrac{A}2\int_{\frac1a}^a\dfrac{f(x)}x dx \\ =&A\int_1^a\dfrac{f(x)}xdx =A\int_{\frac1a}^1\dfrac{f(x)}xdx \end{aligned} \]

证： 令$x=\dfrac1t$，则

分部积分

定理 5. (分部积分)
 设$u(x),v(x)$在$[a,b]$上具有连续的一阶导数$u'(x), v'(x)$，则有

\[\int_a^b u(x)v'(x)dx=u(x)v(x)\big|_a^b - \int_a^b u'(x)v(x)dx \]

或

\[\int_a^b u(x)d(v(x))=u(x)v(x)\big|_a^b - \int_a^b v(x)d(u(x)) \]

证： $ (u(x)v(x))'=u'(x)v(x)+u(x)v'(x)$是$[a,b]$上连续函数，则可积且有原函数。

\[\begin{aligned} \int_a^b (u(x)v(x))'dx =&u(x)v(x)|_a^b \\ =&\int_a^b u(x)v'(x)dx+\int_a^b u'(x)v(x)dx \end{aligned} \]

移项后即得证。

例 14. [例4.5.6] $\displaystyle\int_0^{\sqrt 3}x\arctan(x)dx$

例 15. $\displaystyle\int_0^3\arcsin\sqrt{\dfrac{x}{1+x}}dx$

8.

\[\begin{aligned} &\int_0^{\sqrt 3}x\arctan(x)dx =\int_0^{\sqrt 3}\arctan(x)d(\dfrac{x^2}2) \\ =&\dfrac{x^2}2\arctan(x)\bigg|_0^{\sqrt 3} -\int_0^{\sqrt 3}\dfrac{x^2}2d(\arctan(x)) \\ =&\dfrac{3}2\arctan{\sqrt 3}-\dfrac12\int_0^{\sqrt 3}\dfrac{x^2}{1+x^2}dx \\ =&\dfrac32\arctan{\sqrt 3} -\dfrac12\int_0^{\sqrt 3}\left(1-\dfrac1{1+x^2}\right)dx \\ =&\arctan{\sqrt 3}-\dfrac{\sqrt 3}2 =\dfrac23\pi-\dfrac{\sqrt 3}2 \end{aligned} \]

10. 令$t=\sqrt{\dfrac{x}{1+x}}$，则$x=\dfrac{t^2}{1-t^2}$，

\[\begin{aligned} &\int_0^3\arcsin\sqrt{\dfrac{x}{1+x}}dx =\int_0^{\frac{\sqrt 3}2}\arcsin(t)d\left(\dfrac{t^2}{1-t^2}\right) \\ =&\dfrac{t^2}{1-t^2}\arcsin(t)\bigg|_0^{\frac{\sqrt 3}2} -\int_0^{\frac{\sqrt 3}2}\dfrac{t^2}{1-t^2}d(\arcsin(t)) \\ =&3\dfrac{\pi}3-\int_0^{\frac{\sqrt 3}2}\dfrac{t^2}{(1-t^2)^{\frac32}}dt \end{aligned} \]

令$t=\sin u$，有

\[\begin{aligned} =&\pi-\int_0^{\frac{\pi}3}\dfrac{\sin^2u}{\cos^2u}du \\ =&\pi-\int_0^{\frac{\pi}3}(\dfrac{1}{\cos^2u}-1)du \\ =&\dfrac{4\pi}3-\sqrt 3 \end{aligned} \]

递推公式

例 16. $\displaystyle\int_0^{\frac{\pi}2}\sin^m(x)dx$ (例：4.5.7)

例 17. $\displaystyle\int_0^1x^k\ln^mxdx , k>0, m\in\mathbb N$ (例4.5.8)

例 18. $\displaystyle\lim_{n\to+\infty}(\dfrac{(2n)!!}{(2n-1)!!})^2\dfrac1{2n+1}$ (Wallis公式)

15. $J_m=\int_0^{\frac{\pi}2}\sin^m(x)dx$

\[\begin{aligned} J_m=\int_0^{\frac{\pi}2}\sin^{m-1}(x)d(-\cos x) \\ =-\sin^{m-1}(x)\cos x|_0^{\frac{\pi}2}+(m-1)\int_0^{\frac{\pi}2}\sin^{m-2}x\cos^2xdx \\ =(m-1)J_{m-2}-(m-1)J_m \end{aligned} \]
\[J_m=\dfrac{m-1}m J_{m-2} \]

16. $\displaystyle\int_0^1x^k\ln^mxdx , k>0, m\in\mathbb N$

\[\begin{aligned} H_m=\int_0^1x^k\ln^mxdx \\ =\dfrac1{k+1}x^{k+1}\ln^mx|_{0+}^1-\dfrac{m}{k+1}\int_{0}^1x^k\ln^{m-1}xdx \end{aligned} \]

则有

\[H_m=-\dfrac{m}{k+1}H_{m-1} \]

17.

\[\int_0^{\frac{\pi}2}\sin^{2n+1}xdx<\int_0^{\frac{\pi}2}\sin^{2n}xdx<\int_0^{\frac{\pi}2}\sin^{2n-1}xdx \]

则有

\[\dfrac{(2n)!!}{(2n+1)!!}<\dfrac{(2n-1)!!}{(2n)!!}\dfrac{\pi}2<\dfrac{(2n-2)!!}{(2n-1)!!} \]

即

\[(\dfrac{(2n)!!}{(2n-1)!!})^2\dfrac{1}{2n+1}<\dfrac{\pi}2<(\dfrac{(2n)!!}{(2n-1)!!})^2\dfrac1{2n} \]

式子两边表达式的差为

\[\begin{aligned} (\dfrac{(2n)!!}{(2n-1)!!})^2\dfrac{1}{2n+1}-(\dfrac{(2n)!!}{(2n-1)!!})^2\dfrac1{2n} \\ =(\dfrac{(2n)!!}{(2n-1)!!})^2\dfrac1{2n(2n+1)}<\dfrac1{2n}\dfrac{\pi}2 \end{aligned} \]
\[\lim_{n\to+\infty}(\dfrac{(2n)!!}{(2n-1)!!})^2\dfrac1{2n+1}=\dfrac{\pi}2 \]

例 19. $f(x)$连续，$f(0)\neq 0$，求

\[\lim_{x\to0}\dfrac{\int_0^x(x-t)f(t)dt}{x\int_0^xf(x-t)dt} \]

例 20. $f(x)$连续，$f'(0)$存在，求

\[\lim_{h\to0}\dfrac1{h^2}\int_0^h(f(x+h)-f(x-h))dx \]

例 21. 计算$\displaystyle\int_0^1\left(\int_x^1\arctan(t^2)dt\right)dx$

15. 用洛必达法则。

先求$\displaystyle\left(\int_0^xf(x-t)dt\right)'=?$。 令$u=x-t$

\[\begin{aligned} \int_0^xf(x-t)dt =&\int_x^0f(u)d(x-u) \\ =&-\int_x^0f(u)du=\int_0^xf(u)du \end{aligned} \]

有 $\displaystyle\left(\int_0^xf(x-t)dt\right)'=f(x)$

\[\lim_{x\to0}\dfrac{\int_0^x(x-t)f(t)dt}{x\int_0^xf(x-t)dt} =\lim_{x\to0}\dfrac{x\int_0^xf(t)dt-\int_0^xtf(t)dt}{x\int_0^xf(x-t)dt} \]

由洛必达法则，有

\[\begin{aligned} =&\lim_{x\to0}\dfrac{\int_0^xf(t)dt+xf(x)-xf(x)}{\int_0^xf(u)du+xf(x)} \\ =&\lim_{x\to0}\dfrac{\int_0^xf(t)dt}{\int_0^xf(u)du+xf(x)} \end{aligned} \]

(洛必达法则不能用)

\[\begin{aligned} =&\lim_{x\to0}\dfrac{xf(\xi)}{xf(x)+xf(\xi)} , \xi\in(0,x) \\ =&\dfrac{f(0)}{2f(0)}=\dfrac12 \end{aligned} \]

或

\[\begin{aligned} \lim_{x\to0}\dfrac{\int_0^xf(t)dt}x =&\lim_{x\to0}\dfrac{\int_0^xf(t)dt-0}{x-0} \\ =&\left(\int_0^xf(t)dt\right)'|_{x=0}=f(0) \end{aligned} \]
\[\begin{aligned} \lim_{x\to0}\dfrac{\int_0^xf(t)dt}{\int_0^xf(u)du+xf(x)} =&\lim_{x\to0}\dfrac{\dfrac1x\int_0^xf(t)dt}{\dfrac1x\int_0^xf(u)du+f(x)} \\ =&\dfrac{f(0)}{f(0)+f(0)}=\dfrac12 \end{aligned} \]

16. 洛必达法则，

\[\begin{aligned} &\lim_{h\to0}\dfrac1{h^2}\int_0^h(f(x+h)-f(x-h))dx \\ =&\lim_{h\to0}\dfrac{\int_h^{2h}f(u)du-\int_{-h}^0f(t)dt}{h^2} \\ =&\lim_{h\to0}\dfrac{2f(2h)-f(h)-f(-h)}{2h} \end{aligned} \]

不能用洛必达法则（为什么呢？）

\[\begin{aligned} =\lim_{h\to0}\bigg(\dfrac{2(f(0)+2hf'(0)+o(h))}{h^2} \\ -\dfrac{(f(0)+hf'(0)+o(h))}{h^2} \\ -\dfrac{(f(0)-hf'(0)+o(h))}{h^2}\bigg) \\ =\lim_{h\to0}\dfrac{4hf'(0)+o(h)}{2h}=2f'(0) \end{aligned} \]

18. 记$F(x)=\displaystyle\int_x^1\arctan(t^2)dt$，则

\[\begin{aligned} I=&\int_0^1F(x)dx=F(x)x|_0^1-\int_0^1xd(F(x)) \\ =&x\int_x^1\arctan(t^2)dt\bigg|_0^1 -\int_0^1x(-\arctan(x^2))dx \\ =&\int_0^1\arctan(x^2)d(\dfrac{x^2}2) \end{aligned} \]
\[\begin{aligned} =&\left.\dfrac{x^2}2\arctan(x^2)\right|_0^1-\int_0^1\dfrac{x^2}2\dfrac{1}{1+x^4}2xdx \\ =&\dfrac{\pi}8-\int_0^1\dfrac{1}4\dfrac1{1+x^4}d(x^4+1) \\ =&\dfrac{\pi}8-\dfrac{\ln 2}4 \end{aligned} \]

求$\displaystyle\int_a^b\left(f(x)\int_c^xg(y)dy\right)dx$型的积分

若$F'(x)=f(x)$，则

\[\begin{aligned} \int_a^b\left(f(x)\int_c^xg(y)dy\right)dx =&\int_a^b\left(\int_c^xg(y)dy\right)d(F(x)) \\ =&\left.F(x)\int_c^xg(y)dy\right|_a^b-\int_a^bF(x)g(x)dx \end{aligned} \]

问题. 求$\displaystyle\int_0^1x^2\left(\int_1^xe^{-y^2}dy\right)dx=? $

目录

本节读完

例 22. 谢谢

22.

例 23. $\displaystyle\int_0^a\sqrt{a^2-x^2}dx, a>0$

取$x=a\sin t$，$dx=a\cos t dt$，

\[\begin{aligned} x=0, \Rightarrow t=0 \\ x=a, \Rightarrow t=\dfrac{\pi}2 \end{aligned} \]
\[\begin{aligned} &\int_0^a\sqrt{a^2-x^2}dx=\int_0^{\frac{\pi}2}a\cos t a \cos t dt \\ =&\int_0^{\frac{\pi}2}a^2\cos^2tdt =\int_0^{\frac{\pi}2}a^2(\dfrac{1+\cos2t}{2}dt \\ =&\left.\dfrac{a^2}2(t+\dfrac12\sin(2t))\right|_0^{\frac{\pi}2} =\dfrac14\pi a^2 \end{aligned} \]

当$f(x)$在$[0,1]$上连续，则有

\[\int_0^{\pi}xf(\sin x)dx=\dfrac{\pi}2\int_0^{\pi}f(\sin x)dx \]

例 24. $\displaystyle\int_{0}^{\pi}\dfrac{x\sin x}{1+\cos^2x}dx$

例 25. $\displaystyle\int_0^{\frac{\pi}4}\dfrac{x}{\cos^2x+\sin x\cos x}dx$

取$g(x)=x$，$F(x)=f(\sin x)$。由前面的定理可证。

6. $\displaystyle\int_{0}^{\pi}\dfrac{x\sin x}{1+\cos^2x}dx$

令$g(x)=x$，$f(x)=\dfrac{\sin x}{1+\cos^2x}$，则

\[\begin{aligned} g(\pi-x)+g(x)=\pi \\ f(\pi-x)=f(x) \end{aligned} \]

所以，

\[\begin{aligned} =\dfrac{\pi}2\int_0^{\pi}f(x)dx =\dfrac{\pi}2\int_0^{\pi}\dfrac{\sin x}{1+\cos^2x}dx \\ =-\dfrac{\pi}2\int_0^{\pi}\dfrac{1}{1+\cos^2x}d(\cos x)=\dfrac{\pi^2}4 \end{aligned} \]

7. $\displaystyle\int_0^{\frac{\pi}4}\dfrac{x}{\cos^2x+\sin x\cos x}dx$

\[f(x)=\dfrac{1}{\cos^2x+\sin x\cos x}=\dfrac{2}{\cos(2x)+1+\sin(2x)} \]
\[\begin{aligned} f(\dfrac{\pi}4-x)=\dfrac{2}{\cos(\frac{\pi}2-2x)+1+\sin(\frac{\pi}2-2x)} \\ =\dfrac{2}{\cos(2x)+1+\sin(2x)}=f(x) \end{aligned} \]

$g(x)=x$，则$g(\dfrac{\pi}4-x)+g(x)=\dfrac{\pi}4$

\[\int_0^{\frac{\pi}4}f(x)g(x)dx=\dfrac{\pi}8\int_0^{\frac{\pi}4}f(x)dx \]
\[=\dfrac{\pi}8\int_0^{\frac{\pi}4}\dfrac{\sec^2x}{1+\tan x}dx =\dfrac{\pi}8\ln(1+\tan x)|_0^{\frac{\pi}4} =\dfrac{\pi}8\ln2 \]

例 26. $f(x)$在$[0,1]$上连续， $\displaystyle \int_0^{\frac{\pi}2}f(|\cos(x)|)dx=2$， 求 $\displaystyle\int_0^{2\pi}f(|\cos(x)|)dx $

例 27. 求$\displaystyle\int_0^1f(x)dx$，若

\[f(x)=\displaystyle\int_x^{x+2\pi}(1+e^{\sin t}-e^{-\sin t})dt+\dfrac1{1+x}\int_0^1f(t)dt \]

6. $\displaystyle\int_0^{2\pi}f(|\cos(x)|)dx$

$=2\int_0^{\pi}f(|\cos x|)dx$ (以$\pi$为周期)

$=2\int_{-\frac{\pi}2}^{\frac{\pi}2}f(|\cos x|)dx$ (一整周期)

$=4\int_{0}^{\frac{\pi}2}f(|\cos x|)dx$ (偶函数)

$=4\times2=8$

7. 函数$e^{\sin t}-e^{-\sin t}$以$2\pi$为周期，且为奇函数

$\displaystyle\int_x^{x+2\pi}(e^{\sin t}-e^{-\sin t})dt=\displaystyle\int_0^{2\pi}(e^{\sin t}-e^{-\sin t})dt=0$

\[f(x)=2\pi+\dfrac1{1+x}\int_0^1f(t)dt \]
\[\int_0^1f(x)dx=2\pi+\int_0^1\dfrac{1}{1+x}dx\int_0^1f(t)dt \]
\[=2\pi+\ln2\int_0^1f(t)dt \]
\[\int_0^1f(x)dx=\dfrac{2\pi}{1-\ln 2} \]

分段

例 28. $\displaystyle f(x)=\begin{cases}x, 0\leq x\leq 1 \\ 2-x, 1<x\leq 2\end{cases}$， 求

\[\displaystyle\int_{2n}^{2n+2}f(x-2n)e^{-x}dx \]

绝对值

例 29. $\displaystyle\int_{-2}^2|x^2+2|x|-3|dx$

13. 令$t=x-2n$，则

\[\begin{aligned} \int_{2n}^{2n+2}f(x-2n)e^{-x}dx=\int_0^2f(t)e^{-t-2n}dt \\ =e^{-2n}(\int_0^1+\int_1^2)f(t)e^{-t}dt \\ =e^{-2n}(\int_0^1te^{-t}dt+\int_1^2(2-t)e^{-t}dt) \\ =e^{-2n}\dfrac{(e-1)^2}{e^2} \end{aligned} \]

14. $\int_{-2}^2|x^2+2|x|-3|dx$

\[\begin{aligned} =\int_{-2}^0|x^2-2x-3|dx+\int_0^3|x^2+2x-3|dx \\ =\int_{-2}^{-1}(x^2-2x-3)dx+\int_{-1}^0-(x^2-2x-3)dx \\ +\int_0^1-(x^2+2x-3)dx+\int_1^3(x^2+2x-3)dx \\ =\dfrac{49}3 \end{aligned} \]

例 30. 证明$\displaystyle\lim_{n\to+\infty}\int_0^1 e^{x^2}\cos(nx)dx=0$

17.

\[\begin{aligned} \int_0^1e^{x^2}\cos(nx)dx =&\dfrac1n\int_0^1e^{x^2}d(\sin(nx)) \\ =&\dfrac1ne^{x^2}\sin(nx)|_0^1-\dfrac1n\int_0^1\sin(nx)e^{x^2}2xdx \\ =&\dfrac{e\sin(n)}{n}-\dfrac1n\int_0^12xe^{x^2}\sin(nx)dx \end{aligned} \]

所以有

\[\begin{aligned} |\int_0^1e^{x^2}\cos(nx)dx| \leq\dfrac{e}{n}+\dfrac1n\int_0^1|2xe^{x^2}\sin(nx)|dx \\ \leq\dfrac{e}{n}+\dfrac1n\int_0^12edx =\dfrac{e}{n}+\dfrac{2e}{n}=\dfrac{3e}n \end{aligned} \]

即

\[\lim_{n\to+\infty}\int_0^1 e^{x^2}\cos(nx)dx=0 \]

例 31. $f(x)$以$T$为周期的函数，且在$[0,T]$上可积。则

\[\lim_{x\to+\infty}\dfrac1x\int_0^xf(t)dt=\dfrac1T\int_0^Tf(x)dx \]

例 32. 求$\displaystyle\lim_{x\to+\infty}\dfrac1x\int_0^x|\sin t|dt$

例 33. $f(x)$连续，且$\displaystyle\lim_{x\to0}\dfrac{f(x)}x=2$，$\phi(x)=\displaystyle\int_0^1f(xt)dt$，求$\phi'(x)$

19. $\forall x>T$，$\exists n$，满足$x=nT+x', x'\in[0,T)$

\[\begin{aligned} \dfrac1x\int_0^xf(t)dt=\dfrac1{nT+x'}\int_0^{nT+x'}f(t)dt \\ =\dfrac1{nT+x'}\int_0^{nT}f(t)dt+\dfrac1{nT+x'}\int_{nT}^{nT+x'}f(t)dt \\ =\dfrac{n}{nT+x'}\int_0^Tf(t)dt+\dfrac{1}{nT+x'}\int_0^{x'}f(t)dt \end{aligned} \]

由

\[|\int_0^{x'}f(t)dt|\leq\int_0^{x'}|f(t)|dt\leq\int_0^T|f(t)|dt \]

有界。取极限即得结论

20.

\[\begin{aligned} \lim_{x\to+\infty}\dfrac1x\int_0^x|\sin t|dt =\dfrac1{\pi}\int_0^{\pi}|\sin t|dt \\ =\dfrac1{\pi}\int_0^{\pi}\sin(t)dt =\dfrac2{\pi} \end{aligned} \]

21. 令$xt=u$，则

\[\phi(x)=\int_0^1f(xt)dt=\int_0^xf(u)\dfrac1xdu=\dfrac1x\int_0^xf(u)du \]
  • $x\neq0$时，有
\[\phi'(x)=\dfrac{1}{x^2}(f(x)x-\int_0^xf(u)du)=\dfrac{f(x)}x-\dfrac1{x^2}\int_0^xf(u)du \]
  • $x=0$时，
\[\begin{aligned} \phi'(0)=\lim_{x\to0}\dfrac{\phi(x)-\phi(0)}{x-0} \\ =\lim_{x\to0} \end{aligned} \]

例 34. 设$f(x)$在$[0,2]$连续，且$\displaystyle\int_0^2f(x)dx=0$，则$\exists\xi\in(0,2)$，满足

\[f(2-\xi)+f(\xi)=0 \]

例 35. $f(x)$在$[0,1]$上连续，$\displaystyle\int_0^1f(x)dx=\displaystyle\int_0^1xf(x)dx$，则$\exists\xi\in(0,1)$，满足

\[\int_0^{\xi}f(x)dx=0 \]

22. （把f(x)当作是函数$\int f(t)dt$的导函数）

\[\begin{aligned} F(x)=\int_0^xf(2-t)dt+\int_0^xf(t)dt \\ =\int_2^{2-x}f(u)d(2-u)+\int_0^xf(t)dt \\ =\int_{2-x}^2f(u)du+\int_0^xf(t)dt \\ \end{aligned} \]

显然$F(x)$可导，且

\[\begin{aligned} F(0)=\int_2^2f(u)du+\int_0^0f(t)dt=0 \\ F(2)=\int_0^2f(u)du+\int_0^2f(t)dt=0 \end{aligned} \]

则，$\exists\xi\in(0,2)$，满足

\[F'(\xi)=f(2-\xi)+f(\xi)=0 \]

23. 取

\[G(x)=x\int_0^xf(t)dt-\int_0^xtf(t)dt \]

则$G(x)$可导，且$G(0)=0, G(x)=0$

例 36. $f(x)$在$[0,\pi]$上连续，$\displaystyle\int_0^{\pi}f(x)dx=0$，$\displaystyle\int_0^{\pi}f(x)\cos xdx=0$，则$\exists\xi_1\neq\xi_2\in(0,\pi)$，满足

\[f(\xi_1)=f(\xi_2)=0 \]

例 37. $f(x)$在$[0,1]$上有二阶连续函数，$f'(0)=f'(1)=0$，证明： $\exists\xi\in(0,1)$，满足

\[\int_0^1f(x)dx=\dfrac12(f(0)+f(1))+\dfrac1{24}f''(\xi) \]

24.

\[F(x)=\int_0^xf(t)dt, x\in[0,\pi] \]

则$F(0)=0, F(\pi)=\int_0^{\pi}f(x)dx=0$。又

\[\begin{aligned} 0=\int_0^{\pi}f(x)\cos xdx=\int_0^{\pi}\cos xd(F(x)) \\ =F(x)\cos x|_0^{\pi}+\int_0^{\pi}F(x)\sin(x)dx \\ =\int_0^{\pi}F(x)\sin(x)dx =\pi F(\xi)\sin\xi , \xi\in(0,\pi) \end{aligned} \]

由$\sin(\xi)\neq0,\xi\in(0,\pi)$知，$F(\xi), \xi\in(0,\pi)$。

这样，在$(0,\xi)$与$(\xi,\pi)$间，各有一个导数为$0$的点。

25. $f(x)$有二阶连续导数，做Taylor展开

\[\begin{aligned} f(x)=f(0)+f'(0)x+\dfrac{f''(\xi_1)}2 x^2 , \xi_1\in(0,x) \\ f(x)=f(1)+f'(1)(x-1)+\dfrac{f''(\xi_2)}2 (x-1)^2 , \xi_2\in(0,x) \\ \end{aligned} \]

则

\[\begin{aligned} \int_0^1f(x)dx=(\int_0^{\frac12}+\int_{\frac12}^1)f(x)dx \\ =\dfrac12 f(0)+\dfrac12\int_0^{\frac12}f''(\xi_1)x^2dx \\ +\dfrac12 f(1)+\dfrac12\int_{\frac12}^1f''(\xi_2)(x-1)^2dx \\ \end{aligned} \]
\[\begin{aligned} =&\dfrac12(f(0)+f(1)) \\ &+\dfrac12(\int_0^{\frac12}f''(\xi_1)x^2dx+\int_{\frac12}^1f''(\xi_2)(x-1)^2dx) \end{aligned} \]

$f''(x)$连续，则有界。设$m\leq f''\leq M$，则

\[\begin{aligned} \int_0^{\frac12}f''(\xi_1)x^2dx+\int_{\frac12}^1f''(\xi_2)(x-1)^2dx \\ \leq M(\int_0^{\frac12}x^2dx+\int_{\frac12}^1(x-1)^2dx)=\dfrac{M}{12} \end{aligned} \]

同样可得

\[\geq m(\int_0^{\frac12}x^2dx+\int_{\frac12}^1(x-1)^2dx)=\dfrac{m}{12} \]

由连续函数的介值性知，$\exists\xi\in(0,1)$，满足

\[\dfrac{f''(\xi)}{12}=\int_0^{\frac12}x^2dx+\int_{\frac12}^1(x-1)^2dx \]

例 38. $f(x)$在$[a,b]$上连续，且$f''(x)\geq0, x\in(a,b)$，则有

\[\int_a^bf(x)dx\leq\dfrac{f(a)+f(b)}2(b-a) \]

26.

\[\begin{aligned} f(a)=f(x)+f'(x)(a-x)+\dfrac{f''(\xi)}2(a-x)^2 \\ \geq f(x)+f'(x)(a-x) \\ f(b)=f(x)+f'(x)(b-x)+\dfrac{f''(\xi)}2(b-x)^2 \\ \geq f(x)+f'(x)(b-x) \\ \end{aligned} \]

所以

\[f(a)+f(b)\geq 2f(x)+f'(x)(a-x)+f'(x)(b-x) \]
\[\begin{aligned} \int_a^b(f(a)+f(b))dx\geq2\int_a^bf(x)dx \\ +\int_a^bf'(x)(a+b-2x)dx \\ =2\int_a^bf(x)dx+f(x)(a+b-2x)|_a^b \\ -\int_a^bf(x)d(a+b-2x) \\ =4\int_a^bf(x)dx+(a-b)f(b)-(b-a)f(a) \end{aligned} \]

移项后，即可得结论

例 39. $\displaystyle\int_{-1}^1\dfrac{x}{\sqrt{5-4x}}dx$

9. $\displaystyle\int_{-1}^1\dfrac{x}{\sqrt{5-4x}}dx=\dfrac14\int_{-1}^1\dfrac{4x-5+5}{\sqrt{5-4x}}dx$

\[\begin{aligned} =&-\dfrac14\int_{-1}^1\sqrt{5-4x}dx+\dfrac54\int_{-1}^1\dfrac1{\sqrt{5-4x}}dx \\ =&\dfrac1{24}(5-4x)^{\frac32}\bigg|_{-1}^1-\dfrac58\sqrt{5-4x}\bigg|_{-1}^1 =\dfrac16 \end{aligned} \]

9. $\displaystyle\int_{-1}^1\dfrac{x}{\sqrt{5-4x}}dx$，令$t=\sqrt{5-4x}$，$dx=-\frac12 tdt$

\[\begin{aligned} =\int_{3}^1\dfrac{\frac14(5-t^2)}{t}(-\dfrac12t)dt =\dfrac18\int_1^3(5-t^2)dt \\ =\dfrac18(5t-\dfrac{t^3}3)|_1^3 =\dfrac16 \end{aligned} \]