这一讲的位置

上一讲证明了连续函数可积，但积分还只是一个定义：要算 $\int_0^1 x^2\,dx$，得回到上下和。这一讲补上两件事，让积分真正可用：

1. 代数性质：数乘、加法、单调、区间可加，以及由它们推出的 $|\int f|\le\int|f|$。
2. 微积分基本定理：积分与导数互为逆运算。第一版把变上限积分求导，第二版用原函数算积分。

最后把积分推广到两类「不合法」的情形：区间无界、函数无界，也就是反常积分。课程的收尾是弧长公式——积分第一次被用来定义一个几何量。

时间轴

积分的代数性质

设 $f,g$ 在 $[a,b]$ 上 Riemann 可积。

1. 数乘：$cf$ 可积，且 $\int_a^b cf=c\int_a^b f$。
2. 加法：$f+g$ 可积，且 $\int_a^b(f+g)=\int_a^b f+\int_a^b g$。
3. 单调性：若 $f\le g$，则 $\int_a^b f\le\int_a^b g$。
4. 区间可加：若 $c\in[a,b]$，则 $f$ 在 $[a,c]$、$[c,b]$ 上都可积，且

$$\int_a^b f=\int_a^c f+\int_c^b f .$$

单调性的直观版本：$f\le g$ 意味着在每个子区间上 $\sup f\le\sup g$、$\inf f\le\inf g$，于是逐点比较上下和，再取确界即可。

证明加法

用可积判据：给定 $\varepsilon>0$，要找一个分割使 $U(f+g,P)-L(f+g,P)<\varepsilon$。

因为 $f$ 可积，存在分割 $P_1$ 使

$$U(f,P_1)-L(f,P_1)<\frac{\varepsilon}{2};$$

因为 $g$ 可积，存在分割 $P_2$ 使

$$U(g,P_2)-L(g,P_2)<\frac{\varepsilon}{2}.$$

取 $P$ 为两者分点的并，它是 $P_1,P_2$ 的公共加细。于是

$$U(f,P)-L(f,P)<\frac{\varepsilon}{2}, \qquad U(g,P)-L(g,P)<\frac{\varepsilon}{2}.$$

关键的一步是两个函数的和的上下确界与各自上下确界的关系。在任一子区间上，

$$\sup(f+g)\le\sup f+\sup g, \qquad \inf(f+g)\ge\inf f+\inf g .$$

这两个不等式通常不是等式：若 $f$ 与 $g$ 一正一负、彼此抵消，和的振荡可以比各自振荡之和小得多。但不等式方向足够。

于是对同一个 $P$，

$$\begin{aligned} U(f+g,P)-L(f+g,P) &=\sum_i\Big(\sup(f+g)-\inf(f+g)\Big)\Delta x_i\\ &\le\sum_i\Big[(\sup f-\inf f)+(\sup g-\inf g)\Big]\Delta x_i\\ &=\big(U(f,P)-L(f,P)\big)+\big(U(g,P)-L(g,P)\big)\\ &<\varepsilon . \end{aligned}$$

由判据，$f+g$ 可积。

积分值的等式要另外说明。对任意分割 $P$，

$$L(f,P)+L(g,P)\le L(f+g,P)\le U(f+g,P)\le U(f,P)+U(g,P).$$

对 $P$ 取上确界与下确界，两端分别给出 $\int f+\int g$，中间被夹住，所以

$$\int_a^b(f+g)=\int_a^b f+\int_a^b g .$$

课件的加法证明与这里的顺序相同；这里补上了等式的夹逼。

关于记号

课程顺带解释了符号：$\int$ 是一个拉长的 $S$，表示求和；$dx$ 与 $\Delta x$ 同源，可以理解为把 $\Delta x_i$ 取极限后的写法。所以

$$\int_a^b f(x)\,dx$$

整体读作「把 $f(x_i)\Delta x_i$ 加起来，再让分割变细」。当然，严格定义仍然是上下确界，不是某个极限。

区间可加

命题。 若 $f$ 在 $[a,b]$ 上可积，$c\in[a,b]$，则 $f$ 在 $[a,c]$ 与 $[c,b]$ 上都可积。

证明。 给定 $\varepsilon>0$。由 $f$ 在 $[a,b]$ 上可积，存在分割 $P$ 使

$$U(f,P)-L(f,P)<\varepsilon .$$

若 $c$ 不是 $P$ 的分点，把它加进去，得到加细 $P^*$。加细只会让差变小，所以仍有

$$U(f,P^*)-L(f,P^*)<\varepsilon .$$

现在 $P^*$ 在 $[a,c]$ 上的限制 $P^*|_{[a,c]}$ 是 $[a,c]$ 的一个分割，在 $[c,b]$ 上同理。上下和在分割点处可以拆开：

$$U(f,P^*)-L(f,P^*) = \Big(U(f,P^*|_{[a,c]})-L(f,P^*|_{[a,c]})\Big) + \Big(U(f,P^*|_{[c,b]})-L(f,P^*|_{[c,b]})\Big).$$

右边两项都非负，而总和小于 $\varepsilon$，所以每一项都小于 $\varepsilon$。对 $[a,c]$ 用判据，$f$ 在此可积；对 $[c,b]$ 同理。

积分的等式由同一拆分得到：把每个上下和都拆成两段，取确界后即得

$$\int_a^b f=\int_a^c f+\int_c^b f .$$

$\square$

这里是「加细」的第二次关键用法：先把 $c$ 塞进分割，才能把和干净地拆成两段。

绝对值不等式

推论。 若 $f$ 与 $|f|$ 在 $[a,b]$ 上可积，则

$$\left|\int_a^b f\right|\le\int_a^b|f| .$$

证明。 对每个 $x$ 都有

$$f(x)\le|f(x)|, \qquad -f(x)\le|f(x)| .$$

由单调性，

$$\int_a^b f\le\int_a^b|f|, \qquad \int_a^b(-f)\le\int_a^b|f| .$$

第二式左边用数乘性质写成 $-\int_a^b f$。于是

$$-\int_a^b|f|\le\int_a^b f\le\int_a^b|f|,$$

即 $\left|\int_a^b f\right|\le\int_a^b|f|$。$\square$

这个不等式在估计积分时几乎每次都会用到：它把「有正有负的代数和」换成「绝对值的和」，代价是失去抵消。

微积分基本定理（第一版）

定理。 设 $f$ 在 $[a,b]$ 上连续，定义

$$F(x)=\int_a^x f(s)\,ds, \qquad x\in[a,b].$$

则 $F$ 在 $(a,b)$ 上可导，且

$$F'(x)=f(x).$$

证明。 固定 $x_0$，先看 $x>x_0$。由区间可加，

$$F(x)-F(x_0) =\int_a^x f-\int_a^{x_0} f =\int_{x_0}^x f .$$

在 $[x_0,x]$ 上，$f$ 连续，所以有最大值 $M$ 与最小值 $m$（极值定理）。于是

$$m(x-x_0)\le\int_{x_0}^x f\le M(x-x_0).$$

除以 $x-x_0>0$：

$$m\le\frac{F(x)-F(x_0)}{x-x_0}\le M .$$

当 $x\to x_0^+$ 时，区间 $[x_0,x]$ 收缩到 $x_0$，由连续性 $M\to f(x_0)$、$m\to f(x_0)$。夹逼给出

$$\lim_{x\to x_0^+}\frac{F(x)-F(x_0)}{x-x_0}=f(x_0).$$

从左侧逼近时，$x-x_0<0$，不等式两边同除以负数要反向，但同样的夹逼给出左极限也等于 $f(x_0)$。两侧一致，故 $F'(x_0)=f(x_0)$。$\square$

证明的结构是夹逼：积分被两个矩形夹住，商被 $\inf$ 与 $\sup$ 夹住，而这两个数都趋于 $f(x_0)$。注意连续性只用在一处：保证 $M,m\to f(x_0)$。

微积分基本定理（第二版）

定理。 设 $F$ 在 $[a,b]$ 上可导，$f=F'$ 在 $[a,b]$ 上 Riemann 可积，则

$$\int_a^b f(x)\,dx=F(b)-F(a).$$

课程把证明留在讲义里，这里补出思路——它只用中值定理与可积判据。

证明。 取任意分割 $P=\{a=x_0<\cdots<x_n=b\}$。在每个子区间上用中值定理：存在 $c_i\in(x_{i-1},x_i)$，使

$$F(x_i)-F(x_{i-1})=F'(c_i)(x_i-x_{i-1})=f(c_i)\Delta x_i .$$

把 $n$ 个等式相加，左边望远镜式相消：

$$F(b)-F(a)=\sum_{i=1}^n f(c_i)\Delta x_i .$$

右边的每个 $f(c_i)$ 都介于该子区间上的 $\inf$ 与 $\sup$ 之间，所以这个和介于 $L(f,P)$ 与 $U(f,P)$ 之间：

$$L(f,P)\le F(b)-F(a)\le U(f,P).$$

这对任意分割成立。$f$ 可积，所以取上下确界后仍有

$$\int_a^b f\le F(b)-F(a)\le\int_a^b f,$$

故 $F(b)-F(a)=\int_a^b f$。$\square$

证明中「$f$ 可积」这一条不能省：$F'$ 未必 Riemann 可积（导数可以有很坏的间断），那时等式可能不成立。

例子

$$\int_0^1 x^2\,dx .$$

取 $F(x)=x^3/3$，则 $F'=x^2$，且 $x^2$ 连续因而可积。由第二版，

$$\int_0^1 x^2\,dx=F(1)-F(0)=\frac13-0=\frac13 .$$

用上下和要算很久的量，现在一步就出来了。这就是基本定理的意义：把积分从「求极限」变成「找原函数」。

反常积分

定积分要求区间有界、函数有界。去掉其中一条，就要用极限补定义。

类型一：区间无界

设 $f$ 在 $[a,b]$ 上可积对每个 $b>a$ 成立。若极限

$$\lim_{b\to\infty}\int_a^b f(x)\,dx$$

存在，则定义

$$\int_a^\infty f(x)\,dx=\lim_{b\to\infty}\int_a^b f(x)\,dx,$$

并称该反常积分收敛；否则称发散。

例 1。

$$\int_1^\infty\frac{dx}{x^2}.$$

取 $F(x)=-1/x$，则 $F'=1/x^2$。对有限 $b$，

$$\int_1^b\frac{dx}{x^2}=F(b)-F(1)=-\frac1b+1 .$$

令 $b\to\infty$：$-1/b\to0$，所以

$$\int_1^\infty\frac{dx}{x^2}=1 .$$

曲线无限延伸，面积却是有限的。

例 2。

$$\int_1^\infty\frac{dx}{x}.$$

取 $F(x)=\log x$，则

$$\int_1^b\frac{dx}{x}=\log b-\log1=\log b\to\infty .$$

所以发散。两个函数都趋于 $0$，但 $1/x^2$ 掉得够快，$1/x$ 不够快——这是后面级数判别法会反复出现的分界。

类型二：函数无界

设 $f$ 在 $[c,b]$ 上可积对每个 $c>a$ 成立（$f$ 在 $a$ 附近可能无界甚至无定义）。若极限

$$\lim_{c\to a^+}\int_c^b f(x)\,dx$$

存在，则定义

$$\int_a^b f(x)\,dx=\lim_{c\to a^+}\int_c^b f(x)\,dx .$$

例 3。

$$\int_0^1\frac{dx}{\sqrt x}.$$

取 $F(x)=2\sqrt x$，则 $F'=1/\sqrt x$。对 $a>0$，

$$\int_a^1\frac{dx}{\sqrt x}=F(1)-F(a)=2-2\sqrt a .$$

令 $a\to0^+$：$\sqrt a\to0$，所以

$$\int_0^1\frac{dx}{\sqrt x}=2 .$$

函数在 $0$ 附近趋于无穷，面积仍然有限。

例 4。

$$\int_0^1\frac{dx}{x}.$$

取 $F(x)=\log x$，则对 $a>0$，

$$\int_a^1\frac{dx}{x}=\log1-\log a=-\log a\to+\infty .$$

发散。$1/\sqrt x$ 的奇点「温和」，$1/x$ 的奇点「太强」。

弧长

课程最后给出一个定义，不作展开。设平面曲线

$$\gamma(s)=\big(f(s),g(s)\big), \qquad s\in[a,b],$$

其中 $f,g$ 可导且导数连续。曲线的弧长定义为

$$L=\int_a^b\sqrt{f'(s)^2+g'(s)^2}\;ds .$$

直观来源：$s$ 增大一点点 $ds$，横坐标变化 $f'(s)\,ds$、纵坐标变化 $g'(s)\,ds$，由勾股定理，曲线段长近似为

$$\sqrt{(f'\,ds)^2+(g'\,ds)^2}=\sqrt{f'^2+g'^2}\;ds .$$

把所有这些小段「加起来」，就是上面的积分。这里仍然只是定义，下一讲会讨论它是否良定义、与参数化是否无关。

本节结论

• 积分满足数乘、加法、单调与区间可加；加法证明的关键是 $\sup(f+g)\le\sup f+\sup g$、$\inf(f+g)\ge\inf f+\inf g$。
• 区间可加要先把 $c$ 加进分割，再把上下和拆成两段。
• $\left|\int_a^b f\right|\le\int_a^b|f|$。
• 第一版基本定理：$F(x)=\int_a^x f$ 可导，$F'=f$，证明靠夹逼。
• 第二版基本定理：若 $F'=f$ 且 $f$ 可积，则 $\int_a^b f=F(b)-F(a)$，证明靠中值定理与望远镜求和。
• 反常积分用极限定义：区间无界或函数无界；$\int_1^\infty x^{-2}=1$、$\int_1^\infty x^{-1}$ 发散；$\int_0^1 x^{-1/2}=2$、$\int_0^1 x^{-1}$ 发散。
• 弧长定义为 $\int_a^b\sqrt{f'^2+g'^2}\,ds$。

理解检验

1. 证明 $f+g$ 可积且 $\int(f+g)=\int f+\int g$。为什么 $\sup(f+g)\le\sup f+\sup g$ 一般不是等式？

参考答案：给定 $\varepsilon>0$，取分割 $P$ 同时使 $U(f,P)-L(f,P)<\varepsilon/2$ 与 $U(g,P)-L(g,P)<\varepsilon/2$（先分别取分割，再取公共加细）。在每段上 $\sup(f+g)\le\sup f+\sup g$、$\inf(f+g)\ge\inf f+\inf g$，故 $U(f+g,P)-L(f+g,P)\le[U(f,P)-L(f,P)]+[U(g,P)-L(g,P)]<\varepsilon$，由判据可积。等式由 $L(f,P)+L(g,P)\le L(f+g,P)\le U(f+g,P)\le U(f,P)+U(g,P)$ 对 $P$ 取确界夹逼得到。不等式一般取不到等号：若 $g=-f$，则 $f+g\equiv0$，$\sup(f+g)=0$，而 $\sup f+\sup g$ 可能为正，因为 $f$ 在区间上可以既有正也有负。

2. 证明：$f$ 在 $[a,b]$ 上可积，则 $f$ 在 $[a,c]$ 上可积。加细在这里起什么作用？

参考答案：给定 $\varepsilon>0$，取分割 $P$ 使 $U(f,P)-L(f,P)<\varepsilon$。若 $c$ 不在 $P$ 中，把 $c$ 加入得到加细 $P^*$；加细使差变小，故 $U(f,P^*)-L(f,P^*)<\varepsilon$。此时 $P^*$ 限制在 $[a,c]$ 上是一个分割，且

$$U(f,P^*)-L(f,P^*)=\big(U_{[a,c]}-L_{[a,c]}\big)+\big(U_{[c,b]}-L_{[c,b]}\big).$$

两项非负而和小于 $\varepsilon$，故第一项小于 $\varepsilon$，由判据 $f$ 在 $[a,c]$ 可积。加细的作用是让 $c$ 成为一个分割点，从而上下和能干净地拆成两段；否则 $c$ 落在某个子区间内部，无法拆分。

3. 证明 $\left|\int_a^b f\right|\le\int_a^b|f|$。

参考答案：对每个 $x$ 有 $f(x)\le|f(x)|$ 与 $-f(x)\le|f(x)|$。由单调性，$\int f\le\int|f|$ 且 $\int(-f)\le\int|f|$。由数乘性质，$\int(-f)=-\int f$。于是 $-\int|f|\le\int f\le\int|f|$，即 $\left|\int f\right|\le\int|f|$。这里用到了 $|f|$ 也可积这一前提，否则右边没有意义。

4. 证明第一版基本定理，并指出连续性用在哪里。

参考答案：设 $F(x)=\int_a^x f$，固定 $x_0$，对 $x>x_0$ 有 $F(x)-F(x_0)=\int_{x_0}^x f$。在 $[x_0,x]$ 上 $f$ 连续，取到最小值 $m$ 与最大值 $M$，于是 $m(x-x_0)\le\int_{x_0}^x f\le M(x-x_0)$，除以 $x-x_0$ 得 $m\le\frac{F(x)-F(x_0)}{x-x_0}\le M$。令 $x\to x_0^+$，由连续性 $M,m\to f(x_0)$，夹逼得右导数等于 $f(x_0)$；左侧同理（注意除数为负时要反向）。连续性只用在这里：保证区间收缩时上下确界都收敛到 $f(x_0)$。若 $f$ 在该点不连续，$F$ 仍可导，但导数等于 $f$ 的「平均值的极限」，不一定等于 $f(x_0)$。

5. 用中值定理证明第二版基本定理，并说明为什么要求 $f=F'$ 可积。

参考答案：取任意分割 $P$，在每个子区间 $[x_{i-1},x_i]$ 上用中值定理，得 $c_i$ 使 $F(x_i)-F(x_{i-1})=f(c_i)\Delta x_i$。相加后左边望远镜相消为 $F(b)-F(a)$，右边是 $\sum f(c_i)\Delta x_i$，它介于 $L(f,P)$ 与 $U(f,P)$ 之间。对任意分割成立，故 $\int f\le F(b)-F(a)\le\int f$，两边相等。要求 $f=F'$ 可积，是因为只有可积时上下积分才相等，夹逼才有意义；若导数不可积，积分可能不存在，等式也就无从谈起。

6. 判断 $\int_1^\infty x^{-2}\,dx$ 与 $\int_1^\infty x^{-1}\,dx$ 的收敛性，并解释差异的来源。

参考答案：$\int_1^b x^{-2}\,dx=[-1/x]_1^b=1-1/b\to1$，收敛；$\int_1^b x^{-1}\,dx=[\log x]_1^b=\log b\to\infty$，发散。两者都趋于 $0$，差别在速度：$x^{-2}$ 的尾部面积按 $1/b$ 衰减，总和有限；$x^{-1}$ 的对数增长虽然极慢，却无界。用分割的话说，$x^{-1}$ 在每个远处的单位区间上贡献约 $1/k$，而调和级数 $\sum1/k$ 发散——积分与级数在这里给出同样的分界。

与前后课的关系

第一版基本定理说明「连续函数的积分是它的一个原函数」，第二版说明「原函数在两端点的差就是积分」。两版合起来，微积分从「用矩形逼近」变成了「求导与求值的互逆运算」。

下一讲转向函数列的收敛：逐点收敛与一致收敛的区别，以及为什么一致收敛才能保证极限与积分、导数交换次序。那正是本节所有估计（夹逼、绝对值不等式、区间可加）要反复使用的地方。再往后，反常积分与级数收敛判别会共用同一套比较工具，而弧长会在曲线与曲面的讨论中再次出现。

Where this lecture sits

The previous lecture proved that continuous functions are integrable, but the integral was still only a definition: computing $\int_0^1 x^2\,dx$ meant returning to upper and lower sums. This lecture supplies two things that make integrals usable:

1. Algebraic properties: scalar multiples, sums, monotonicity, additivity over subintervals, and the resulting bound $|\int f|\le\int|f|$.
2. The fundamental theorem of calculus: integration and differentiation are inverse operations. The first version differentiates the area function; the second evaluates integrals with an antiderivative.

The lecture then extends the integral to two "illegal" situations — unbounded intervals and unbounded functions — which are the improper integrals. It closes with a definition of arc length, the first time an integral defines a geometric quantity.

Timeline

Algebraic properties of the integral

Let $f,g$ be Riemann integrable on $[a,b]$.

1. Scalar multiples: $cf$ is integrable and $\int_a^b cf=c\int_a^b f$.
2. Sums: $f+g$ is integrable and $\int_a^b(f+g)=\int_a^b f+\int_a^b g$.
3. Monotonicity: if $f\le g$, then $\int_a^b f\le\int_a^b g$.
4. Additivity over subintervals: if $c\in[a,b]$, then $f$ is integrable on $[a,c]$ and on $[c,b]$, and

$$\int_a^b f=\int_a^c f+\int_c^b f .$$

Monotonicity is immediate from the picture: $f\le g$ gives $\sup f\le\sup g$ and $\inf f\le\inf g$ on every subinterval, so upper and lower sums compare termwise; take suprema and infima.

Proof for sums

Use the integrability criterion: for a given $\varepsilon>0$, find a partition with $U(f+g,P)-L(f+g,P)<\varepsilon$.

Since $f$ is integrable, there is $P_1$ with

$$U(f,P_1)-L(f,P_1)<\frac{\varepsilon}{2};$$

since $g$ is integrable, there is $P_2$ with

$$U(g,P_2)-L(g,P_2)<\frac{\varepsilon}{2}.$$

Let $P$ be the common refinement (union of division points). Then

$$U(f,P)-L(f,P)<\frac{\varepsilon}{2}, \qquad U(g,P)-L(g,P)<\frac{\varepsilon}{2}.$$

The decisive step relates the extrema of a sum to those of the summands. On any subinterval,

$$\sup(f+g)\le\sup f+\sup g, \qquad \inf(f+g)\ge\inf f+\inf g .$$

These are usually not equalities: if $f$ and $g$ have opposite signs and cancel, the oscillation of the sum can be much smaller than the sum of the oscillations. But the inequalities point the right way.

Hence, for the same $P$,

$$\begin{aligned} U(f+g,P)-L(f+g,P) &=\sum_i\Big(\sup(f+g)-\inf(f+g)\Big)\Delta x_i\\ &\le\sum_i\Big[(\sup f-\inf f)+(\sup g-\inf g)\Big]\Delta x_i\\ &=\big(U(f,P)-L(f,P)\big)+\big(U(g,P)-L(g,P)\big)\\ &<\varepsilon . \end{aligned}$$

The criterion gives integrability of $f+g$.

The equality of integrals needs a separate squeeze. For any partition $P$,

$$L(f,P)+L(g,P)\le L(f+g,P)\le U(f+g,P)\le U(f,P)+U(g,P).$$

Taking suprema and infima over $P$, the ends give $\int f+\int g$, so the middle is squeezed:

$$\int_a^b(f+g)=\int_a^b f+\int_a^b g .$$

On the notation

The lecture explains the symbols: $\int$ is a stretched $S$, standing for summation; $dx$ shares its origin with $\Delta x$ and is meant as a limiting version of $\Delta x_i$. Thus

$$\int_a^b f(x)\,dx$$

reads as "sum $f(x_i)\Delta x_i$ and let the partition get finer". Of course the rigorous definition is still a supremum and infimum, not literally a limit.

Additivity over subintervals

Proposition. If $f$ is integrable on $[a,b]$ and $c\in[a,b]$, then $f$ is integrable on $[a,c]$ and on $[c,b]$.

Proof. Given $\varepsilon>0$, integrability on $[a,b]$ gives a partition $P$ with

$$U(f,P)-L(f,P)<\varepsilon .$$

If $c$ is not a division point, insert it to obtain the refinement $P^*$. Refinement only decreases the gap, so

$$U(f,P^*)-L(f,P^*)<\varepsilon .$$

Now restrict $P^*$ to $[a,c]$; this is a partition of $[a,c]$, and similarly for $[c,b]$. Sums split at the division point:

$$U(f,P^*)-L(f,P^*) = \Big(U(f,P^*|_{[a,c]})-L(f,P^*|_{[a,c]})\Big) + \Big(U(f,P^*|_{[c,b]})-L(f,P^*|_{[c,b]})\Big).$$

Both terms are nonnegative and their sum is less than $\varepsilon$, so each is less than $\varepsilon$. The criterion gives integrability on each piece.

The equality of integrals follows from the same splitting: split every upper and lower sum into two parts and take suprema and infima:

$$\int_a^b f=\int_a^c f+\int_c^b f .$$

$\square$

This is the second crucial use of refinement: $c$ must first become a division point before the sums can be split cleanly.

The absolute value inequality

Corollary. If $f$ and $|f|$ are integrable on $[a,b]$, then

$$\left|\int_a^b f\right|\le\int_a^b|f| .$$

Proof. For every $x$,

$$f(x)\le|f(x)|, \qquad -f(x)\le|f(x)| .$$

Monotonicity gives

$$\int_a^b f\le\int_a^b|f|, \qquad \int_a^b(-f)\le\int_a^b|f| .$$

The left side of the second is $-\int_a^b f$ by linearity. Hence

$$-\int_a^b|f|\le\int_a^b f\le\int_a^b|f|,$$

which is the claim. $\square$

This inequality is used in almost every estimate: it replaces a signed sum by a sum of absolute values, at the cost of losing cancellation.

The fundamental theorem, version 1

Theorem. Let $f$ be continuous on $[a,b]$ and define

$$F(x)=\int_a^x f(s)\,ds, \qquad x\in[a,b].$$

Then $F$ is differentiable on $(a,b)$ and

$$F'(x)=f(x).$$

Proof. Fix $x_0$ and first take $x>x_0$. Additivity gives

$$F(x)-F(x_0) =\int_a^x f-\int_a^{x_0} f =\int_{x_0}^x f .$$

On $[x_0,x]$, continuity plus the extreme value theorem give a maximum $M$ and a minimum $m$. Thus

$$m(x-x_0)\le\int_{x_0}^x f\le M(x-x_0).$$

Dividing by $x-x_0>0$,

$$m\le\frac{F(x)-F(x_0)}{x-x_0}\le M .$$

As $x\to x_0^+$ the interval shrinks to $x_0$, and continuity gives $M\to f(x_0)$ and $m\to f(x_0)$. Squeezing,

$$\lim_{x\to x_0^+}\frac{F(x)-F(x_0)}{x-x_0}=f(x_0).$$

For $x<x_0$ the denominator is negative, so the inequalities reverse, but the same squeeze gives the left-hand limit $f(x_0)$. The two sides agree, hence $F'(x_0)=f(x_0)$. $\square$

The structure is a squeeze: the integral sits between two rectangles, the quotient between $\inf$ and $\sup$, and both converge to $f(x_0)$. Continuity is used exactly once — to make $M,m\to f(x_0)$.

The fundamental theorem, version 2

Theorem. Let $F$ be differentiable on $[a,b]$ and suppose $f=F'$ is Riemann integrable on $[a,b]$. Then

$$\int_a^b f(x)\,dx=F(b)-F(a).$$

The lecture leaves the proof to the notes; here is the argument, which uses only the mean value theorem and the criterion.

Proof. Take any partition $P=\{a=x_0<\cdots<x_n=b\}$. On each subinterval the mean value theorem gives $c_i\in(x_{i-1},x_i)$ with

$$F(x_i)-F(x_{i-1})=F'(c_i)(x_i-x_{i-1})=f(c_i)\Delta x_i .$$

Summing over $i$, the left side telescopes:

$$F(b)-F(a)=\sum_{i=1}^n f(c_i)\Delta x_i .$$

Each $f(c_i)$ lies between the infimum and supremum on its subinterval, so the sum lies between $L(f,P)$ and $U(f,P)$:

$$L(f,P)\le F(b)-F(a)\le U(f,P).$$

This holds for every partition. Since $f$ is integrable, passing to suprema and infima gives

$$\int_a^b f\le F(b)-F(a)\le\int_a^b f,$$

so $F(b)-F(a)=\int_a^b f$. $\square$

The hypothesis "$f$ is integrable" cannot be dropped: a derivative need not be Riemann integrable (derivatives can have bad discontinuities), and then the identity may fail.

Example

$$\int_0^1 x^2\,dx .$$

Take $F(x)=x^3/3$, so $F'=x^2$ and $x^2$ is continuous, hence integrable. Version 2 gives

$$\int_0^1 x^2\,dx=F(1)-F(0)=\frac13-0=\frac13 .$$

What would take a long computation with upper and lower sums now takes one step. That is the point of the theorem: it turns integration from "take a limit" into "find an antiderivative".

Improper integrals

A definite integral requires a bounded interval and a bounded function. Dropping either condition forces a limit.

Type 1: unbounded interval

Suppose $f$ is integrable on $[a,b]$ for every $b>a$. If

$$\lim_{b\to\infty}\int_a^b f(x)\,dx$$

exists, define

$$\int_a^\infty f(x)\,dx=\lim_{b\to\infty}\int_a^b f(x)\,dx,$$

and call the improper integral convergent; otherwise divergent.

Example 1.

$$\int_1^\infty\frac{dx}{x^2}.$$

Take $F(x)=-1/x$, so $F'=1/x^2$. For finite $b$,

$$\int_1^b\frac{dx}{x^2}=F(b)-F(1)=-\frac1b+1 .$$

As $b\to\infty$, $-1/b\to0$, so

$$\int_1^\infty\frac{dx}{x^2}=1 .$$

The curve runs on forever, yet the area is finite.

Example 2.

$$\int_1^\infty\frac{dx}{x}.$$

Take $F(x)=\log x$. Then

$$\int_1^b\frac{dx}{x}=\log b-\log1=\log b\to\infty .$$

Divergent. Both functions tend to $0$, but $1/x^2$ decays fast enough and $1/x$ does not — the same boundary that recurs in series tests.

Type 2: unbounded function

Suppose $f$ is integrable on $[c,b]$ for every $c>a$ (possibly unbounded or undefined near $a$). If

$$\lim_{c\to a^+}\int_c^b f(x)\,dx$$

exists, define

$$\int_a^b f(x)\,dx=\lim_{c\to a^+}\int_c^b f(x)\,dx .$$

Example 3.

$$\int_0^1\frac{dx}{\sqrt x}.$$

Take $F(x)=2\sqrt x$, so $F'=1/\sqrt x$. For $a>0$,

$$\int_a^1\frac{dx}{\sqrt x}=F(1)-F(a)=2-2\sqrt a .$$

As $a\to0^+$, $\sqrt a\to0$, so

$$\int_0^1\frac{dx}{\sqrt x}=2 .$$

The function blows up near $0$, but the area is still finite.

Example 4.

$$\int_0^1\frac{dx}{x}.$$

Take $F(x)=\log x$. For $a>0$,

$$\int_a^1\frac{dx}{x}=\log1-\log a=-\log a\to+\infty .$$

Divergent. The singularity of $1/\sqrt x$ is mild; the singularity of $1/x$ is too strong.

Arc length

The lecture ends with a definition, not developed further. Let

$$\gamma(s)=\big(f(s),g(s)\big), \qquad s\in[a,b],$$

be a plane curve with $f,g$ differentiable and derivatives continuous. Its arc length is defined as

$$L=\int_a^b\sqrt{f'(s)^2+g'(s)^2}\;ds .$$

The intuition: when $s$ increases by a tiny $ds$, the horizontal change is $f'(s)\,ds$ and the vertical change is $g'(s)\,ds$; by Pythagoras the piece of curve is approximately

$$\sqrt{(f'\,ds)^2+(g'\,ds)^2}=\sqrt{f'^2+g'^2}\;ds .$$

Adding all these small pieces gives the integral. It is still only a definition; the next lecture asks whether it is well defined and independent of the parametrization.

What this lecture establishes

• The integral is linear, monotone and additive over subintervals; the sum proof hinges on $\sup(f+g)\le\sup f+\sup g$ and $\inf(f+g)\ge\inf f+\inf g$.
• Additivity requires inserting $c$ into the partition before splitting the sums.
• $\left|\int_a^b f\right|\le\int_a^b|f|$.
• Version 1: $F(x)=\int_a^x f$ is differentiable with $F'=f$, proved by squeezing.
• Version 2: if $F'=f$ and $f$ is integrable, then $\int_a^b f=F(b)-F(a)$, proved by the mean value theorem and a telescoping sum.
• Improper integrals are limits, for unbounded intervals or unbounded functions: $\int_1^\infty x^{-2}=1$, $\int_1^\infty x^{-1}$ diverges, $\int_0^1 x^{-1/2}=2$, $\int_0^1 x^{-1}$ diverges.
• Arc length is $\int_a^b\sqrt{f'^2+g'^2}\,ds$.

Comprehension check

1. Prove that $f+g$ is integrable with $\int(f+g)=\int f+\int g$. Why is $\sup(f+g)\le\sup f+\sup g$ usually strict?

Reference answer: Given $\varepsilon>0$, choose a partition $P$ making both $U(f,P)-L(f,P)<\varepsilon/2$ and $U(g,P)-L(g,P)<\varepsilon/2$ (take partitions for each function, then their common refinement). On every subinterval $\sup(f+g)\le\sup f+\sup g$ and $\inf(f+g)\ge\inf f+\inf g$, so $U(f+g,P)-L(f+g,P)\le[U(f,P)-L(f,P)]+[U(g,P)-L(g,P)]<\varepsilon$, and the criterion gives integrability. The equality follows by squeezing: $L(f,P)+L(g,P)\le L(f+g,P)\le U(f+g,P)\le U(f,P)+U(g,P)$, then taking suprema and infima. The supremum inequality is usually strict: if $g=-f$, then $f+g\equiv0$ and $\sup(f+g)=0$, while $\sup f+\sup g$ can be positive because $f$ takes both signs on the interval.

2. Prove that integrability on $[a,b]$ implies integrability on $[a,c]$. What role does refinement play?

Reference answer: Given $\varepsilon>0$, take $P$ with $U(f,P)-L(f,P)<\varepsilon$. If $c\notin P$, insert it to get the refinement $P^*$; refinement decreases the gap, so $U(f,P^*)-L(f,P^*)<\varepsilon$. Restricting $P^*$ to $[a,c]$ gives a partition there, and

$$U(f,P^*)-L(f,P^*)=\big(U_{[a,c]}-L_{[a,c]}\big)+\big(U_{[c,b]}-L_{[c,b]}\big).$$

Both terms are nonnegative and their sum is below $\varepsilon$, so the first is below $\varepsilon$, and the criterion applies on $[a,c]$. Refinement is what makes $c$ a division point; otherwise $c$ lies inside a subinterval and the sums cannot be split.

3. Prove $\left|\int_a^b f\right|\le\int_a^b|f|$.

Reference answer: For every $x$, $f(x)\le|f(x)|$ and $-f(x)\le|f(x)|$. Monotonicity gives $\int f\le\int|f|$ and $\int(-f)\le\int|f|$, and linearity turns the second into $-\int f\le\int|f|$. Hence $-\int|f|\le\int f\le\int|f|$, which is the claim. The argument presupposes that $|f|$ is integrable; otherwise the right side is undefined.

4. Prove version 1 of the fundamental theorem and say where continuity is used.

Reference answer: Let $F(x)=\int_a^x f$ and fix $x_0$. For $x>x_0$, $F(x)-F(x_0)=\int_{x_0}^x f$. On $[x_0,x]$, continuity gives a minimum $m$ and maximum $M$, so $m(x-x_0)\le\int_{x_0}^x f\le M(x-x_0)$, and dividing by $x-x_0$ gives $m\le\frac{F(x)-F(x_0)}{x-x_0}\le M$. As $x\to x_0^+$, continuity gives $M,m\to f(x_0)$, so the right derivative is $f(x_0)$; the left side is analogous, with inequalities reversed because the denominator is negative. Continuity is used precisely here: it makes the extrema converge to $f(x_0)$ as the interval shrinks. Without continuity at the point, $F$ is still differentiable, but its derivative is a limit of averages and need not equal $f(x_0)$.

5. Prove version 2 using the mean value theorem, and explain why $f=F'$ must be integrable.

Reference answer: For any partition $P$, the mean value theorem on each subinterval gives $c_i$ with $F(x_i)-F(x_{i-1})=f(c_i)\Delta x_i$. Summing telescopes the left side to $F(b)-F(a)$ and leaves $\sum f(c_i)\Delta x_i$, which lies between $L(f,P)$ and $U(f,P)$. Since this holds for every partition, $\int f\le F(b)-F(a)\le\int f$, hence equality. Integrability of $f=F'$ is required because the squeeze only makes sense when the lower and upper integrals coincide; if the derivative is not integrable the integral may not exist and the identity has no meaning.

6. Determine convergence of $\int_1^\infty x^{-2}\,dx$ and $\int_1^\infty x^{-1}\,dx$, and explain the difference.

Reference answer: $\int_1^b x^{-2}\,dx=[-1/x]_1^b=1-1/b\to1$, convergent; $\int_1^b x^{-1}\,dx=[\log x]_1^b=\log b\to\infty$, divergent. Both integrands tend to $0$; the difference is speed. The tail area of $x^{-2}$ decays like $1/b$ and has finite total; the logarithm grows without bound, slowly but surely. In partition language, $x^{-1}$ contributes roughly $1/k$ on the unit interval far out, and the harmonic series $\sum1/k$ diverges — the integral and the series draw the same boundary.

Relation to neighbouring lectures

Version 1 says the integral of a continuous function is one of its antiderivatives; version 2 says the difference of an antiderivative at the endpoints is the integral. Together they turn calculus from "approximate by rectangles" into "differentiate and evaluate".

The next lecture turns to sequences of functions: pointwise versus uniform convergence, and why uniform convergence is what allows limits to pass through integrals and derivatives. Every estimate in this lecture — squeezing, the absolute value bound, additivity over subintervals — is used there. Later, improper integrals and series tests share the same comparison tools, and arc length returns in the study of curves and surfaces.