8.2 Properties of the Integral

8.9   Definition (Operations on functions.) Let $f\colon S\to\mbox{{\bf R}}$ and $g\colon T\to\mbox{{\bf R}}$ be functions where $S,T$ are sets. Let $c \in \mbox{{\bf R}}$. We define functions $f\pm g$, $fg$, $cf$, ${f\over g}$ and $\vert f\vert$ as follows:

$$\begin{array}{l} (f+g)(x)=f(x)+g(x) \mbox{ for all } x\in S\c... ... f\vert(x)=\vert f(x)\vert \mbox{ for all } x\in S. \end{array}$$

Remark: These operations of addition, subtraction, multiplication and division for functions satisfy the associative, commutative and distributive laws that you expect them to. The proofs are straightforward and will be omitted.

8.10   Definition (Partition-sample sequence.) Let $[a,b]$ be an interval. By a partition-sample sequence for $[a,b]$ I will mean a pair of sequences $(\{P_n\},\{S_n\})$ where $\{P_n\}$ is a sequence of partitions of $[a,b]$ such that
$\{\mu (P_n)\}\to 0$, and for each $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$, $S_n$ is a sample for $P_n$.

8.11   Theorem (Sum theorem for integrable functions.) Let $f,g$ be integrable functions on an interval $[a,b]$. Then $f\pm g$ and $cf$ are integrable on $[a,b]$ and

$$\int_a^b (f\pm g)=\int_a^b f \pm \int_a^b g,$$

and

$$\int_a^b cf=c\int_a^b f.$$

Proof: Suppose $f$ and $g$ are integrable on $[a,b]$. Let $(\{P_n\},\{S_n\})$ be a partition-sample sequence for $[a,b]$. If $P_n=\{x_0,\cdots ,x_m\}$ and $S_n=\{s_1,\cdots ,s_m\}$, then

$$\begin{eqnarray*} \sum (f\pm g,P_n,S_n)&=&\sum_{i=1}^m (f\pm g)(s_i)(x_i-x_{i-1}... ... g(s_i)(x_i-x_{i-1})\\ &=& \sum (f,P_n,S_n)\pm \sum(g,P_n,S_n). \end{eqnarray*}$$

Since $f$ and $g$ are integrable, we have

$$\{\sum(f,P_n,S_n)\}\to\int_a^b f \mbox{ and } \{\sum(g,P_n,S_n)\}\to\int_a^b g.$$

By the sum theorem for sequences,

$$\{\sum(f\pm g),P_n,S_n)\}=\{\sum(f,P_n,S_n)\pm\sum(g,P_n,S_n)\}\to\int_a^b f\pm \int_a^bg.$$

Hence $f\pm g$ is integrable and $${ \int_a^b (f\pm g)=\int_a^bf\pm\int_a^b g}$$. The proof of the second statement is left as an exercise.

8.12   Notation ( $${\int_a^b f(t)\;dt}$$) If $f$ is integrable on an interval $[a,b]$ we will sometimes write $${\int_a^b f(x)\;dx}$$ instead of $${\int_a^bf}$$. The ``$x$" in this expression is a dummy variable, but the ``$d$" is a part of the notation and may not be replaced by another symbol. This notation will be used mainly in cases where no particular name is available for $f$. Thus

$$\int_1^2 t^3+3t\;dt \mbox{ or } \int_1^2x^3+3x\;dx \mbox{ or } \int_1^2(x^3+3x)dx$$

means $${\int_1^2}F$$ where $F$ is the function on $[1,2]$ defined by $F(t)=t^3+3t$ for all $t\in [1,2]$. The ``$d$" here stands for difference, and $dx$ is a ghost of the differences $x_i-x_{i-1}$ that appear in the approximations for the integral. The $dx$ notation is due to Leibniz.

8.13   Example. Let

$$f(x) = (x-1)^2 - \displaystyle { {1 \over x} + {3\over \sqrt x} = x^2 - 2x + x^0 - {1\over x} + 3 x ^{-{1\over 2}}}.$$

This function is integrable over every closed bounded subinterval of $(0,\infty)$, since it is a sum of five functions that are known to be integrable. By several applications of the sum theorem for integrals we get

$$\begin{eqnarray*} \int_1^2 f &=& \int_1^2 (x^2 - 2x + 1 - {1\over x}+ 3 x ^{-{1\... ... - \ln(2) +6(\sqrt 2 - 1) = -{17 \over 3} - \ln(2) + 6\sqrt {2}. \end{eqnarray*}$$

8.14   Exercise. A Calculate the following integrals.

a)

$${ \int_1^a (2-x)^2dx }$$. Here $a > 1$.

b)

$${ \int_1^4 \sqrt{x} - {1\over x^2}\;dx}$$.

c)

$${ \int_1^{27} x^{-{1\over 3}} dx}$$.

d)

$${ \int_0^{27} x^{-{1\over 3}} dx}$$.

e)

$${ \int_1^2{x + 1 \over x} dx}$$.

f)

$${ \int_a^b M\;dx}.$$ Here $a\leq b$, and $M$ denotes a constant function.

8.15   Theorem (Inequality theorem for integrals.) Let $f$ and $g$ be integrable functions on the interval $[a,b]$ such that

$$f(x) \leq g(x) \mbox{ for all }x \in [a,b].$$

Then

$$\int_a^b f \leq \int_a^b g.$$

8.16   Exercise. Prove the inequality theorem for integrals. A

8.17   Corollary. Let $f$ be an integrable function on the interval $[a,b]$. Suppose $\vert f(x)\vert \leq M$ for all $x\in [a,b]$. Then

$$\left\vert \int_a^b f \right\vert \leq M(b-a).$$

Proof: We have

$$-M \leq f(x) \leq M \mbox{ for all }x\in [a,b].$$

Hence by the inequality theorem for integrals

$$\int_a^b -{M} \leq \int_a^b f \leq \int_a^b {M}.$$

Hence

$$-M(b-a) \leq \int_a^b f \leq M(b-a).$$

It follows that

$$\left\vert \int_a^b f \right\vert \leq M(b-a).\mbox{ $\diamondsuit$}$$

8.18   Theorem. Let $a,b,c$ be real numbers with $a<b<c$, and let $f$ be a function from $[a,c]$ to $\mbox{{\bf R}}$. Suppose $f$ is integrable on $[a,b]$ and $f$ is integrable on $[b,c]$. Then $f$ is integrable on $[a,c]$ and $${\int_a^c f=\int_a^b f+\int_b^c f}$$.

Proof: Since $f$ is integrable on $[a,b]$ and on $[b,c]$, it follows that $f$ is bounded on $[a,b]$ and on $[b,c]$, and hence $f$ is bounded on $[a,c]$. Let $(\{P_n\},\{S_n\})$ be a partition-sample sequence for $[a,c]$. For each $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$ we define a partition $P_n ^\prime$ of $[a,b]$ and a partition $P_n^{\prime\prime}$ of $[b,c]$, and a sample $S_n ^\prime$ for $P_n ^\prime$, and a sample $S_n ^{\prime\prime}$ for $P_n^{\prime\prime}$ as follows:

$$\mbox{ Let } P_n=\{x_0,x_1,\cdots ,x_m\},\;\; S_n=\{s_1,s_2,\cdots ,s_m\}.$$

Then there is an index $j$ such that $x_{j-1}\leq b\leq x_j$.

Let

$$P_n^\prime = \{x_0,\cdots ,x_{j-1},b\},\quad P_n ^{\prime\prime}=\{b,x_j,\cdots ,x_m\}$$ (8.19)

$$S_n^\prime = \{s_1,\cdots ,s_{j-1},b\},\quad S_n ^{\prime\prime}=\{b,s_{j+1},\cdots ,s_m\}$$ (8.20)

We have

$$\sum(f,P_n^\prime,S_n^\prime) + \sum(f,P_n^{\prime\prime},S_n^{\prime\prime})\mbox{{}}$$

$$= \sum_{i=1}^{j-1}f(s_i)(x_i-x_{i-1})+f(b)(b-x_{j-1})+f(b)(x_j-b)\mbox{{}}$$

$$\mbox{} +\sum_{i=j+1}^m f(s_i)(x_i-x_{i-1})\mbox{{}}$$

$$= \sum_{i=1}^mf(s_i)(x_i-x_{i-1})+f(b)(x_j-x_{j-1})-f(s_j)(x_j-x_{j-1})\mbox{{}}$$

$$= \sum(f,P_n,S_n)+\Delta_n,$$ (8.21)

where

$$\Delta_n=\Big( f(b)-f(s_j)\Big)(x_j-x_{j-1}).$$

Let $M$ be a bound for $f$ on $[a,c]$. Then

$$\vert f(b)-f(s_j)\vert\leq \vert f(b)\vert+\vert f(s_j)\vert\leq M+M=2M.$$

Also,

$$(x_j-x_{j-1})\leq\mu (P_n).$$

Now

$$0 \leq \vert\Delta_n\vert = \vert f(b)-f(s_j)\vert \cdot \vert x_j-x_{j-1}\vert \leq 2M\mu (P_n).$$

Since

$$\lim \{ 2M \mu (P_n) \} = 0,$$

it follows from the squeezing rule that $\{\vert\Delta_n\vert\} \to 0$ and hence $\{ \Delta_n \} \to 0$.

From equation (8.21) we have

$$\sum(f,P_n,S_n)=\sum(f,P_n^\prime ,S_n^\prime)+\sum(f,P_n^{\prime\prime} ,S_n^{\prime\prime})-\Delta_n.$$ (8.22)

Since $\mu(P_n^\prime)\leq\mu(P_n)$ and $\mu(P_n^{\prime\prime})\leq\mu(P_n)$, we see that $(\{P_n^\prime \},\{S_n^\prime\})$ is a partition-sample sequence on $[a,b]$, and $(\{P_n^{\prime\prime}\},\{S_n^{\prime\prime}\})$ is a partition-sample sequence on $[b,c]$. Since $f$ was given to be integrable on $[a,b]$ and on $[b,c]$, we know that

$$\{\sum(f,P_n^\prime ,S_n^\prime )\}\to\int_a^bf$$

and

$$\{\sum(f,P_n^{\prime\prime},S_n^{\prime\prime})\}\to\int_b^cf.$$

Hence it follows from (8.22) that

$$\{\sum(f,P_n,S_n)\}\to\int_a^bf+\int_b^cf$$

i.e., $f$ is integrable on $[a,c]$ and

$$\int_a^c f=\int_a^b f+\int_b^c f.\mbox{ $\diamondsuit$}$$

8.23   Corollary. Let $a_1,a_2,\cdots ,a_n$ be real numbers with $a_1\leq a_2\cdots\leq a_n$, and let $f$ be a bounded function on $[a_1,a_n]$. If the restriction of $f$ to each of the intervals $[a_1,a_2],[a_2,a_3],\cdots ,[a_{n-1},a_n]$ is integrable, then $f$ is integrable on $[a_1,a_n]$ and

$$\int_{a_1}^{a_n}f=\int_{a_1}^{a_2}f+\int_{a_2}^{a_3}f+\cdots +\int_{a_{n-1}}^{a_n}f.$$

8.24   Definition (Spike function.) Let $[a,b]$ be an interval. A function $f:[a,b] \to \mbox{{\bf R}}$ is called a spike function, if there exist numbers $c$ and $k$, with $c \in [a,b]$ such that

$$f(x)=\cases{0 &if $x\in [a,b]\setminus \{c\}$\cr k &if $x=c$.\cr}$$

8.25   Theorem (Spike functions are integrable.) Let $a,b,c,k$ be real numbers with $a<c$ and $a\leq b\leq c$.Let

$$f(x)=\cases{0 &if $x\in [a,c]\setminus \{b\}$\cr k &if $x=b$.\cr}$$

Then $f$ is integrable on $[a,c]$ and $${\int_a^c f=0}$$.

Proof: Case 1: Suppose $k \geq 0.$ Observe that $f$ is increasing on the interval $[a,b]$ and decreasing on the interval $[b,c]$, so $f$ is integrable on each of these intervals. The set of points under the graph of $f$ is the union of a horizontal segment and a vertical segment, and thus is a zero-area set. Hence

$$\int_a^b f=A_a^b f=0\qquad \int_b^c f=A_b^c f=0.$$

By the previous theorem, $f$ is integrable on $[a,c]$, and

$$\displaystyle {\int_a^c f=\int_a^b f+\int_b^c f=0+0=0}$$

.

Case 2: Suppose $k < 0$. Then by case 1 we see that $-f$ is integrable with integral equal to zero, so by the sum theorem for integrals $\int f = 0$ too. $\diamondsuit$

8.26   Corollary. Let $a,b,c,k$ be real numbers with $a<c$ and $a\leq b\leq c$. Let $f\colon [a,c]\to\mbox{{\bf R}}$ be an integrable function and let $g\colon [a,c]\to\mbox{{\bf R}}$ be defined by

$$g(x)=\cases{ f(x) &if $x\in [a,c]\setminus\{b\}$ \cr k &if $x=b$.\cr}$$

Then $g$ is integrable on $[a,c]$ and $${\int_a^c g=\int_a^c f}$$.

8.27   Corollary. Let $f$ be an integrable function from an interval $[a,b]$ to $\mbox{{\bf R}}$. Let $a_1 \cdots a_n$ be a finite set of distinct points in $\mbox{{\bf R}}$, and let $k_1\cdots k_n$ be a finite set of numbers. Let

$$g(x)=\cases{ f(x) &if $x\in [a,b]\setminus\{a_1,\cdots ,a_n\}... ... k_j &if $x=a_j$ \mbox{ for some $j$ with $1\leq j\leq n$.}\cr}$$

Then $g$ is integrable on $[a,b]$ and $${\int_a^b f=\int_a^b g}$$. Thus we can alter an integrable function on any finite set of points without changing its integrability or its integral.

8.28   Exercise. Prove corollary 8.26, i.e., explain why it follows from theorem 8.25.

8.29   Definition (Piecewise monotonic function.) A function $f$ from an interval $[a,b]$ to $\mbox{{\bf R}}$ is piecewise monotonic if there are points $a_1,a_2,\cdots ,a_n$ in $[a,b]$ with $a<a_1<a_2\cdots <a_n<b$ such that $f$ is monotonic on each of the intervals $[a,a_1],[a_1,a_2],\cdots ,[a_{n-1},a_n],[a_n,b]$.

8.30   Example. The function whose graph is sketched below is piecewise monotonic.

8.31   Theorem. Every piecewise monotonic function is integrable.

Proof: This follows from corollary 8.23. $\diamondsuit$

8.32   Exercise. A Let

$$f(x)=\cases{ x &if $0\leq x<1$\cr x-1 &if $1\leq x\leq 2$.\cr}$$

Sketch the graph of $f$. Carefully explain why $f$ is integrable, and find $${\int_0^2 f}$$.

8.33   Example. Let $g(x) = \vert (x-1)(x-2) \vert.$ Then

$$g(x) = \cases{ x^2 - 3x + 2 & for $x \in [0,1]$\cr -x^2 + 3x - 2 & for $x \in [1,2]$\cr x^2 - 3x + 2 & for $x \in [2,3].$ }$$

Hence $g$ is integrable on $[0,3]$, and

$$\begin{eqnarray*}\int_0^3 g &=& \int_0^1(x^2-3x+2)dx -\int_1^2(x^2 - 3x + 2)dx ... ... + 2\right)\\ &=& {13 \over 3} + {-9 \over 2} + 2 = {11\over 6} \end{eqnarray*}$$

8.34   Exercise. A Calculate the following integrals. Simplify your answers if you can.

a)

$${ \int_0^2\vert x^3-1\vert dx }$$.

b)

$${ \int_a^b(x-a)(b-x) dx }$$. Here $0<a<b$.

c)

$${ \int_a^b\vert(x-a)(b-x)\vert dx }$$. Here $0<a<b$.

d)

$${ \int_0^1 (t^2 - 2)^3 dt}.$$