The Definite Integral

Let $y=f(x)$ be a function continuous for $a\leqq x\leqq b$. We plot its graph; for instance, it might be as shown in the graph on the following page.

Now let us define a number which we shall call the (definite) integral from a to b of $f(x)$, and for which we write

$${\int }_a^{b}f(x)dx,$$

as follows:

Consider the area bounded by the graph of $y=f(x)$, the x-axis, the line $x=a$ and the line $x=b$. If a part of this area lies above the x-axis, we shall consider it as positive. If a part lies below the x-axis, we shall consider it as negative. Thus we have the arrangement of signs shown in the diagram.

Then we define

$${\int }_a^{b}f(x)dx$$

to be simply the algebraic sum of these areas, taken with the corresponding signs.

If $b\leqq a$, we define

$${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx,$$

since

$${\int }_b^{a}f(x)dx$$

is then defined in the above manner. In particular, we have

$${\int }_a^{a}f(x)dx=-{\int }_a^{a}f(x)dx,$$

or

$${\int }_a^{a}f(x)dx=0.$$

Also, if $a,b,c$ are any numbers such that $f(x)$ is continuous on the smallest interval containing them all, then

$${\int }_a^{c}f(x)dx={\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx.$$

If $a\leqq b\leqq c$, this is obvious from the definition. It is left as an exercise to verify it in the other cases.

Now suppose $x$ is in the interval $a\leqq t\leqq b$. Then

$${\int }_a^{x}f(t)dt$$

has been defined, where we regard $x$ as fixed for the moment, and denote our variable by $t$. Let us consider this as a function $F(x)$. Then the difference quotient for $F(x)$ is

\begin{aligned} \frac{F(x+h) - F(x)}{h} &= \frac{\displaystyle \int_{a}^{x+h} f(t)dt - \int_{a}^{x} f(t)dt}{h} \\ &= \frac{1}{h} \int_{x}^{x+h} f(t)\,dt \end{aligned}

by the above remark.

Now let $M$ be the maximum of $f(t)$ on the interval between $x$ and $x+h$, $m$ its minimum. Then we assert that if ,

$$Mh\geqq {\int }_x^{x+h}f(t)dt\geqq mh,$$

or

$$M\geqq \frac{1}{h}{\int }_x^{x+h}f(t)dt\geqq m.$$

The first inequality is reversed if , but the second one is still valid. We shall show this in the case where . The idea for the remaining cases will then be clear. Because $f(t)$ is continuous, we can assume $h$ is small enough so that for all $t$ between $x$ and $x+h$. Then our statement amounts to saying that the area under the graph of $f(t)$ between $x$ and $x+h$ lies between the area of a rectangle of width $h$ whose height is the smallest value of $f(t)$ in this interval and that of a rectangle of width $h$ and height the maximum value of $f(t)$ in the interval, as shown in the diagram on the following page.

But since $f(t)$ is continuous at $x,f(t)$ approaches $f(x)$ for all $t$ between $f(x)$ and $f(x+h)$, as $h$ approaches 0. Since $M$ and $m$ are values for $f(t)$, they also approach $f(x)$. Thus we have

$$f(x)=\underset{h\to 0}{lim}M\geqq \underset{h\to 0}{lim}\frac{1}{h}{\int }_x^{x+h}f(t)dt\geqq \underset{h\to 0}{lim}m=f(x);$$

but then all these are equal, and F'(x) = f(x).

Now we have a procedure for evaluating

$${\int }_a^{b}f(t)dt.$$

First we find any function $G(x)$ such that G'(x) = f(x) for $a\leqq x\leqq b$; such a function is called an indefinite integral, or an antiderivative, of $f(x)$. Then we know that

$$F(x)={\int }_a^{x}f(t)dt$$

has the property that F'(x) = f(x) = G'(x). Therefore $F(x)=G(x)+c$, where $c$ is some constant. But $F(a)=0$, so $c=-G(a)$. Then

$${\int }_a^{b}f(t)dt=F(b)=G(b)+c=G(b)-G(a).$$

We often write this simply as

$${\int }_a^{b}f(x)dx=G(x){|}_a^b=G(b)-G(a).$$

This is known as the fundamental theorem of the calculus.

The result allows us to solve the problem of finding areas which occupied the Greek mathematicians for centuries. They had some special techniques which will extend the class of functions we can handle.

Note. Once again we have a linear operation; if $F(x)$ is an indefinite integral of $f(x),G(x)$ one for $g(x)$, then

(F(x) + G(x))' = F'(x) + G'(x) = f(x) + g(x),

or $F(x)+G(x)$ is an indefinite integral for $f(x)+g(x)$. Likewise, $cF(x)$ is an indefinite integral for $cf(x)$. For definite integrals, with $F(x),f(x),G(x),g(x)$ and $c$ as above, we have

\begin{aligned} \int_{a}^{b} (f(t) + g(t))dt &= (F(b) + G(b)) - (F(a) + G(a))\\[4pt] &= (F(b) - F(a)) + (G(b) - G(a)) \\[4pt] &= \int_{a}^{b} f(t)dt + \int_{a}^{b} g(t)\,dt \end{aligned}

and

\begin{aligned} \int_{a}^{b} cf(t)dt &= cF(b) - cF(a) \\ &= c(F(b) - F(a)) \\ &= c \int_a^{b} f(t)\,dt. \end{aligned}

Observe also that any two indefinite integrals for $f(x)$ differ only by a constant.

Examples

By examining the rules for differentiation, you will see that the indefinite integrals below are correct; we write

$$\int f(x)dx=F(x)$$

to mean simply that $F(x)$ is an indefinite integral of $f(x)$, or that F'(x) = f(x):

1. ${\displaystyle \int e^{x}dx=e^x;}$
2. ${\displaystyle \int \cos xdx=\sin x;}$
3. ${\displaystyle \int \sin xdx=-\cos x;}$
4. ${\displaystyle \int {\sec }^{2}xdx=\tan x;}$
5. ${\displaystyle \int x^{b}dx=\frac{1}{b+1}{x}^{b+1},}$ $b$ a positive rational number;
6. ${\displaystyle \int a^{x}dx=\frac{1}{\log a}\cdot a^x,}$

In rules 1) and 6), we are still discussing functions of which we have no real knowledge. Likewise we could have written

$$\int \frac{1}{x}dx=\log x,$$

but this and the rules above are to be regarded for the present only as pieces of information of use in developing technique. We shall in the next section define $\log x$ in terms of an integral.

Please observe that when we write

$${\int }_a^{b}f(x)dx,$$

the expression is meaningless to us unless $f(x)$ is defined and continuous throughout the interval $a\le x\le b$. Thus

$${\int }_0^4\frac{1}{x}dx$$

is undefined, as is

$${\int }_{-1}^1\sqrt{x}dx.$$

The definite integral can be defined in such a way that more functions fall into its domain (though neither of these two does), but we shall not pursue this question further.

Exercises