156. Definite Integrals and Areas.
It will be remembered that, in Ch. VI, § 145, we assumed that, if is a continuous function of , and is the graph of , then the region shown in Fig. 47 has associated with it a definite number which we call its area. It is clear that, if we denote and by and , and allow to vary, this area is a function of , which we denote by .

Making this assumption, we proved in § 145 that , and we showed how this result might be used in the calculation of the areas of particular curves. But we have still to justify the fundamental assumption that there is such a number as the area .
We know indeed what is meant by the area of a rectangle, and that it is measured by the product of its sides. Also the properties of triangles, parallelograms, and polygons proved by Euclid enable us to attach a definite meaning to the areas of such figures. But nothing which we know so far provides us with a direct definition of the area of a figure bounded by curved lines. We shall now show how to give a definition of which will enable us to prove its existence.
Let us suppose continuous throughout the interval , and let us divide up the interval into a number of sub-intervals by means of the points of division , , , …, , where Further, let us denote by the interval , and by the lower bound (§ 102) of in , and let us write say.
It is evident that, if is the upper bound of in , then . The aggregate of values of is therefore, in the language of § 80, bounded above, and possesses an upper bound which we will denote by . No value of exceeds , but there are values of which exceed any number less than .
In the same way, if is the upper bound of in , we can define the sum
It is evident that, if is the lower bound of in , then . The aggregate of values of is therefore bounded below, and possesses a lower bound which we will denote by . No value of is less than , but there are values of less than any number greater than .
It will help to make clear the significance of the sums and if we observe that, in the simple case in which increases steadily from to , is and is . In this case is the total area of the rectangles shaded in Fig. 48, and is the area bounded by a thick line. In general and will still be areas, composed of rectangles, respectively included in and including the curvilinear region whose area we are trying to define.

We shall now show that no sum such as can exceed any sum such as . Let , be the sums corresponding to one mode of subdivision, and , those corresponding to another. We have to show that and .
We can form a third mode of subdivision by taking as dividing points all points which are such for either , or , . Let , be the sums corresponding to this third mode of subdivision. Then it is easy to see that For example, differs from in that at least one interval which occurs in is divided into a number of smaller intervals so that a term of is replaced in by a sum where , , … are the lower bounds of in , , …. But evidently , , …, so that the sum just written is not less than . Hence and the other inequalities (1) can be established in the same way. But, since , it follows that which is what we wanted to prove.
It also follows that . For we can find an as near to as we please and an as near to as we please, and so would involve the existence of an and an for which .
So far we have made no use of the fact that is continuous. We shall now show that , and that the sums , tend to the limit when the points of division are multiplied indefinitely in such a way that all the intervals tend to zero. More precisely, we shall show that, given any positive number , it is possible to find so that whenever for all values of .
There is, by Theorem II of § 106, a number such that whenever every is less than . Hence But and all the three terms on the right-hand side are positive, and therefore all less than . As is a constant, it must be zero. Hence and , , as was to be proved.
We define the area of as being the common limit of and , that is to say . It is easy to give a more general form to this definition. Consider the sum where denotes the value of at any point in . Then plainly lies between and , and so tends to the limit when the intervals tend to zero. We may therefore define the area as the limit of .
157. The definite integral.
Let us now suppose that is a continuous function, so that the region bounded by the curve , the ordinates and , and the axis of , has a definite area. We proved in Ch. VI, § 145, that if is an ‘integral function’ of , i.e. if then the area in question is .
As it is not always practicable actually to determine the form of , it is convenient to have a formula which represents the area and contains no explicit reference to . We shall write
The expression on the right-hand side of this equation may then be regarded as being defined in either of two ways. We may regard it as simply an abbreviation for , where is some integral function of , whether an actual formula expressing it is known or not; or we may regard it as the value of the area , as directly defined in § 156.
The number is called a definite integral; and are called its lower and upper limits; is called the subject of integration or integrand; and the interval the range of integration. The definite integral depends on and and the form of the function only, and is not a function of . On the other hand the integral function is sometimes called the indefinite integral of .
The distinction between the definite and the indefinite integral is merely one of point of view. The definite integral is a function of , and may be regarded as a particular integral function of . On the other hand the indefinite integral can always be expressed by means of a definite integral, since
But when we are considering ‘indefinite integrals’ or ‘integral functions’ we are usually thinking of a relation between two functions, in virtue of which one is the derivative of the other. And when we are considering a ‘definite integral’ we are not as a rule concerned with any possible variation of the limits. Usually the limits are constants such as and ; and is not a function at all, but a mere number.
It should be observed that the integral , having a differential coefficient , is a fortiori a continuous function of .
Since is continuous for all positive values of , the investigations of the preceding paragraphs supply us with a proof of the actual existence of the function , which we agreed to assume provisionally in § 128.
158. Area of a sector of a circle. The circular functions.
The theory of the trigonometrical functions , , etc., as usually presented in text-books of elementary trigonometry, rests on an unproved assumption. An angle is the configuration formed by two straight lines , ; there is no particular difficulty in translating this ‘geometrical’ definition into purely analytical terms. The assumption comes at the next stage, when it is assumed that angles are capable of numerical measurement, that is to say that there is a real number associated with the configuration, just as there is a real number associated with the region of Fig. 47. This point once admitted, and may be defined in the ordinary way, and there is no further difficulty of principle in the elaboration of the theory. The whole difficulty lies in the question, what is the which occurs in and ? To answer this question, we must define the measure of an angle, and we are now in a position to do so. The most natural definition would be this: suppose that is an arc of a circle whose centre is and whose radius is unity, so that . Then , the measure of the angle, is the length of the arc . This is, in substance, the definition adopted in the text-books, in the accounts which they give of the theory of ‘circular measure’. It has however, for our present purpose, a fatal defect; for we have not proved that the arc of a curve, even of a circle, possesses a length. The notion of the length of a curve is capable of precise mathematical analysis just as much as that of an area; but the analysis, although of the same general character as that of the preceding sections, is decidedly more difficult, and it is impossible that we should give any general treatment of the subject here.

We must therefore found our definition on the notion not of length but of area. We define the measure of the angle as twice the area of the sector of the unit circle.
Suppose, in particular, that is and that is , where . The area is a function of , which we may denote by . If we write for , is the point , and we have Differentiating with respect to , we find Thus the analytical equivalent of our definition would be to define by the equation and the whole theory of the circular functions could be worked out from this starting point, just as the theory of the logarithm is worked out from a similar definition in Ch. IX. See Appendix III.
Example LXIII. Calculation of the definite from the indefinite integral.
1. Show that and in particular that
2. , .
3. , .
[There is an apparent difficulty here owing to the fact that
is a many valued function. The difficulty may be avoided by observing that, in the equation
must denote an angle lying between
and
. For the integral vanishes when
and increases steadily and continuously as
increases. Thus the same is true of
, which therefore tends to
as
. In the same way we can show that
as
. Similarly, in the equation
where
,
denotes an angle lying between
and
. Thus, if
and
are both numerically less than unity, we have
4. , .
5. if , except when , when the value of the integral is , which is the limit of as .
6. , .
7. , if . [For the form of the indefinite integral see Ex. LIII. 3, 4. If then the subject of integration has an infinity between and . What is the value of the integral when is negative and ?]
8. , if and are positive. What is the value of the integral when and have opposite signs, or when both are negative?
9. Fourier’s integrals. Prove that if and are positive integers then is always equal to zero, and are equal to zero unless , when each is equal to .
10. Prove that and are each equal to zero except when , when each is equal to ; and that according as is odd or even.
159. Calculation of the definite integral from its definition as the limit of a sum.
In a few cases we can evaluate a definite integral by direct calculation, starting from the definitions of §§ 156 and 157. As a rule it is much simpler to use the indefinite integral, but the reader will find it instructive to work through a few examples.
Example LXIV
1. Evaluate by dividing into equal parts by the points of division , , , …, , and calculating the limit as of
[This sum is
which tends to the limit
as
. Verify the result by graphical reasoning.]
2. Calculate in the same way.
3. Calculate , where , by dividing into parts by the points of division , , , …, , , where . Apply the same method to the more general integral .
4. Calculate and by the method of Ex. 1.
5. Prove that as .
[This follows from the fact that
which tends to the limit
as
, in virtue of the direct definition of the integral.]
6. Prove that . [The limit is .]
160. General properties of the definite integral.
The definite integral possesses the important properties expressed by the following equations.
(1)
This follows at once from the definition of the integral by means of the integral function , since . It should be observed that in the direct definition it was presupposed that the upper limit is greater than the lower; thus this method of definition does not apply to the integral when . If we adopt this definition as fundamental we must extend it to such cases by regarding the equation (1) as a definition of its right-hand side.
(2)
(3)
(4)
(5)
The reader will find it an instructive exercise to write out formal proofs of these properties, in each case giving a proof starting from () the definition by means of the integral function and () the direct definition.
The following theorems are also important.
(6) If when , then .
We have only to observe that the sum of § 156 cannot be negative. It will be shown later (Misc. Ex. 41) that the value of the integral cannot be zero unless is always equal to zero: this may also be deduced from the second corollary of § 121.
(7) If when , then
This follows at once if we apply (6) to and .
(8)
where lies between and .
This follows from (7). For we can take to be the least and the greatest value of in . Then the integral is equal to , where lies between and . But, since is continuous, there must be a value of for which (§ 100).
If is the integral function, we can write the result of (8) in the form so that (8) appears now to be only another way of stating the Mean Value Theorem of § 125. We may call (8) the First Mean Value Theorem for Integrals.
(9) The Generalised Mean Value Theorem for integrals. If is positive, and and are defined as in (7), then and where is defined as in (8).
This follows at once by applying Theorem (6) to the integrals The reader should formulate for himself the corresponding result which holds when is always negative.
(10)The Fundamental Theorem of the Integral Calculus. The function has a derivative equal to .
This has been proved already in § 145, but it is convenient to restate the result here as a formal theorem. It follows as a corollary, as was pointed out in § 157, that is a continuous function of .
Example LXV
1. Show, by means of the direct definition of the definite integral, and equations (1)-(5) above, that
being an integer. [The truth of these equations will appear geometrically intuitive, if the graphs of the functions under the sign of integration are sketched.]
2. Prove that is equal to or to according as is odd or or even. [Use the formula , the last term being or .]
3. Prove that is equal to or to according as is odd or even.
4. If , and is a positive integer not greater than , then If then the value of each of the last two integrals is zero. [Use Ex. LXIII. 9.]
5. If , and is a positive integer not greater than , then If then the value of the last integral is zero. [Use Ex. LXIII. 10.]
6. Prove that if and are positive then
[Use
Ex. LXIII. 8 and Ex. 1 above.]
7. If when , then .
Prove that
9. If then
[The first inequality follows from the fact that
, the second from the fact that
.]
10. Prove that
11. Prove that if , and hence that
12. Prove that
[Put
: then replace
by
and by
.]
13. If and are positive acute angles then If , then the integral lies between and .
14. Prove that
[If
is the sum considered at the end of
§ 156, and
the corresponding sum formed from the function
, then
.]
15. If , then
161. Integration by parts and by substitution.
It follows from § 138 that This formula is known as the formula for .
Again, we know (§ 133) that if is the integral function of , then Hence, if , , we have which is the formula for the transformation of a definite integral by substitution.
The formulae for integration by parts and for transformation often enable us to evaluate a definite integral without the labour of actually finding the integral function of the subject of integration, and sometimes even when the integral function cannot be found. Some instances of this will be found in the following examples. That the value of a definite integral may sometimes be found without a knowledge of the integral function is only to be expected, for the fact that we cannot determine the general form of a function in no way precludes the possibility that we may be able to determine the difference between two of its particular values. But as a rule this can only be effected by the use of more advanced methods than are at present at our disposal.
Example LXVI
1. Prove that
2. More generally, where
3. Prove that
4. Prove that if and are positive then
[Integrate by parts and use
Ex. LXIII. 8.]
5. If then
[Integrate repeatedly by parts.]
6. Prove by integration by parts that if where and are positive integers, then , and deduce that
7. Prove that if then . Hence evaluate the integral for all positive integral values of .
[Put
and integrate by parts.]
8. Deduce from the last example that lies between and .
9. Prove that if then . [Write for and integrate by parts.]
10. Deduce that is equal to according as is odd or even.
11. The Second Mean Value Theorem. If is a function of which has a differential coefficient of constant sign for all values of from to , then there is a number between and such that
[Let
. Then
by the generalised Mean Value Theorem of
§ 160:
i.e. which is equivalent to the result given.]
12. Bonnet’s form of the Second Mean Value Theorem. If is of constant sign, and and have the same sign, then where lies between and . [For , where lies between and , and so is the value of for a value of such as . The important case is that in which .]
Prove similarly that if and have the same sign, then where lies between and . [Use the function . It will be found that the integral can be expressed in the form The important case is that in which .]
13. Prove that if . [Apply the first formula of Ex. 12, and note that the integral of over any interval whatever is numerically less than .]
14. Establish the results of Ex. LXV. 1 by means of the rule for substitution. [In (i) divide the range of integration into the two parts , , and put in the first. In (ii) use the substitution to obtain the first equation: to obtain the second divide the range into two equal parts and use the substitution . In (iii) divide the range into equal parts and use the substitutions , , ….]
15. Prove that
16. Prove that
17. Prove that
[Put
.]
18. Prove that
19. Show by means of the transformation that
20. Show by means of the substitution that when is a positive integer and , and evaluate the integral when , , .
21. If and are positive integers then
[Put
, and use Ex. 6.]