156. Definite Integrals and Areas.

It will be remembered that, in Ch. VI, § 145, we assumed that, if f(x) is a continuous function of x, and PQ is the graph of y=f(x), then the region PpqQ shown in Fig. 47 has associated with it a definite number which we call its area. It is clear that, if we denote Op and Oq by a and x, and allow x to vary, this area is a function of x, which we denote by F(x).

Making this assumption, we proved in § 145 that F(x)=f(x), 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 F(x).

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 F(x) which will enable us to prove its existence.1

Let us suppose f(x) continuous throughout the interval [a,b], and let us divide up the interval into a number of sub-intervals by means of the points of division x0, x1, x2, …, xn, where a=x0<x1<<xn1<xn=b. Further, let us denote by δν the interval [xν,xν+1], and by mν the lower bound (§ 102) of f(x) in δν, and let us write s=m0δ0+m1δ1++mnδn=mνδν, say.

It is evident that, if M is the upper bound of f(x) in [a,b], then sM(ba). The aggregate of values of s is therefore, in the language of § 80, bounded above, and possesses an upper bound which we will denote by j. No value of s exceeds j, but there are values of s which exceed any number less than j.

In the same way, if Mν is the upper bound of f(x) in δν, we can define the sum S=Mνδν.

It is evident that, if m is the lower bound of f(x) in [a,b], then Sm(ba). The aggregate of values of S is therefore bounded below, and possesses a lower bound which we will denote by J. No value of S is less than J, but there are values of S less than any number greater than J.

It will help to make clear the significance of the sums s and S if we observe that, in the simple case in which f(x) increases steadily from x=a to x=b, mν is f(xν) and Mν is f(xν+1). In this case s is the total area of the rectangles shaded in Fig. 48, and S is the area bounded by a thick line. In general s and S 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 s can exceed any sum such as S. Let s, S be the sums corresponding to one mode of subdivision, and s, S those corresponding to another. We have to show that sS and sS.

We can form a third mode of subdivision by taking as dividing points all points which are such for either s, S or s, S. Let s, S be the sums corresponding to this third mode of subdivision. Then it is easy to see that (1)ss,ss,SS,SS. For example, s differs from s in that at least one interval δν which occurs in s is divided into a number of smaller intervals δν,1, δν,2, , δν,p, so that a term mνδν of s is replaced in s by a sum mν,1δν,1+mν,2δν,2++mν,pδν,p, where mν,1, mν,2, … are the lower bounds of f(x) in δν,1, δν,2, …. But evidently mν,1mν, mν,2mν, …, so that the sum just written is not less than mνδν. Hence ss; and the other inequalities (1) can be established in the same way. But, since sS, it follows that ssSS, which is what we wanted to prove.

It also follows that jJ. For we can find an s as near to j as we please and an S as near to J as we please,2 and so j>J would involve the existence of an s and an S for which s>S.

So far we have made no use of the fact that f(x) is continuous. We shall now show that j=J, and that the sums s, S tend to the limit J when the points of division xν are multiplied indefinitely in such a way that all the intervals δν tend to zero. More precisely, we shall show that, given any positive number ϵ, it is possible to find δ so that 0Js<ϵ,0SJ<ϵ whenever δν<δ for all values of ν.

There is, by Theorem II of § 106, a number δ such that Mνmν<ϵ/(ba), whenever every δν is less than δ. Hence Ss=(Mνmν)δν<ϵ. But Ss=(SJ)+(Jj)+(js); and all the three terms on the right-hand side are positive, and therefore all less than ϵ. As Jj is a constant, it must be zero. Hence j=J and 0js<ϵ, 0SJ<ϵ, as was to be proved.

We define the area of PpqQ as being the common limit of s and S, that is to say J. It is easy to give a more general form to this definition. Consider the sum σ=fνδν where fν denotes the value of f(x) at any point in δν. Then σ plainly lies between s and S, and so tends to the limit J when the intervals δν tend to zero. We may therefore define the area as the limit of σ.

 

157. The definite integral.

Let us now suppose that f(x) is a continuous function, so that the region bounded by the curve y=f(x), the ordinates x=a and x=b, and the axis of x, has a definite area. We proved in Ch. VI, § 145, that if F(x) is an ‘integral function’ of f(x), i.e. if F(x)=f(x),F(x)=f(x)dx, then the area in question is F(b)F(a).

As it is not always practicable actually to determine the form of F(x), it is convenient to have a formula which represents the area PpqQ and contains no explicit reference to F(x). We shall write (PpqQ)=abf(x)dx.

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 F(b)F(a), where F(x) is some integral function of f(x), whether an actual formula expressing it is known or not; or we may regard it as the value of the area PpqQ, as directly defined in § 156.

The number abf(x)dx is called a definite integral; a and b are called its lower and upper limits; f(x) is called the subject of integration or integrand; and the interval [a,b] the range of integration. The definite integral depends on a and b and the form of the function f(x) only, and is not a function of x. On the other hand the integral function F(x)=f(x)dx is sometimes called the indefinite integral of f(x).

The distinction between the definite and the indefinite integral is merely one of point of view. The definite integral abf(x)dx=F(b)F(a) is a function of b, and may be regarded as a particular integral function of f(b). On the other hand the indefinite integral F(x) can always be expressed by means of a definite integral, since F(x)=F(a)+axf(t)dt.

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 0 and 1; and 01f(x)dx=F(1)F(0) is not a function at all, but a mere number.

It should be observed that the integral axf(t)dt, having a differential coefficient f(x), is a fortiori a continuous function of x.

Since 1/x is continuous for all positive values of x, the investigations of the preceding paragraphs supply us with a proof of the actual existence of the function logx, 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 cosx, sinx, 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 OA, OP; 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 x associated with the configuration, just as there is a real number associated with the region PpqQ of Fig. 47. This point once admitted, cosx and sinx 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 x which occurs in cosx and sinx? 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 AP is an arc of a circle whose centre is O and whose radius is unity, so that OA=OP=1. Then x, the measure of the angle, is the length of the arc AP. 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 AOP as twice the area of the sector AOP of the unit circle.

Suppose, in particular, that OA is y=0 and that OP is y=mx, where m>0. The area is a function of m, which we may denote by ϕ(m). If we write μ for (1+m2)12, P is the point (μ,mμ), and we have ϕ(m)=12mμ2+μ11x2dx. Differentiating with respect to m, we find ϕ(m)=12(1+m2),ϕ(m)=120mdt1+t2. Thus the analytical equivalent of our definition would be to define arctanm by the equation arctanm=0mdt1+t2; 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 abxndx=bn+1an+1n+1, and in particular that 01xndx=1n+1.

2. abcosmxdx=sinmbsinmam, absinmxdx=cosmacosmbm.

3. abdx1+x2=arctanbarctana, 01dx1+x2=14π.

[There is an apparent difficulty here owing to the fact that arctanx is a many valued function. The difficulty may be avoided by observing that, in the equation 0xdt1+t2=arctanx, arctanx must denote an angle lying between 12π and 12π. For the integral vanishes when x=0 and increases steadily and continuously as x increases. Thus the same is true of arctanx, which therefore tends to 12π as x. In the same way we can show that arctanx12π as x. Similarly, in the equation 0xdt1t2=arcsinx, where 1<x<1, arcsinx denotes an angle lying between 12π and 12π. Thus, if a and b are both numerically less than unity, we have abdx1x2=arcsinbarcsina.]

4. 01dx1x+x2=2π33, 01dx1+x+x2=π33.

5. 01dx1+2xcosα+x2=α2sinα if π<α<π, except when α=0, when the value of the integral is 12, which is the limit of 12αcscα as α0.

6. 011x2dx=14π, 0aa2x2dx=14πa2(a>0).

7. 0πdxa+bcosx=πa2b2, if a>|b|. [For the form of the indefinite integral see Ex. LIII. 3, 4. If |a|<|b| then the subject of integration has an infinity between 0 and π. What is the value of the integral when a is negative and a>|b|?]

8. 012πdxa2cos2x+b2sin2x=π2ab, if a and b are positive. What is the value of the integral when a and b have opposite signs, or when both are negative?

9. Fourier’s integrals. Prove that if m and n are positive integers then 02πcosmxsinnxdx is always equal to zero, and 02πcosmxcosnxdx,02πsinmxsinnxdx are equal to zero unless m=n, when each is equal to π.

10. Prove that 0πcosmxcosnxdx and 0πsinmxsinnxdx are each equal to zero except when m=n, when each is equal to 12π; and that 0πcosmxsinnxdx=2nn2m2,0πcosmxsinnxdx=0, according as nm 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 abxdx by dividing [a,b] into n equal parts by the points of division a=x0, x1, x2, …, xn=b, and calculating the limit as n of (x1x0)f(x0)+(x2x1)f(x1)++(xnxn1)f(xn1).

[This sum is ban[a+(a+ban)+(a+2ban)++{a+(n1)ban}]=ban[na+ban{1+2++(n1)}]=(ba){a+(ba)n(n1)2n2}, which tends to the limit 12(b2a2) as n. Verify the result by graphical reasoning.]

2. Calculate abx2dx in the same way.

3. Calculate abxdx, where 0<a<b, by dividing [a,b] into n parts by the points of division a, ar, ar2, …, arn1, arn, where rn=b/a. Apply the same method to the more general integral abxmdx.

4. Calculate abcosmxdx and absinmxdx by the method of Ex. 1.

5. Prove that nr=0n11n2+r214π as n.

[This follows from the fact that nn2+nn2+12++nn2+(n1)2=r=0n1(1/n)1+(r/n)2, which tends to the limit 01dx1+x2 as n, in virtue of the direct definition of the integral.]

6. Prove that 1n2r=0n1n2r214π. [The limit is 011x2dx.]

 

160. General properties of the definite integral.

The definite integral possesses the important properties expressed by the following equations.3

(1)abf(x)dx=baf(x)dx

This follows at once from the definition of the integral by means of the integral function F(x), since F(b)F(a)={F(a)F(b)}. 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 baf(x)dx when a<b. 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) aaf(x)dx=0

(3)abf(x)dx+bcf(x)dx=acf(x)dx.

(4)abkf(x)dx=kabf(x)dx.

(5)ab{f(x)+ϕ(x)}dx=abf(x)dx+abϕ(x)dx

The reader will find it an instructive exercise to write out formal proofs of these properties, in each case giving a proof starting from (α) the definition by means of the integral function and (β) the direct definition.

The following theorems are also important.

(6) If f(x)0 when axb, then abf(x)dx0.

We have only to observe that the sum s of § 156 cannot be negative. It will be shown later (Misc. Ex. 41) that the value of the integral cannot be zero unless f(x) is always equal to zero: this may also be deduced from the second corollary of § 121.

(7) If Hf(x)K when axb, then H(ba)abf(x)dxK(ba).

This follows at once if we apply (6) to f(x)H and Kf(x).

(8) abf(x)dx=(ba)f(ξ)

where ξ lies between a and b.

This follows from (7). For we can take H to be the least and K the greatest value of f(x) in [a,b]. Then the integral is equal to η(ba), where η lies between H and K. But, since f(x) is continuous, there must be a value of ξ for which f(ξ)=η (§ 100).

If F(x) is the integral function, we can write the result of (8) in the form F(b)F(a)=(ba)F(ξ), 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 ϕ(x) is positive, and H and K are defined as in (7), then Habϕ(x)dxabf(x)ϕ(x)dxKabϕ(x)dx; and abf(x)ϕ(x)dx=f(ξ)abϕ(x)dx, where ξ is defined as in (8).

This follows at once by applying Theorem (6) to the integrals ab{f(x)H}ϕ(x)dx,ab{Kf(x)}ϕ(x)dx. The reader should formulate for himself the corresponding result which holds when ϕ(x) is always negative.

(10)The Fundamental Theorem of the Integral Calculus. The function F(x)=axf(t)dt has a derivative equal to f(x).

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 F(x) is a continuous function of x.

Example LXV

1. Show, by means of the direct definition of the definite integral, and equations (1)-(5) above, that

(i)aaϕ(x2)dx=20aϕ(x2)dx,aaxϕ(x2)dx=0(ii)012πϕ(cosx)dx=012πϕ(sinx)dx=120πϕ(sinx)dx(iii)0mπϕ(cos2x)dx=m0πϕ(cos2x)dx

m 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 0πsinnxsinxdx is equal to π or to 0 according as n is odd or or even. [Use the formula (sinnx)/(sinx)=2cos{(n1)x}+2cos{(n3)x}+, the last term being 1 or 2cosx.]

3. Prove that 0πsinnxcotxdx is equal to 0 or to π according as n is odd or even.

4. If ϕ(x)=a0+a1cosx+b1sinx+a2cos2x++ancosnx+bnsinnx, and k is a positive integer not greater than n, then 02πϕ(x)dx=2πa0,02πcoskxϕ(x)dx=πak,02πsinkxϕ(x)dx=πbk. If k>n then the value of each of the last two integrals is zero. [Use Ex. LXIII. 9.]

5. If ϕ(x)=a0+a1cosx+a2cos2x++ancosnx, and k is a positive integer not greater than n, then 0πϕ(x)dx=πa0,0πcoskxϕ(x)dx=12πak. If k>n then the value of the last integral is zero. [Use Ex. LXIII. 10.]

6. Prove that if a and b are positive then 02πdxa2cos2x+b2sin2x=2πab.

[Use Ex. LXIII. 8 and Ex. 1 above.]

7. If f(x)ϕ(x) when axb, then abfdxabϕdx.

Prove that 0<012πsinn+1xdx<012πsinnxdx,0<014πtann+1xdx<014πtannxdx.

9. If n>1 then .5<012dx1x2n<.524.

[The first inequality follows from the fact that 1x2n<1, the second from the fact that 1x2n>1x2.]

10. Prove that 12<01dx4x2+x3<16π.

11. Prove that (3x+8)/16<1/43x+x3<1/43x if 0<x<1, and hence that 1932<01dx43x+x3<23.

12. Prove that .573<12dx43x+x3<.595.

[Put x=1+u: then replace 2+3u2+u3 by 2+4u2 and by 2+3u2.]

13. If α and ϕ are positive acute angles then ϕ<0ϕdx1sin2αsin2x<ϕ1sin2αsin2ϕ. If α=ϕ=16π, then the integral lies between .523 and .541.

14. Prove that |abf(x)dx|ab|f(x)|dx.

[If σ is the sum considered at the end of § 156, and σ the corresponding sum formed from the function |f(x)|, then |σ|σ.]

15. If |f(x)|M, then |abf(x)ϕ(x)dx|Mab|ϕ(x)|dx.

 

161. Integration by parts and by substitution.

It follows from § 138 that abf(x)ϕ(x)dx=f(b)ϕ(b)f(a)ϕ(a)abf(x)ϕ(x)dx. This formula is known as the formula for .

Again, we know (§ 133) that if F(t) is the integral function of f(t), then f{ϕ(x)}ϕ(x)dx=F{ϕ(x)}. Hence, if ϕ(a)=c, ϕ(b)=d, we have cdf(t)dt=F(d)F(c)=F{ϕ(b)}F{ϕ(a)}=abf{ϕ(x)}ϕ(x)dx; 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 F(x) in no way precludes the possibility that we may be able to determine the difference F(b)F(a) 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 abxf(x)dx={bf(b)f(b)}{af(a)f(a)}.

2. More generally, abxmf(m+1)(x)dx=F(b)F(a), where F(x)=xmf(m)(x)mxm1f(m1)(x)+m(m1)xm2f(m2)(x)+(1)mm!f(x).

3. Prove that 01arcsinxdx=12π1,01xarctanxdx=14π12.

4. Prove that if a and b are positive then 012πxcosxsinxdx(a2cos2x+b2sin2x)2=π4ab2(a+b).

[Integrate by parts and use Ex. LXIII. 8.]

5. If f1(x)=0xf(t)dt,f2(x)=0xf1(t)dt, ,fk(x)=0xfk1(t)dt, then fk(x)=1(k1)!0xf(t)(xt)k1dt.

[Integrate repeatedly by parts.]

6. Prove by integration by parts that if um,n=01xm(1x)ndx, where m and n are positive integers, then (m+n+1)um,n=num,n1, and deduce that um,n=m!n!(m+n+1)!.

7. Prove that if un=014πtannxdx then un+un2=1/(n1). Hence evaluate the integral for all positive integral values of n.

[Put tannx=tann2x(sec2x1) and integrate by parts.]

8. Deduce from the last example that un lies between 1/{2(n1)} and 1/{2(n+1)}.

9. Prove that if un=012πsinnxdx then un={(n1)/n}un2. [Write sinn1xsinx for sinnx and integrate by parts.]

10. Deduce that un is equal to 246(n1)357n,12π135(n1)246n, according as n is odd or even.

11. The Second Mean Value Theorem. If f(x) is a function of x which has a differential coefficient of constant sign for all values of x from x=a to x=b, then there is a number ξ between a and b such that abf(x)ϕ(x)dx=f(a)aξϕ(x)dx+f(b)ξbϕ(x)dx.

[Let axϕ(t)dt=Φ(x). Then abf(x)ϕ(x)dx=abf(x)Φ(x)dx=f(b)Φ(b)abf(x)Φ(x)dx=f(b)Φ(b)Φ(ξ)abf(x)dx, by the generalised Mean Value Theorem of § 160: i.e. abf(x)ϕ(x)dx=f(b)Φ(b)+{f(a)f(b)}Φ(ξ), which is equivalent to the result given.]

12. Bonnet’s form of the Second Mean Value Theorem. If f(x) is of constant sign, and f(b) and f(a)f(b) have the same sign, then abf(x)ϕ(x)dx=f(a)aXϕ(x)dx, where X lies between a and b. [For f(b)Φ(b)+{f(a)f(b)}Φ(ξ)=μf(a), where μ lies between Φ(ξ) and Φ(b), and so is the value of Φ(x) for a value of x such as X. The important case is that in which 0f(b)f(x)f(a).]

Prove similarly that if f(a) and f(b)f(a) have the same sign, then abf(x)ϕ(x)dx=f(b)Xbϕ(x)dx, where X lies between a and b. [Use the function Ψ(ξ)=ξbϕ(x)dx. It will be found that the integral can be expressed in the form f(a)Ψ(a)+{f(b)f(a)}Ψ(ξ). The important case is that in which 0f(a)f(x)f(b).]

13. Prove that |XXsinxxdx|<2X if X>X>0. [Apply the first formula of Ex. 12, and note that the integral of sinx over any interval whatever is numerically less than 2.]

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 [a,0], [0,a], and put x=y in the first. In (ii) use the substitution x=12πy to obtain the first equation: to obtain the second divide the range [0,π] into two equal parts and use the substitution x=12π+y. In (iii) divide the range into m equal parts and use the substitutions x=π+y, x=2π+y, ….]

15. Prove that abF(x)dx=abF(a+bx)dx.

16. Prove that 012πcosmxsinmxdx=2m012πcosmxdx.

17. Prove that 0πxϕ(sinx)dx=12π0πϕ(sinx)dx.

[Put x=πy.]

18. Prove that 0πxsinx1+cos2xdx=14π2.

19. Show by means of the transformation x=acos2θ+bsin2θ that ab(xa)(bx)dx=18π(ba)2.

20. Show by means of the substitution (a+bcosx)(abcosy)=a2b2 that 0π(a+bcosx)ndx=(a2b2)(n12)0π(abcosy)n1dy, when n is a positive integer and a>|b|, and evaluate the integral when n=1, 2, 3.

21. If m and n are positive integers then ab(xa)m(bx)ndx=(ba)m+n+1m!n!(m+n+1)!.

[Put x=a+(ba)y, and use Ex. 6.]

  1. The argument which follows is modelled on that given in Goursat’s Cours d’Analyse (second edition), vol. i, pp. 171 et seq.; but Goursat’s treatment is much more general.↩︎
  2. The s and the S do not in general correspond to the same mode of subdivision.↩︎
  3. All functions mentioned in these equations are of course continuous, as the definite integral has been defined for continuous functions only.↩︎

155. Differentials Main Page 162. Alternative proof of Taylor’s Theorem