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CHAPTER XIV
INTEGRAL CALCULUS
DEFINITIONS AND STANDARD FORMS
I. If values of the continuous function y = f (x) be given for equidistant intervals, and if (x) be a function such that
    f(x+h)f(x)=(x),  
then
where
I.
E
a
(x) = f (a + nh) — f (a),
b = a + (n — I) h
(p. 97).

This gives an expression for the sum of the values of (x) for values of x differing by the constant finite difference h. We may obtain in a similar manner the limit of the sum of the values of
(x) when the difference h tends to zero.
Let    (x) = df (x) = Lt f (x + h) —    f (x)
dx    j,_>-o    h

ck (a) = f(a+h)—f(a)+ h

where 7), tends to zero as h tends to zero.
h(a)=f(a+h)—f (a)±hl1.
Similarly,
h~(a+h) =f(a+2h)-f(a+h)+hp7h, h(a+2h)=f(a+3h)-f(a+2h)+hn3, ........................
hq(a+n — rh)=f (a + nh) — f (a + n — rh)+h~7n. On summing:
h[0(a)+(a+h)+6(a+2h)+...+6(a+n1h)]
=f(a+nh)f(a)+h++7)3+...+nn)• If the n small quantities are all numerically less than 77 , then
h(n,+~7z+~7s+...+7)n)<hn~.
Then we may write

224    INTEGRAL CALCULUS
If now b — a = nh, so that the product of n and h is always finite, then the limit of hnl) as h -3 o is zero.
Lth(m+12+7),+...+~In)=o, whenh —> o.
I.e. Lth[O (a)+cb (a+h)+ck (a+2h)+...+ck (a+n—Ih)]
h--~0
= f (a + nh) — f (a)
= f (b) — f (a).
The limit

Lt h[O(a)+(a+h)+cb(a+2h)+...+(a+n—Ih)]
h 
b
is denoted by the symbol {    (x) dx and is called the definite
.a
integral of ck (x) with respect to x, between the limits x = a and x = b.
Corresponding to the symbol D for the operation of differentiathe symbol I is sometimes used to denote integration with respect to a variable.
b
2. In finite differences we can find the sum E F (x) only when we
a
know (or can obtain) another function f (x) such that Af (x) = F (x).
b
Similarly we can evaluate ( cb (x) dx in general only if the differ
a
ential coefficient of a known function f (x) is (x). In other words,
corresponding to the relations
A f (x) = F (x),
and    f (x) = EF (x),
we have    Df (x) = (x),
and    f (x) = 10 (x).
Just as    DE    I,

so    DI -i.
Again    A { f (x) + c} = F (x),
where c is a constant independent of x, and
f (x) + c = EF (x). Also    D {f (x) + c} = c6 (x).
.'. f (x) + c = Ic6 (x).

INDEFINITE INTEGRAL    225
Therefore the process IDf (x) does not reproduce f (x) but f (x) + c, and I and D are not commutative.
It is evident that corresponding to E = 0-1 we have the analogous relation I D-1, where it is to be remembered that
Ick (x) = f (x) + c.

but only with the initial problem of determining this inverse function, we are said to integrate cl. (x) with respect to x. In that event we find the indefinite integral and write the integral function


as ( (x) dx. In the same way as dx represents the operation of finding the differential coefficient of a function of x with respect to x, so J dx represents the operation of finding the integral. The
symbol J is meaningless by itself, and we must be careful always to associate with this symbol the dx which renders it intelligible.
In finite differences the number of functions which are immediintegrable is strictly limited : in the infinitesimal calculus a far larger number of functions exists such that, given i (x), we can find D 1~ (x). Every function of x is not integrable (in the calculus sense), and it is only by application of the known properties of the differential calculus that it is possible to evaluate J (x) dx.
226    INTEGRAL CALCULUS
we complete the set of inner rectangles P„ K1M1117o, P1 K2 M2M i Pn_1K„M„Mn_1 it is evident that the sum of these rectangles is slightly less than the area cut off by the curve, the ordinates Po Mo, P„ M„ and the x-axis.
Again, if we complete the set of outer rectangles of which QoP1M1Mo is the first and Qn_1P„M„M„_1 is the last, the sum of
Mo M, d12 1113
Fig. 31.
these outer rectangles will be slightly greater than the area of the curve PoP„ M,,, M0 .
Now the difference between the set of outer and of inner rectis evidently the sum of the small rectangles QOPo K1P1, etc., and this sum is h (P„M„ — PoMo), since the rectangles are all of base h. Since P„M„ — Porto is finite, the limit of h (P,,M,, — P01110) as n tends to infinity is zero. In other words, as h -~ o the differbetween the area PoP„M„Mo and the sum of the rectangles PoK1MiMo, P1K2M2M1i ...    117„Mn—1 tends to zero.
But P0K1M1M0=M0M1.P0M0=hf (a)
P1 K2 M2 M1=M1M2.P1M, = hf (a + h)
Pn—1KnM„Mn—1=Mri 1Mn Pn 1Mn—1=hf (a + n — 1h).

GEOMETRICAL INTERPRETATION OF AN IrNTEGRAL 227
Area P0P„M„M0 = Lt h a+~~ i>h    t,
f (x) = J f (x) dx
h—>0    a    a
where    b = a + nh.
We may therefore define the definite integral as the area of the curve y = f (x) between the curve, the ordinates x = a, x = b and the x-axis.
5. Alternatively we may proceed as follows :
Let PQ be the curve y = f (x) and let the area PHML be z. If H be the point (x, y), so that a small increase in the length OM,
D H     K
P
O L    M N X Fig. 32.
namely MN, may be denoted by Ox, then DN = y + Ay and the area PDNL = z + Oz.
It is evident from Fig. 32 that the rectangle
CDNM= DN.MN= (y+ Ay) Ax;

the area    HDNM = Az;

and the rectangle    HKNM = y:,x.

.. (y + Ay) Ox > Oz > yLx,

or    y + Ay > Az/Ax > y,

when Ox -~ o, y + Ay y, since Ay tends to zero as Ox tends to
zero.
Oz dz
c Lto Ox dx'
dz and the area z = Jydx.
Y
Q
Also
15-2

228    INTEGRAL CALCULUS
The arguments in this paragraph and in para. 4 postulate a concave curve. Similar arguments apply to a convex curve. An ordinary curve having points of inflexion can be broken up into portions concave or convex as the case may be.
6. The definition of a definite integral enables us to represent in a convenient form the limits of the sums of certain series when the number of terms tends to infinity.
Example 1.
Obtain as a definite integral
-/ I    I
    nLtw [ V ff2 - 12 ~V + n2 - 22 + ... + A/n2 -    (n -    1)2 .
The expression in brackets may be written as
E= - •      +-     +...+I      I     ,
(i) follows directly from the fact that if O. (x) = df (x) dx '
then    d dx(x) = a d(xx) = a (x), Ja(x) dx = af (x)=aJ (x) dx.

STANDARD FORMS    229
(ii) dx(fudx+ fvdx+Jwdx+...)=u+v+w+..., Judx+ Jvdxf Iwdxf...= J(u+vfwf...)dx.
By the use of these theorems and the simpler standard results that have been obtained by direct differentiation of well-known forms, various integrals can be written down at once.
For example,
dxn+1    d xn+l
dx = (n + I) xn or dx n + I = xn'


xn+l
.. Jxndx =                (1). n + I
d xn+1
Since dx {n I + c = xn, where c is any constant, the in
xn+l
definite integral J x'dx is n + + c. Strictly speaking, in evaluating indefinite integrals the arbitrary constant should always be added to the result. The constant of integration will be omitted in the
following examples, but wherever there is an indefinite integral the presence of the constant is to be inferred.


d
dx log x x,


dx
x = log x     (ii).
From the two theorems above it is evident that j (axn+b)dx=a Jxndx+b)dx
xn+1

=an+I+bx
unless n is — I,
and    J(x+b)dx=a I x+b jdx
    = a log x + bx     (iv).

230    INTEGRAL CALCULUS dex x=
~
d    '
    f exdx = ex     (v).
d dx sin x = cos x and dx cos x = — sin x,
J cos x dx = sin x
J sin x dx = — cos x
dx tan x = sect x and dx cot x = — cosec2 x,
J sec2 x dx = tan x
J cosec2 x dx = — cot x
    (vi).
    (vii).

dx sin-1 x =
I
— x2
and dx cos-1 x = —    — x2'
dx
= sin-1 x or — cos-1 x     (viii).
..    Vi — x2
These two apparently different results are the same, the difference being in the constant of integration. Let cos-1 x = a so that

7T
cosa=x; then sin(n2—a =x.
IT
.. sin-'x=n --a 2
IT n-=cos-1x
2
= constant — cos-' x.
The above are the principal standard forms, and by the use of these forms in conjunction with methods which will be outlined later a large number of different forms of functions can be in(see Chapter xv).

ILLUSTRATIVE EXAMPLES 231 8. Some simple functions that can be integrated directly from the standard forms are given below.
Example 2.
Find r dx x2 + a2. If we differentiate tan-1 x we obtain - + I . Let us see the effect of differentiating tan-1 kx.
dx
d
tan-1 kx = ---d kx) tan-I kx d dx x)
I
I +k2x2k.


This is almost in the form required if we replace k by I . Then a
x

    tan-1 - = a    I
dx
a _ x2 + a2 '
    dx _ I    x
-I
x2+a2 a tan -a'
We cannot immediately recognize x2    a2 as the differential coefficient of a known function. If, however, we express - 2 I a2 as
I    I 1 2a x — a x + aJ'
we see at once that each of the component fractions is the derivative of a logarithmic function of x.
1 dx
= log (x — a), x — a
f dx = log (x + a); x + a

d tan I x = I dx    a
I    I
=a    ,
x2 a    x2 + a2
I +
a2

232    INTEGRAL CALCULUS
f dx    I / I    I x2—a2 2a, x—a x+a I [log (x — a) — log (x + a)]
Ito x—a = 2a gx+a'
Similarly
f dx    i    I    I a2—x2 2a.~~a—x+a+x~dx = 2a [— log (a — x) + log (a + x)]


=Ito a+x
2a ga — x'
Example 4.
Find rsin nOdO and [cos nO de.
d0 sin n0 = d (no) sin nod do )
= cos nO.n = n cos nO,


(cos nO dO = I sin nO. n


(sin nO dO = — cos nO. n
Note: sin"' B and cos"' 0 are not immediately integrable. If, however, we express these functions in terms of multiple angles we can at once write down their integrals.
Sint 0 = (I — cos 29).


f sine OdO = f (2 — 2 cos 20) de


= 0 — 2.1 sin 20
= 2B—4 sin 20.
sin 30 = 3 sin 0 — 4 sin3 0,
,', sin3 0 = * sin B — i sin 30.


fsin30,do= /(*sin0— f sin30)do
_ — i cos o + cos 30,
Similarly
E.g.
Again
and so on.

EXAMPLES    233 EXAMPLES 14
1. Denoting fur dx by lux express the operator I in terms of the operator A and show that, as far as first differences,
I=E+I —A
-- .
2 12
3. 5.
7. 9•
Integrate with respect to x:
3xn ;    e2x.
4. sin x; cos x; cosec2x. 6. sin 2x; cOs 3x; sec2 4x.

3X—2 ax2+bx+c x2;    xn
(x2 + a2)2 ; a + bx + cx2 ; (x + I)—3.
ax; ax + b; ax + bx + c.
I    I    1
x + 1' (I — x)2' (a + 3x)5
1    x2—x+1 (x + I) (x — I)' x (x2 + 1)
8.
a    I
VI—x2; "VI - -a-2X
I r. Express cos 3X in terms of powers of cos x and hence integrate cos3 X.
 234    INTEGRAL CALCULUS
29. Prove by differentiation that log l(x    + Vaz —    az)1 is the in
tegral of
'\/x2 — a2 .
20. Integrate    /
(i) 2 1/x — 2x2; (ii) 1/x (x5 + 3) ,
and verify the result by differentiation.
21. v dt ' f dt and s = o when t = o.
Prove that    (i) v = u + ft, where u is constant;