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CHAPTER XVI
DEFINITE INTEGRALS: AREAS:
MISCELLANEOUS THEOREMS
b
 23 I3 8    I 7
3    3    3 3 3
 DEFINITE INTEGRALS    263
Ja b (x) dx = f (b) — f (a)
— =—[f(a)—f(b)l
J O (x) dx.
b
b c (ii)    ja c (x) dx = Ja 0 (x) dx + Jb (x) dx.
As before, if    r
dx f (x)= (x), then J (x) dx = f (x).
f:O (x) dx = f (c) f (a)
=f (c) — f (h)+f (b) —f (a)
b
=l b 0 (x) dx + Ja 0 (x) dx. (iii)    jo (x) dx = jack o (a — x) dx.
This is an example of the change of limits brought about by the substitution of a different variable for the original variable x.
If    (a — x) = y so that — dx = dy,
then    0 (a — x) dx = — 0 (y) dy.
Also when    x = o, y = a,
and when    x = a, y = o. a Jo (x)dx= rf Jo 0 (y)dy=.1 0 (a—x).—dx
a
o    'a
_— f a( a    o0(a—x)dx.
It should be noted that if the upper limit is the independent variable, the integral is not a definite integral, but simply another form of the indefinite integral.
For example,
fa 0 (x) dx = f (x) — f (a)
= f (x) + a constant = J (x) dx.

264    INTEGRAL CALCULUS
3. There is no new principle involved in the evaluation of definite integrals. Care must be taken, however, that if a substitution be made for the independent variable, the limits of integration are changed accordingly. This is particularly to be noted when the substitution turns an algebraic expression into a trigonometrical expression, or vice versa.
The following illustrative examples show the procedure to be
employed.
Example 2.
1I    x2
. o (3x + 2)2
dx.


Put 3X + 2 = y: then when x = o, y = 2; and when x = 1, y = 5. Also 3dx = dy.
The integral becomes
J y 2
    3    )2dy
2    y2    3
5y2—4y+4d
"= 27 I2    y2    y
I
1(I—4+4)dy
27 .12    y y
5
=27 [y—4 logy—yl2
I
27 [(5 -4 log 5 — 45) — (2 — 4 loge — zl
= 1 (3—41og+=-I /21—41ogD. 27    2 5/ 27 `5
Example 3.    a dx
0 a2 + x2.


r dx is a standard form and its value is I tan—1 x. !a2+x2    a    a
Therefore the definite integral
= I tan—1 x
a a    a o
= I tan—1 I — I tan—1 0 a    a

I    77'    I    71'
a 4 a    4a

ILLUSTRATIVE EXAMPLES    265


There are two points to be noted when evaluating a definite integral for which the indefinite integral is an inverse trigonometrical function. They are
(I) In no circumstances can the result be expressed in degrees, since the ordinary rules for the differentiation and integration of trigonofunctions hold only when the angles are measured in radians.
(z) The value of the definite integral is usually the smallest positive angle. If, for example, tan x is made to vary continuously from
o to I and x commences at the value o, it will end at the value': 4 similarly if it commences at the value nor it will end at the value
IT
n7r+ —.
4
The reasons for these restrictions will be more apparent when geometrical applications of definite integrals are considered. (See para. 6, later.)
Example 4.
fa    dx
Jo (a2 + x2)1.


Let x = a tan 0; then dx = a sec2 040. When x = o,    tan 0 = o and 0 = o;
and when x = a,
4Therefore the integral becomes
tan q = l and ¢ =



4 a sec
2 0 dO
IO (a2 tan2 + a2)g. I4 a sect 0d-_ /4 I
I
o a' secs    o 2
a cos
= rQ sin 0 14
0dq
= -I- I sin - — sin o
a    4
I r I
L    0 a2 V2
I
a2 V2
Note. The substitution x =y will also simplify this integral.

266    INTEGRAL CALCULUS
Example 5.
Joy sin2 xdx.
    fsing xdx    1-.2 sing xdx = fTr 4 (I — cos 2x) dx
.~o    . o = [4 (x — 4 sin 2x)" =4 [9r — 4 sin 27r — 0] =
TT,
since    sin 27r = 0. Example 6.
Prove that    2 sin" xdx = f 2 cos' xdx.
.o    .o
From (iii) above (para. 2),
foaf(x)dx= oaf(a—x)dx
and    sin x = cos (2 — x) .
IT    IT
.'. J 2 sin' xdx = f 2 sin's (- - x) dx = f2 cos" xdx.
    o    0    2    0
Example 7.
Evaluate f i x sin xdx. .o
The function x sin x is the product of two functions of x and we must integrate the expression by parts. We may obtain the indefinite integral by this method and insert the limits after the integration has been performed.
ix sin x dx = — x cos x — f (— cos x) dx =—xcosx+sinx.
n
fou x sin x dx = [— x cos x + sin x2
IT    IT    IT
_ - - cos — +sin - - ocoso+sino
2    2    2
= 0 + I — 0 + 0
= I.

INTEGRATION BY PARTS    267
4. When the function whose definite integral is required is the product of two functions of the variable, we may proceed as above, or we may adopt a more specific formula for definite integration.
Let ut and vt be two functions of a variable t, and let a and b be two constants independent of t.
Then
Juvt dt = ut J vt dt — J dtt ( J vtdt) dt.
re    r
We may write J vtdt as J vkdk; for if J vtdt = Vt then a
Jt vkdk = [Vk]t
= Vt — Va
a    a
= J vt dt — constant
=Jvtdt,
the constant being simply the constant of integration. (See para. 2.)
.'. JUVdl = ut Javkdk— Jdtt ( Javkdk) dt.
Jautvtdt = [ut J Vkdk — J dtt \Javkdk) dt]b
= ub Jbvkdk — ua Javkdk — Jt Ut vkdk) dt a    a    a dt    a
= ub J bvkdk — Ja t (I    dh) dt, dt .a
a
since    vkdk = Va — Va = o.
a
Alternatively, since
b
— f vkdk = — (Vb — Vt) = Vt — constant = Jvtdt,
.t
we may obtain
Jbutvtdt = ua Jbvkdk + Jb du_t ( vkdk) dt. a    a    a dt .t (See Actuarial Note, j.LA. vol. xLIV, pp. 403—5.)

268    INTEGRAL CALCULUS
Applying the first of these formulae to the evaluation of the integral in Ex. 7, we have
2    2
J'o x sin x dx =2j0sin x dx — ro'dx( jox sin k dk) dx,
where the "u" function is x, and the " v" function is sin x. Now
Jo sin k dk = [— coskI = — cos x + cos o = — cosx+1. o
. •    2
J x sin x dx = 2 r— cos xi 2 o — 12 o (— cosx + I) dx o    L 2 — cost+coso — L— sinx F xJ2
2
=2 Lo+11—L—sing+-2
    7T    . IT    IT
= - + sin - - -
    2    2 2
IT
= sin -
2
= i, as before.
5. The following examples are illustrative of the methods emin the evaluation of certain types of integrals.
n
The value of the function x when x = o is evidently zero (since e
e° = I), and the limit of the function when x -> co is also zero (see Ex. 3, Chap. xiii). These properties of the function enable the definite integral ( xne-xdx to be readily evaluated when n is a
0
positive integer. Thus
r    dCn
J xn e-x dx = xn (— e-x) — dx (— a-x) dx = — xne-x + n j xn-1 a-xdx.

INTEGRATION BY PARTS    269
rxne-ldx    00.
= — xne-x + n xn-le-xdx
_o    .o
= n o xn-le-xdx since [— xne x~ o is zero.
Similarly and so on.
J~
xn_1e-xdx = (n —
I)
. xn-2e-xdx, o
00 xne-xdx = n (n — I) (n — 2) ..,    e-xdx
0
=n(n—I)(n—2)... [_e]
Lt e-x = Lt I = o
x,co    x-.0 e''

and e-x = I when x = o.
xne-xdx=n (n — 1) (n — 2) ... I.
Again, consider the integral    xn-1 (I — xn-1 dx, where m and
0
n are positive and m is an integer. If we put I — x = z, then — dx = dz. The new variable z takes the values I and o when x has the values o and I respectively, and the form of the integral is


j now J — (I — z)n-1 zm-ld2'.
This is the same as J z'n-1 (I — z)n-1 dz.
0
= 91 (1 — x)m-1 + m      n _I xn (I — x)Yn-2dx
= n (I — x)m-1 + m 7t      I I xn (I — x)'"-2dx.
F
0

Changing the variable to x—which does not alter the value of
r1
the integral—the integral becomes J x'n-1 (I — x)n-1dx.
0
To evaluate the integral we proceed in the usual manner.
J xn-1 (I — x)m-1 dx = x- (I — x)m-1 — d (I — x)m-1 xn dx
n    dx    n

270    INTEGRAL CALCULUS
n
The term n (I — x)m-I vanishes for limits x = o and x = I. I    m — I
I
~G xn-I (I — x)m-I dx =    
n    to xn (I — x)m-2dx
= m — I Ill — 2 i I xn+i (I    dx,
n n+I Jo    '
similarly.
— (m — I) (m — 2) ... 2 . I r I xmttt-2dt n(n+1)... (n+m—2) .lo
(m— i)(m—2)...2.I n(n+1)...(n+yn
The above integrals are of the utmost importance in the higher branches of mathematics. They are called Eulerian Integrals,
'1
xm-I (1 — x)n-I dx .o
being the First Eulerian Integral and
lw xnC-x dx 0 the Second Eulerian Integral. The proofs above have been based
on the assumption that the indices have particular values (e.g. in I' 1
xm-I (I — x)n-Idx, m is an integer). It can be proved, however, that .o
the properties of the integrals are the same if certain of these restrictions are are removed. The First Eulerian Integral ~o x'n_I (i — x)n-Idx is a function of the positive quantities m and n and is written as /8 (m, n);
the Second Eulerian Integral is a function of n alone and is written as F (n + I). These functions are called Beta and Gamma functions
respectively.
We have
(n + m — I)! '
if m and n are positive integers.
Q (m, n) = I I
N    xm-I (I – x)n-I dx
.o
(m— I)(m—2)...2.1 n(n+1)...(n+m— 1) (m — I)! (n — I)!


AREAS OF CURVES    271
Also        F (n + i) _ / xne-adx = n (n — r) ... 3.2.1,
.o
which is n! when n is a positive integer.
Hence    p (m, n) = F (m) F (n)/F (m + n) = p (n, m).
6. Areas of curves.
Ja b
I
t has been shown that the integral    (x) dx represents the
area of the curve y = (x) between the curve, the x-axis and the two ordinates x = a and x = b. Every definite integral denotes an area, and provided that the function in question is integrable we can find areas of those parts of curves cut off by different straight lines and, in many instances, by other curves. In solving problems connected with areas it is always advisable to draw a rough graph of the curve : otherwise the true area required may not be apparent.
Example 8.
Find the area cut off by the curve Y y2 = 4x, the x-axis and the ordinates
x = o and x = 4.
.4 Area =.'o y dx
= I0 4
2 1/x dx,    since y2 = 4X
Note. Since the result represents an area, we should write ro3 square units as our answer. In practice, the words " square units" are omitted, but it should not be forgotten that this qualification always exists. If, for example, squared paper were used and -we chose an inch as our

The curve is the parabola LOK in the 0 diagram, and the area required is that bounded by the curve OL and the straight lines OX, LM, i.e. the part OLM.

Fig. 33.

272    INTEGRAL CALCULUS


unit for x (along OX) and for y (along OY), the area OML would be 104 square inches.
Similarly, if the integral to be evaluated were
1    dx
~0    — x2' the area required would be
rr    1
[sin—ix~    or (sin–1 1 — sin–1 o), 0
the value of which is 2 . The full result would be " 2 square units" or "1 (3'14159 ...) square units," so that if our units were inches the area would be 1.571 square inches, correct to three decimal places.


Example 9.
Find the area of the loop of the curve y2 = x4 (x + 2).
For real values of the variables x cannot be less than — 2. Also when y = o, x = o or — 2. Again for every value of x between o and — 2
Fig. 34.
there will be two values of y, equal in magnitude and opposite in sign.


Also    dy = -d (x2 v x     + 2) dx dx
x2
    -/    +2xVX+2. 2"Vx+2
If this be equated to zero,

x22+4x(x+2)=0;
i.e.    5x2 + 8x = o,
. x=oor—8/5.
ILLUSTRATIVE EXAMPLES 273 Of these x = — 8/5 gives a maximum value to y, and K will be the highest point of the loop.
If we integrate y between the limits — z and o we shall obtain the area cut off by the curve between the axes and the ordinate x = — 2, i.e. the area OKA. The area of the whole loop OKAK1 will be twice this area.
Therefore the area of the loop
0
= 2 Iydx= 2    x2(x+2)1dx.
_2    2
To evaluate the integral, let (x + 2)1 = Z.
Then when x = — 2, z = o, and when x = o, z =
Also    — dx = dz.
2 (x + 2)}
The required area is therefore
2 o (z2 — 2)2 z.2z dz
0
= 2 i'%12 (z4 — 4z2 + 4) 222 dz
.o
= 2 J ` 2 (2z6 — 8z4 + 8z2) dz
J o
x227 8z5 8z3 2
-2[7 — 5 + 3J0
    _ z [2 .21 8.2    8.2 ,
—    7    5 + 3 ]
which simplifies to    256 V 2
105
Note. In the above figure the area AKO corresponds to the positive value of 1/x + 2, and the area AK1 0 to the negative value. The area
0
= 2 / x21/x + 2 dx, where the 1/x + 2 means the positive value of
the square root ; there is therefore no ambiguity of sign when z is substituted for 1/x + 2.
Example 10.
Find the area between the curve y2 (1 — x) = x3 and its asymptote.
The straight line x = r is an asymptote to the curve. x cannot exceed i for real values of y, and the curve gradually approaches the straight line x = r, meeting it only at an infinite distance from the origin.
F    Is

274    INTEGRAL CALCULUS
We require therefore
211ydx=2jolx' (i — x)ldx.
The substitution is x = sin2 B and the integral becomes
2 f 2 2 sin4 0 dB. .o
Expressing sin4 B in terms of multiple angles, we have for the area required
2 f 2 2 ( — cos 20 + cos 40) d6    X, .o
=4 [8_sin2o+sin4oJo _
-4 8
3 2
17
7. Differentiation under the integral sign.
There are various devices for evaluating definite integrals where the function to be integrated is of the form f (x, k), k being in-dependent of x. A method that can often be used to advantage depends upon the process of differentiating under the integral sign.
b
Let u = J f (x, k) dx where a and b are constants independent
a
of k. Suppose that k be changed to k + Ak, so that u becomes u + Au, x remaining unaltered.
Then a+Du=bf (x,k+Ak)dx
a
and    Au = J b f (x, k + Ak) dx — JJ b
(x, k) dx
a = J [f (x, k + Lk) — f (x, k)] dx, a
Du fbf(x,k+L\k)—f (x,k)dx.
.. OkJa    Ak
f(x,k+Ok)—f(x,k)df(x,k)+a, Ak    dk

Y
    1      X
Y,
Fig. 35.
But

DIFFERENTIATION UNDER THE INTEGRAL SIGN 275
where a is a small quantity which vanishes in the limit as Ak -~ o ; ..u_ cbdf (x,k)dx+JLadx.
Ak — a dk    a
When Ak +- o the second integral vanishes, for it cannot be numerically greater than (b — a) al (where al is the greatest value of a), and a1 ultimately vanishes.
Therefore when Ak -+ o we shall have
du _    Au _ ( df (x, k)
dk o Lt o Ak — J a dk      dx'
By successive differentiation it follows that
d"'u 1b dnf (x, k) dx.
dkn    a dkn
If a or b be infinite this proof will not hold, for then we cannot say that, when Ak —# o, (b — a) al vanishes. A complete proof involves higher mathematical analysis and it will be sufficient to assume that in the examples dealt with in this chapter we may differentiate under the integral sign, even if one of the limits be infinite.
The following example is a practical application of the method. Example 11.
i01    dx =o
x"-1
n    n'
Let x = a +
z
z , where a and z are independent; then
z    o.
We may write the expression for x as
i
1 + a/z'
so that when x = 1, a/z = o andlz is infinite.      l
dx=d\a+zl d,1 a+z)
a
(a + z)2 dz.
• 1    0o    z    n-1    a
xn-1dx - -         dz
la    ,    a + z(a+z)2
azn-1 dz.
o (a + z)n T1
We have
when x = o,
r8-2

276 But
INTEGRAL CALCULUS
/
xn-1 dx = 1/n ; .o
    rco    zn-1
.I o (a + z)n+1 dz = Ilan.
Since a is independent of z, we have
    da Joy (a + z)n+1 dz = .j0 da (a +nz)n+1] dz = 0    (a + z)n+2-1 dz.
    .m    zn-1
Since    Jo (a + z)n+1 dz = I/an,
d CO zn-1    d I    I dz =    (    =-( da Jo (a + z)n+1    da an/ n a2).
d    zn-1    •oo — (n + 1) zn-1
da Jo (a + z)n+l dz = Jo    (a + z)n}2      dz,
-°° — (n + I) zn-1    I — I \
    Jo    (a + z)"+2
— dz = ( a2 , ,
    zn-1    I    I
fo (a+z)n}2dz=n(n+1)a2
    From the known integral    z t-1 Jo (a + z)n+l dz we have therefore obtained
zn
the integral I    — -1    dz.
0 (a+ z)n+2
This process may be repeated, and we shall have
ao    n—1
f: (a + z)n}3 dz = n (n +I I) (n + 2) a3'
,00    zn-1    1.2.3
l0 (a     z)n}4dz=n(n+ I) (n + 2) (n + 3) a4'
so that generally
I    zn-1    1.2.3 ... (r — 1)
10 (a + z)n+r dz n (n + I) (n + 2) ... (n + r    I) ar
(r — I)! (n — I)! I (n + r — I)! ar'
if r and n are positive integers.
Also

DOUBLE INTEGRALS    277 8. Double integrals.
The formula for integration by parts is Juvdx=ujvdx_J(u' Jvdx)dx.

If x be the function u the second term on the right-hand side is J (x' J vdx) dx, i.e. f(Jvdx)dx.
Omitting the brackets the term is JJ vdxdx. This is a form of
double integral, and its meaning is simply that we must integrate v with respect to x, and then integrate the result also with respect to x.
Thus, since    Jx9dx = xex — ex,
J J xexdxdx = J(xex — ex) dx = (xex — ex) — ex = xex — 2ex.

A more general form of double integral is the form in which there are two independent variables. If v be a function of x and y,
JI vdxdy denotes the process of integrating v with respect to x and
then integrating the new function with respect to y.
In performing the integration with respect to x, y must be asconstant, and similarly when integrating the result with respect to y, x must be assumed constant.
Example 12.
Evaluate    f[(4x + 3y2) dxdy.

Since this is an abbreviated form of f [ (4x + 3)2) dxl dy, we must first find f (4X + 3y2) dx, where y is assumed to be independent of x. [(4x + 3Y2) dx = 4 2 + 3xy2 + a.

Again, f(2x2 + 3xy2 + a) dy = 2x2y + xy3 + ay + b ;

278    INTEGRAL CALCULUS
ll(4x+3y2)dxdy =2x2y+xy3+ay+b,
where a and b are arbitrary constants.
If we had been required to integrate f f (4x + 3y2) dydx, we should have obtained, firstly,
f(4x + 3y2) dy = 4xy + y3 + c,
[(4xy + y3 + c) dx = 2x2y + xy3 + cx + d,
where c and d are constants not necessarily the same as a and b above. 9. Suppose that, in the above example, we had had to evaluate the integral between limits, so that the problem read thus :
Find the value of 113 (4x + 3y2) dxdy.
Firstly,    .1•4
(4x + 3y2) dx = [2x2 + 3xy2J 3
=2.42+3.4)12—2.323.3y2
= 14 + 3Y2.
Secondly, fI 2 (14 + 33,2) dy = [14Y + y3i 1 2
= 14.2+23- 14.I — 13
= 2I.
Now if the order of integration had been reversed, we should 1 4 '2
s I
have had to evaluate . I (4x + 3y2) dydx. In the usual manner,    [4xY    11
J (4x + 3y 2) dy =    + y,3]
=4.2X+23 -4. Ix — 13 =4x+7.    4
j 3 4 (4x + 7) dx = [2x2 + 7x] 3
=2.42+7.4—2.32—7.3 = 21, as before. and then

DOUBLE INTEGRALS    279
This leads to the general proposition :
If x and y be independent, then
Js Ja    by= JJf(x,y)dydx,
a    a a provided that the limits of x and y are independent of each other,
and that neither f (x, y) nor its integrals become infinite for any values of x and y between the limits of integration.
The proof of this theorem is difficult, the most satisfactory demonstration depending on a double summation. It will be therefore taken for granted that the proposition holds within the limitations imposed. For a rigid proof of the proposition the student should consult any recogtextbook on more advanced Integral Calculus.
10. It should be noted that where one or more of the limits of a double integral is a function of either variable, we may not take the order of integration indifferently. A common form of double integral that occurs in mean value and probability problems is one in which one of the limits for integration involves one of the variables. The integral is of the type
a—x
f (x, y) dy dx, 0 o
where the result of integrating f (x, y) with respect to y and inthe limits produces a function of x. In these problems it is necessary to adhere strictly to the order of the integration.
Example 13.
Show that
Joa Joa-x (x2 + y2) dydx + to -x Joa (x2 + y2) dxdy.
Joa Jo -x (x2 + y2) dy dx = fra [x2y + iy31:_x dx
/Da -    [x2 (a — x) + 13 (a — x)3] dx
I a
= 3 0 (a3 — 3a2 x + 6ax2 — 4 xx3) dx

280    INTEGRAL CALCULUS
I [a3 x — 3a2 x2 + 6ax3 —Ala
3    2    3    4 Jo
= ? ra4 —    
3a4 + 6a4 — 4a4 3L    2    3    4 a4
6
~a x ~a
o a— x    a
Jo    to (x2 + y2) d x d y = r    3x3 L + 2 ] 0
dy
= !a x (3a3 + aye) dy
.o
a3y ay31a-x
—    +
    3    3 Jo
=3 [a2 (a — x) + (a — x)3],
which is a function of x and is obviously not equal to the constant
a4
quantity 6 .
EXAMPLES 16
 3.
4.
5.
6.

EXAMPLES    281
 282    INTEGRAL CALCULUS
 EXAMPLES    283
do ( ibf (x, c) dx) _ IQb df ~d ,    c) dx,
where the limits are independent of c.
x dx    7ra !o I + cos a sin x sin a'
deduce    j (x2 + a2)-i dx
by differentiation under the sign of integration.
42. Prove that
Given

284    INTEGRAL CALCULUS