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On Integration by Series.

When the integral of a proposed function cannot be exactly determined, we must have recourse to approximations. Thus in order to find

fx dx

where X is a function of x, we must develope X in a series according to as cending or descending powers of x, and then multiplying each term by dr integrate them in succession. For example, we know that

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Let P be the integral of X dæ a function of x, and C the arbitrary constant which we must add in order to render the result perfectly general, we have

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So long as this calculation is altogether abstract, C may have any value whatever; but when we wish to apply this integral to the solution of some given problem, the constant C ceases to be arbitrary and must answer certain condi tions.

Thus, for example, if it be required to determine the area PP' M/M A included between the ordinates MP, M'P', which correspond respectively to the abscissas a and b, since we have

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A

A

M M'

But since the required area P+C commences when a = AM = a, A ought to be 0 when we make x = a in P + C, or

Q+ C = 0

Q being the value which the function of æ represented by P assumes when aa, hence we find

C=-Q

whence the area A = P-- Q.

It only now remains to substitute b for x, and we shall have the area included within the prescribed limits. We shall have several examples in what follows.

CHAPTER IV.

APPLICATION OF THE INTEGRAL CALCULUS TO Finding the LENGTHS AND AREAS OF CURVEs, and the SURFACES AND VOLUMES OF SOLIDS OF REVOLUTION.

I. THE RECTIFICATION OF CURVES.

We have seen in p. 747 of the differential calculus, that if s represent the are of a curve,

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dy2
dx2

= √dy2 + dx

and we shall now apply this formula to a few examples.

(1.) To find the length of the arc of the common parabola. The equation is y2 = 4mx, where 4m is the parameter.

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=

2m

2m

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4m

√y2+4m2

¡√y2 + 4m2 + m log (y + √y2 + 4m2) + C

If we suppose the arc to be measured from the vertex; then when y=0, s=0, and .. 0=0+ m log 2m + C .. C =

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m log 2m, and therefore

+ m log y + √y2+4m2

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2m

x6 + ....) dx

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205 5

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.....

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In a similar manner we might obtain the following results:

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We may use any of these results for the rectification of the circle; but those are to be selected which are best adapted for facility of computation. We shall take the result in equation (5), and therefore by equation (A) we have

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.. = 3.1415926536 = semicircumference to radius unity.

Hence the circumference of a circle whose diameter is unity is 3.1415926536, which is true as far as nine decimal places.

(3.) To find the length of the arc of a cycloid.

Here y√2rx x+vers, is the equation of the curve.

dy r-x

=

dx √2rx

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.. ds = dx 1+ = √2r ·

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When x = 0, s = 0 .. C= 0, and when x=2r; then

semicycloidal arc = 2√√4r2 = 4r

and the whole length of the cycloid is = 8r = 4 times the diameter of the generating circle.

(4.) To find the length of the arc of an ellipse.

Here a2y2 + b2x2 = a2b2; whence, if a2 — b2

...

a2

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=1

b2
a2

.. ds = dx

1 + x2( e3 − 1)

a2

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=

.. ds =

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x2 - a2

where xav.

The numerator must now be developed by the binomial theorem, and the several terms of the series being multiplied by dv, divided by √/l--v2, and integrated between any proposed limits, will give the length of the elliptic arc required.

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