A to balance 
balances E0 X 
THE QUADRATURE OF THE PARABOLA 89 
And E 2 0 2 : 0 2 R 2 = QO : 0H. 2 = (n+ 1): 2, 
or QR o= JL 0 o E 2 . 
22 n+1 22 
am R 1 0 2 ), 
ended at H A , 
t is, 
It follows that 0 2 S = SR 2 , and so on. 
Consequently 0 1 R 1 , 0 2 R 2 , 0 3 R 3 ... are divided into 1,2, 3 ... 
equal parts respectively by the lines from Q meeting qE. 
It follows that the difference between the circumscribed and 
inscribed figures is equal to the triangle FqQ, which can be 
made as small as we please by increasing the number of 
divisions in Qq, i.e. in qE. 
Since the area of the segment is equal to § A Eq Q, and it is 
easily proved (Prop. 17) that A EqQ = 4 (triangle with same 
base and equal height with segment), it follows that the area 
) on; 
of the segment = times the latter triangle. 
It is easy to see that this solution is essentially the same as 
that given in The Method (see pp. 29-30, above), only in a more 
. . . + P n + 1 5 
orthodox form ('geometrically speaking). For there Archi 
medes took the sum of all the straight lines, as 0 1 R 1 , 0 2 R 2 ..., 
as making up the segment notwithstanding that there are an 
infinite number of them and straight lines have no breadth. 
1" Pfl+l 
+ AE n O n Q. • 
consisting of 
circumscribed 
er, than 
* 
Here he takes inscribed and circumscribed trapezia propor 
tional to the straight lines and having finite breadth, and then 
compresses the figures together into the segment itself bjr 
increasing indefinitely the number of trapezia in each figure, 
i.e. diminishing their breadth indefinitely. 
The procedure is equivalent to an integration, thus: 
If X denote the area of the triangle FqQ, we have, if n be 
nethod of ex- 
IqQ (Prop. 16). 
d, it has only 
>f parts in Qq 
mscribed and 
I please. This 
II the parts, as 
the number of parts in Qq, 
(circumscribed figure) 
= sum of As QqF, QR l F 1 , QR 2 F 2 ,... 
= sum of As QqF, Q0 1 R 1 , Q0 2 S,... 
-Ui+(”- 1)2 h>- 2>2 +-+- 1 4 
= * . X (Z 2 + 2 2 Z 2 + 3 2 Z 2 + ... + ?i 2 Z 2 ). 
?rZ 2 v ' 
’ 5 
1 -.OX 
a+1 22 
Similarly, we find that 
(inscribed figure) = • X { Z 2 + 2 2 Z 2 + ... + (n — 1 ) 2 Z 2 }.