INVERSE METHOD OF DEFINITE INTEGRALS. 



357 



Take (Fig. 1.) AB = 1, AH = 1 - h, or 

 BH = h both along the axis of x, and make 

 A the origin, then putting x = 0, we have 



1 -i- h 



y — jz — T7^, which is very great, and tends to be 



infinite as h approaches unity, and is represented 

 by AC\ next putting x = l — h = AH, we get the 



corresponding ordinate HE = (jZThf) ' (T+W' 

 which also tends to infinity ; lastly, putting x = l 



we have y = 



l-h 



= BD, which tends to vanish 



(1 + h)i 

 in the ultimate case representing V^. 



Now varying the parameter h so as to make 

 it approach unity, the points C and £1 recede 

 indefinitely from the axis of x, and the point 

 7> approaches it indefinitely. 



Yet the area DBACE remains constant (for 

 the integral between x = 0, and x = \ of 



{l-2^(l-2x) + A''}J 

 relative to x is evidently unity). 



And the altitude GN of the centre of gravity of this area is also 

 constant, for 



ANH H 



h!^f 





and therefore is the same as that of the parallelogram HF, when 

 AF=^AB, for the distance Gg from the axis of y 



x{\-h){\^h) \~h AH 





{l-^h(X-2x) + ¥\l 2 2' 



Hence G tends ultimately to the point g in the axis of y, which 

 shews that the area DBH'E' tends absolutely to vanish, HE' being 



3 A 2 



