GEOMETRICAL ILLUSTRATIONS. 79 



then, from the last equation in Art. 101 it follows, whatever 

 be the value of a and a + h t that F(a + h) F(a) = 0, 



therefore F(a + K) = F(a). 



Hence the function F(x) has always the same value whatever 

 be the value of the variable ; that is, it is constant with 

 respect to x, or in other words does not depend on ar. 



From this it follows, that two functions which have the 

 same differential coefficient with respect to any variable can 

 only differ by a constant. For the differential coefficient 

 of the difference of these functions being always zero, it 

 follows from what we have just proved that this difference 

 is a constant. 



103. The result of Art. 101 admits of the following simple 

 geometrical verification. 



We have already shewn, Art. 43, that if u represent the 

 area contained between the 

 axes of x and y, the ordi- 

 nate y, and any curve, then 



du 



Let u = F(x), and therefore 

 y = F'(x) is the equation to the curve; let OM=a, MN=h', 



then area OAPM=F(a), 



area OAQN= F(a + A), 

 therefore area PQNM = F (a + h) - F (a). 



Now it is obvious that a point R must exist between P and Q, 

 such that, drawing the ordinate RL, 



the rectangle RL.MN= the area PQIV1 

 But RL = F'(a+6h}, 



where 6 is some proper fraction ; therefore 



7\ _ TJT / \ 



