[152r]

 Ockham Park
         Surrey

 Dear M^r De Morgan.  I am indeed extremely
obliged to you for all your late communications.
In two or three days more, I shall have
several observations & to send you in reply
to some of them.
My object in writing today, is to make another
enquiry concerning the substitution of \(\varphi_1(a+h)-\varphi_1a\) 
for \(\varphi(a+h)-\varphi a\) in page 100, for I perceive
on carefully examining the passage, that I do
not quite understand it

 \(\varphi_1x=\varphi x+C\), which ['last side' inserted] means the Primitive Function ['of \(\varphi'x\) ' inserted]
and the Primitive Function means the Function
which differentiated gives \(\varphi'x\)
Therefore \(\varphi_1(a+h)=\varphi(a+h)+C\) 

  And \(\varphi_1a=\varphi a+C\) 
Consequently \(\varphi_1(a+h)-\varphi_1a=\varphi(a+h)+C-(\varphi a+C)\)
                                                         \(=\varphi(a+h)-\varphi a\) 

[152v] This is my version of it.  But you tell me,
\( \varphi_1(a+h)=\varphi(a+h)+C\) 
 \(\varphi_1a=\varphi a\), (which ought to be I say
\(\varphi_1a=\varphi a+C\) )

From which we should have,
  \(\varphi_1(a+h)-\varphi_1a=\varphi(a+h)+C-\varphi a\) 
Consequently \(\varphi_1(a+h)-\varphi_1a\) is not equal

  to \(\varphi(a+h)-\varphi a\) as is required
to be proved, but is \(=\varphi(a+h)-\varphi a+C\) .

I cannot unravel this at all.
Second^ly : [something crossed out] I do not see why the Indefinite
Integral only is \(=\varphi x+C=\) Primitive Function.

 of \(\varphi'x\) 
The argument at the top of page 101 seems
to [something crossed out] me to apply equally to the Definite Integral

As follows : It is proved that

 \(\varphi_1(a+h)-\varphi_1a=\int_a^{a+h}\varphi'x.dx\) 
\( \varphi_1a\) is just as much here an arbitrary Constant
as it is in \(\varphi_1x-\varphi_1a=\int_a^x\varphi'x.dx\) 
Therefore \(\int_a^{a+h}\varphi'x.dx=\varphi_1(a+h)+C_1\)
                                     \(=\varphi(a+h)+C+C_1\)
                                     \(=\varphi(a+h)+\text{an arbitrary Constant}\) 

[153r] just as with \(\varphi_1x-\varphi_1a=\int_a^x\varphi'x.dx\) 

Thirdly : With respect ['to' inserted] the assumption that when
\( a\) is arbitrary, then any function of \(a\), say \(\varphi a\),
is also arbitrary or may be anything we please,
seems to me not always valid.
For instance if \(\varphi a=a^0\), it must be
always \(=1\) .  We may assume \(a=\) anything we
like, but \(\varphi a\) will not in this case be
arbitrary.
It is curious how many little things I [something crossed out] discover in
this Chapter, which in looking back upon
them, I find I have only half-understood.

I shall be exceedingly obliged, if you
can answer these points soon; I think a
word almost may explain them, & they rather
annoy me.

Believe me, with many thanks

 Yours very truly

 A. A. Lovelace