## Folios 106-107: AAL to ADM

Ockham Park

Ripley

Surrey

4^{th}July

Dear M^{r} De Morgan. You are perhaps surprised

that I have not sooner troubled you again.

And you may think it a very __bad__ reason to

give, that I have done nothing. We returned

here on Tuesday, & __now__ I am working away

famously, & hope I have before me 7 or 8 months

of __ditto__. You left me at page 106. I remember

your enquiry if I were s__ure__ that I understood

\( \int_b^{b+k}fx\times\frac{dx}{dt}\) as developped [\textit{sic}] in pages 102, 103. I

answered confidently, that I did. I now enclose

you my own development of this Integration, that

we may be q__uite__ certain of my comprehension of

[something crossed out]} it. On the other page of my

sheet, is the __application__ of it to \(\int udv=uv-\int vdu\)

(page 105); & to \(\int_a^x\frac{1}{v}\frac{dv}{dx}dx\) (page 107).

I have now two questions to propose.

I differ from y__ou__ in my development of \(\int\frac{1}{1-x}dx\)

(see page 107)**[106v] ** I cannot see why the Constant \(C\) is omitted

in t__his__ more than in \(\int\frac{1}{1+x}dx\) .

I subjoin __my__ development: Let \(v=1-x\)

\( \int\frac{1}{1-x}dx=\int\frac{1}{1-x}\times-(-1)dx\) (which is only

another way of writing \(\int\frac{1}{1-x}.1.dx)\)

And as \(\frac{dv}{dx}\) or \(\frac{d(1-x)}{dx}=-1\), we may in the

above substitute \(\int\frac{1}{1-x}dx=\int\frac{1}{1-x}\times-\left(\frac{d(1-x)}{dx}\right)dx\)

Or \(\int\frac{1}{v}dx=\int\frac{1}{v}\times-\frac{dv}{dx}dx\)

\(=\int\frac{1}{v}\frac{dv}{dx}.(-1)dx\) which by

\( \int budx=b\int udx\) (see page 105) is \(=(-1)\int\frac{1}{v}\frac{dv}{dx}dx\)

or \(=-\int\frac{1}{v}\frac{dv}{dx}dx\)

Now since by line 4, \(\int\frac{1}{v}\frac{dv}{dx}dx=\int\frac{1}{v}dv=\)

\(=\log v+C\), it follows that

(--- this same expression) must \(=-(\log v+C)\)

\(=-(\log (1-x)+C)=-\log (1-x)-C\)

\(=\log\frac{1}{1-x}-C\) **[107r] ** Now how do you get rid of \((-C)\) ?

My second question is unconnected with any

of y__our__ books. But I think I may venture to

trouble you with it. In the two equations,

\(V=gT\) (1)

\(S=\frac{1}{2}g.T^2\) (2)

which you will at once r__ecognise__, I want to

know how (2) is derived from (1).

Will you refer to Mechanics (in the Useful

Knowledge Library), page 10, __Note__, which is

as follows, ''Let \(S\) be the space described by the

''falling body. \(V=\frac{dS}{dT}=gT\) . Hence \(dS=gT\,dT\),

''which being integrated gives \(S=\frac{1}{2}g.T^2\) .''

Now can I ['as yet' inserted] understand this application of

Differentiation & Integration?

I conclude that \(\frac{dS}{dT}\) here means

Diff. co of \(S\) with respect to \(T\), \(S\) being (by

Definition & Hypothesis) a function of \(T\), & of \(V\)

I know that \(V=gT\)

And that \(V=\frac{S}{T}\) But I neither see how

\( V=\frac{dS}{dT}\), nor how the subsequent Integration applies.**[107v] ** The object, I need not say, is the solution of

\( S\) .

I mean to work very hard at my Chapter on

Integration &c, now. And I hope this

summer & autumn will see me progressing

at no small rate.

How is the Baby? And does M^{r} De

Morgan enjoy Highgate? __I__ ['am' inserted] enjoying the country

not a little, I assure you.

Yours most truly

A. A. Lovelace

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