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Date: 01/08/98 at 01:30:50
From: B.L.T.
Subject: Integration

Dr. Math,

I was recently doing some Calculus homework and I realized a possible 
problem in integration. I began to wonder what the integrated area is 
if the anti-derivative is undefined. I'll show you what I mean:
		INT(x^-1) from a to b

The process I would follow would be to raise the power by 1 and divide 
by the new power, thus:

My math knowledge tells me that anything divided by 0 is undefined.
Is this true? If so, how does one find the area of the function?	
Thanks a lot for your time,

Date: 01/12/98 at 12:34:42
From: Doctor Joe
Subject: Re: Integration

Dear B.L.T.,

I must praise you for your sharp observation regarding the rule that 
you have told me: Increase power and divide the expresson by the new 
increased power. But lo and behold! your rule works for only powers of 
linear factors with the power not equal to -1. Thus, the example that 
you show precisely falls under this category!

Nonetheless, we have a very neat answer. This is classical calculus 
work as derived by Euler. All derivations are due to Euler and I 
deserve no glory whatsoever in this work.

We first look at the following constant called the natural (Euler) 
number e (which may just appear somewhere on your Casio Scientific 

e is defined as the limit of the function (1+1/x)^x as x tends to 
positive infinity.

Here, I assume that you have some brief knowledge of limits (at least 
pictorially you know what it means to be the limit).

If you use the calculator, you will discover that e is approximately 

Anyway, we look at the following:

 d(e^t)/dt = lim (e^(t+h)-e^t)/h

           = lim e^t (e^h -1)/h

           = e^t lim  lim {{(1+1/x)^x}^h-1}/h
                 h->0 x->+inf

           = e^t lim    lim {{(1+1/x)^(xh)}-1}/h
                 x->+inf h->0

           = e^t lim    lim (1/h)(h + h(xh-1)(1/x)+ ...+ terms in h)
                 x->+inf h->0
           = e^t lim    lim 1 + (xh-1)/x + terms in h
                 x->+inf h->0
           = e^t

It follows that d(e^t)/dt = e^t.

Now, if x = e^y, it follows from implicit differentiation that

        1 = e^y dy/dx.

This is the crucial part. Since the exponential function x(y) = e^y 
is a one-to-one and onto map from [0,inf) to [1,inf), we may define 
an inverse function denoted by ln

such that ln: [1,inf) --> [0,inf)

and moreover, ln(e^y) = y and e^(ln x) = x.

This ln function is actually none other than the natural log (i.e. log 
based e).

Thus, 1 = e^y dy/dx means dy/dx = e^(-y) = 1/x
which further implies that d(ln x)/dx = 1/x.

Now, the rest is up to you.

-Doctor Joe,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Calculus

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