Area Under the Curve y = x^0

Date: 02/14/2005 at 05:55:48
From: Tasha
Subject: Calculus: calculating area

Sketch the graph of y = x^0, where x e [0,1], and calculate the exact
area.

Basically I sketched the graph of y = x^0....then I pinpointed the
condition where x e [0,1], but then the area must be 1 since the top
and bottom are both 1....?

I don't know if I'm doing it right...so I'm writing to check my answer
and that's all I've done so far!  Thank you for your time.

Date: 02/17/2005 at 16:50:57
From: Doctor Ricky
Subject: Re: Calculus: calculating area

Hey Tasha,

Thanks for writing Dr. Math!

So you are looking to calculate the area under the curve y=x^0 between 
[0,1].  First, you must notice that when x = 0, y is undefined since 
0^0 is undefined (I will include a quick proof of this at the end).  
Therefore, we will need to use an improper integral.  For a visual, 
let's see what we're calculating.

          2 _ |
              |
          1 _ |___      y = x^0         /// is the area we're finding
              |///|
      |---|---|---|---|
     -2  -1   0   1   2

The solution you came up with calculates the area of the 1x1 square 
surrounding the area, which is correct.  The area under the curve is 
1.  Now we can calculate it directly using our improper integral.

Since x cannot equal 0, therefore our improper integral is:

          / 1                 / 1
    lim   |   x^0 dx =  lim   |   1 dx    
    k->0  / k           k->0  / k

We can substitute 1 for x^0 now since x doesn't equal 0 (due to our 
limit).  Therefore, we have:

          / 1                 |1
    lim   |   1 dx =  lim   x |  = lim  1 - k = 1  
    k->0  / k         k->0    |k   k->0

We can now see that our total area under the curve 

    y = x^0 , x e [0,1] 

is in fact 1.

In case you are curious whether 0^0 really is undefined, you can read 
about it in the Dr. Math Archives at:

    http://mathforum.org/dr.math/faq/faq.0.to.0.power.html 

Read the information there and make your own decision on whether 
0^0 = 1, is indeterminate or undefined.  I personally like to think of 
it as undefined rather than equal to one because it just seems a 
little more interesting.  As I stated above, here's a proof that I 
like to use when people aren't sure how it could be undefined.  

Enjoy!


     *Proof that 0^0 is undefined*
     -----------------------------
     Note that (a^m)/(a^n) = a^(m-n) , a,m,n are real numbers, m,n 
                                       not zero.
     Assume m = n.

     Therefore (a^m)/(a^m) = 1 <=> a^(m-m) = a^0 = 1

     Assume a = 0.

     This means that (0^m)/(0^m) = 0^(m-m) = 0^0.

     However, since m is not zero, 0^m = 0.

     Therefore, 0/0 = 0^0.

     However, since 0/0 is undefined due to division by 0, 
     then 0^0 is also undefined since they are equal.

Thanks again for writing to Dr. Math and feel free to ask any other 
questions you have about this question or anything else!

- Doctor Ricky, The Math Forum
  http://mathforum.org/dr.math/