Expert: Paul Klarreich - 3/19/2006

Question
Thank you for reading my email.  I am an adult doing home
study of beginning calculus.

Through the operation of integration, we are told that we can
get the formula for the area under a curve (e.g., y=x^2 is the
formula for the area under the curve y=2x).  However, nowhere
have I found any proof of this.  All calculus text books and
internet sites, that I have seen, seem to dance around the issue
of proving integration to be true.  They show Riemann sums as a
"proof" that apparently satisfies students.  But Riemann sums do
not prove integration, as far as I can see.

Someone must have proved integration.  If the proof is in some
advanced form of mathematics, then why not just say, "You, the
student, will just have to take it on faith that you get the formula
for the area under a curve through integration.  If in the future
you take such-and-such advanced mathematics, then you'll be
shown the proof.”?

Students have had the importance of proof drummed into their
heads in algebra and geometry.  Now in calculus it's sorely
lacking.

Is the proof of integration missing in all beginning calculus
texts, or am I the only one who doesn't see it?

Thank you for your time,

Wayne

Answer
You wrote:
.....................
Wayne Asks in Category Calculus:
Subject:  calculus/integation
Question:  Thank you for reading my email. I am an adult doing home study of beginning calculus.

Through the operation of integration, we are told that we can get the formula for the area under a curve (e.g., y=x^2 is the formula for the area under the curve y=2x). However, nowhere have I found any proof of this. All calculus text books and internet sites, that I have seen, seem to dance around the issue of proving integration to be true. They show Riemann sums as a "proof" that apparently satisfies students. But Riemann sums do
not prove integration, as far as I can see.

Someone must have proved integration. If the proof is in some advanced form of mathematics, then why not just say, "You, the student, will just have to take it on faith that you get the formula for the area under a curve through integration. If in the future you take such-and-such advanced mathematics, then you'll be shown the proof.?

Students have had the importance of proof drummed into their heads in algebra and geometry. Now in calculus it's sorely lacking.

Is the proof of integration missing in all beginning calculus texts, or am I the only one who doesn't see it?

Thank you for your time,

Wayne
---------------------
Hi, Wayne,
I am happy to see that your curiosity extends past 'how do I do this problem?'

Yes, indeed, someone did prove 'integration'.  Actually, two people did -- Newton and Leibniz, at about the same time, but Newton had a better press agent.

Most good calculus texts will have a fairly decent proof of the FUNDAMENTAL THEOREM OF CALCULUS, which is the thing you are talking about.  It's the one that says that:

Integration, which is the limit of the sum of pieces of an area, and

Differentiation, which is the limit of the slope of the secant line,

are inverse processes.  

Of course, you must have done some examples like: (You didn't do these? DO THEM!  They are fun to do, even though they take work, and they give you some insight.)

Using an upper sum, and the formula for the sum of the first n squares, i.e.
n(2n+1)(n+1)
------------
    6

show that the area under the graph of y=x^2 from x = 0 to x = b is equal to  b^3/3.

From which you progress easily to

'show that .... x = a to x = b is equal to b^3/3 - a^3/3'.

Which makes you feel that the THEOREM:

Lim     SUM(k=0 to n)[ f(x) delta-x] = F(b) - F(a)
n->inf

where (the fundamental fact is here) the derivative of F(x) is f(x).

Now the books are right, in a sense.  The details of the proof, if you want to do a really good job, are difficult.  So I will sketch a proof and indicate where you might have to wait for details.

---------------------
What does the proof depend on?  It's a theorem that connects a function to its derivative (backwards, maybe) so it will rest on another fundamental theorem that connects a function and its derivative.  Which theorem is that?  You guessed it -- the Mean Value Theorem, which says that:

WARNING: THE MATERIAL BELOW MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.


If f(x) is 'nicely behaved' (you look up the details) then given any a,b, there exists a number c in (a,b) where:
       f(b) - f(a)
f'(c) = -----------
         b - a

OR, in the form we will REALLY want:

         G(x[n]) - G(x[n-1])
g(r[n]) = -------------------
            x[n] - x[n-1]

In this second version,

A. the [] denote subscripts,
B. r[n] is some number in the interval  (x[n-1],x[n]),
C. G is the antiderivative of g.

----------------------------------
OK, an outline of the proof.  You have chopped your interval (a,b) into n subintervals, and the dividing points are  x[0], x[1], ..., x[n], where a=x[0] and b=x[n].

Here is your Riemann sum:

S = sum(k = 1 to n) g(r[k]) delta-x[k]

and  delta-x[k] is just  (x[k] - x[k-1])

S = sum(k = 1 to n) g(r[k]) (x[k] - x[k-1])

How is the r[k] chosen?  There are lots of ways:

Method A. Choose r[k] = x[k-1], the smallest (leftmost) x.  This is a LEFT sum.

Method B. Choose r[k] = x[k], the largest x.  This is a RIGHT sum.

Method C. Choose r[k] = that x that yields the largest g(x).  This is an UPPER sum.

Method D. Choose r[k] = that x that yields the smallest g(x).  This is a LOWER sum.

Method E. Choose r[k] = that x that makes the calculation easiest.  This is a FUN sum.

OK, OK, I made that last one up.  But I left one out, which I will include after this comment:

COMMENT: It can be proved, using limit theorems and some careful arguments, that when we finally form the limit sum, these all come out to the same limit.  The limit sum requires that we do two things:

1. Make the number of intervals approach infinity.
2. Make the size of EVERY interval approach zero.

So in that case, we can really choose the r[k] any way we like, so long as it is in the k-th interval, between x[k-1] and x[k].

If we can choose it any way we like, WE LIKE TO CHOOSE IT ACCORDING TO THE CONDITIONS OF THE MEAN VALUE THEOREM.

Method MVT: Choose  r[k] to be that value of x in (x[k-1],x[k]) that makes:

         G(x[k]) - G(x[k-1])
g(r[k]) = -------------------
            x[k] - x[k-1]

where G is the antiderivative of g.

Multiply that out:

g(r[k])(x[k] - x[k-1]) = G(x[k]) - G(x[k-1])

Then the sum will look like this:

S = sum(k = 1 to n) g(r[k]) (x[k] - x[k-1])

S = sum(k = 1 to n) ( G(x[k]) - G(x[k-1]) )

Now what does that sum really look like?  Like this:

G(x[1]) - G(x[0]) + G(x[2]) - G(x[1]) +...+ G(x[n-1]) - G(x[n-2]) + G(x[n]) - G(x[n-1])

Can you see that almost all the terms in this expression (it's called a telescoping sum) disappear, and what remains is:

G(x[n]) - G(x[0])

But do you remember what  x[0] is?  It's a!  And  x[n] is b, so the sum is:

G(b) - G(a)

And that's the theorem.

Now I hasten to state that this is not the only way to prove the theorem.  I like this proof, but a calculus book I happen to have here uses a completely different approach.

So don't say the calculus books all leave it out.  They don't.