**FTC2**

(Jimmy L.’s proof based on Braxton Collier’s insight)

[rev. 11/16/2002, 10/17/2004]

Given:

i)
*f* is a
continuous function on [*a*, *b*]

ii)
*H*(*x*) =

Prove: *H* ′(*x*) = *f*
(*x*)

Proof

i)
area of a region whose upper boundary is the graph of *f* defined from *a* to (*x* + ∆*x*) =
*H*(*x* + ∆*x*)

ii)
area of another region (whose upper boundary is also
defined by the graph of *f*) from *a* to *x*
= *H*(*x*)

iii)
*H*(*x* + ∆*x*) – *H*(*x*) = the difference between the two
areas = area of the strip that has a width of ∆x and *f* defined from *x* to (*x* + ∆*x*) as an upper boundary

iv)
This difference in areas also ≈ *f* (*x*) ∆*x*

Thus,

(a)
*H*(*x* + ∆*x*) – *H*(*x*) ≈ *f* (*x*)
∆*x*

(b)

[This is a formalization of what we meant by saying “approximately
equal” in the previous step.]

(c)

[Dividing a nonzero number into the limits of both sides of an equation is
permissible, provided the limits both exist.]

(d)

[Cancellation.]

(e)
*H* ′(*x*) = *f*
(*x*), Q.E.D.

[Limit
of a difference quotient of function *H*
is *H* ′, and the right-hand limit is
simply *f* (*x*) since plays no role there.]