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.]