Reproduction of the Fundamental Theorem ofCalculus

John Elijah Spraul

October 15, 2015

Of all the theorems available, the Fundamental Theorem of Calculus isthe one I chose to recreate. I picked it because it is both short and interesting.Here it goes.

Theorem 1 (The Fundamental Theorem of Calculus1). If f is continuouson the interval [a, b] and f(t) = F ′(t), then∫ b

a

f(t) dt = F (b)− F (a) (1)

To understand the Fundamental Theoreom of Calculus, think of f(t) =F ′(t) as the rate of change of the quantity F (t). To calculate the total changein F (t) between times t = a and t = b, we divide the interval a ≤ t ≤ b inton subintervals, each of length 4t. For each small interval, we estimate thechange in F (t), written 4F , and add these. In each subinterval we assumethe rate of change of F (t) is approximately constant, so that we can say

4F ≈ Rate of change of F × Time elapsed.

For the first subinterval, from t0 to t1, the rate of change of F (t) is approxi-mately F ′(t0), so

4F0 ≈ F ′(t0)4t.

Similarly, for the second interval

4F1 ≈ F ′(t1)4t.

1This result is sometimes called the First Fundamental Theorem of Calculus, to distin-guish it from the Second Fundamental Theorem of Section 6.4.

1

Summing over all the subintervals, we get

Total change in F (t)between t = a and t = b

=n−1∑i=0

4F ≈n−1∑i=0

F ′(ti)4t.

We have approximated the change in F (t) as a left-hand sum.However, the total change in F (t) between the times t = a and t = b

is simply F (b) − F (a). Taking the limit as n goes to infinity converts theRiemann sum to a definite integral and suggests the following interpretationof the Fundamental Theorem of Calculus:2

F (b)− F (a) =Total change in F (t)

between t = a and t = b=

∫ b

a

F ′(t) dt.

In words, the definite integral of a rate of change gives the total change.This argument does not, however, constitute a proof of the Fundamental

Theorem. The errors in the various approximations must be investigatedusing the definition of limit. For a proof using the Mean Value Theorem, seethe online supplement at www.wiley.com/college/hugheshallet.

Example 1. If F ′(t) = f(t) and f(t) is velocity in miles/hour, with t in

hours, what are the units of∫ b

af(t) dt and F (b)− F (a)?

Solution: Since the units of f(t) are miles/hour and the units of t are

hours, the units of∫ b

af(t) dt are (miles/hour) × hours = miles. Since F mea-

sures the change in position, the units of F (b)−F (a) are miles. As expected,

the units of∫ b

af(t) dt and F (b)− F (a) are the same.

2We could equally well have used a right-hand sum, since the definite integral is theircommon limit.

2