AP Calculus

Special Focus: The Fundamental
Theorem of Calculus
x
∫ 2
Figure 1. F( x) = −
dt
t
. Here we illustrate the case x = –5 with F(–5) = ln(5/2).
As an illustration of these questions, consider Figure 1 above. That the input variable
of the function F is located in the upper limit of integration is a stumbling block for
students, so the graph can help students see that the variable x is sliding along the t-
axis, and there is a definite domain (in this example, the domain of x is ( −∞, 0 ) ). As x
varies in this example, the values of F(x) are 0 (at x = –2), negative (for − 2 < x < 0 ), and
positive (for x < −2 ). Requiring f to be continuous helps students to recall the definition
of the definite integral as a limit of Riemann sums, which are guaranteed to converge for
continuous functions.
A computer algebra system or a graphing calculator allows one to graph the function
F(x) and then ask what properties of F(x) are apparent. The students should be able to
determine from the graph that the function F(x) seems to be differentiable.
We can prove the antiderivative part of the Fundamental Theorem from first principles
using the definition of the derivative. This is a nice approach, since the properties of the
definite integral are used to simplify the result below. For convenience, assume f ( x) > 0
on the open interval containing a. First notice that
F x + h − F x
F′ ( x) = lim ( ) ( ) = lim
h→0 h
h→0
∫
x+
h
a
∫
x
f ( t) dt − f ( t)
dt
a
= lim
h
h→0
∫
x+
h
x
f ( t)
dt
.
h
To complete the proof, let h > 0 and bound the area representing ∫ f ( t)
dt using
x
the minimum and maximum values of the function f on the domain [ x, x + h]
, arriving
at the inequality
h ⋅ ⎛ min f t f t dt h
⎝ ⎜ ( ) ⎞
[ ] ⎠
⎟ ≤ x h
∫ ( ) ≤ ⋅ max
x
⎝
⎜
+ ⎛
t∈ x, x+ h t∈[
x, x+
h]
x+
h
⎞
f ( t) .
⎠
⎟
AP® Calculus: 2006–2007 Workshop Materials 99