AP Calculus

Special Focus: The Fundamental
Theorem of Calculus
Functions Defined by a Definite Integral
Mark Howell
Gonzaga College High School
Washington, D.C.
You are about to embark on a journey toward a great achievement in the history of
mathematics: the Fundamental Theorem of Calculus. In preparation for that journey, it
is essential that you become familiar with a new type of function, where the independent
variable is a limit of integration.
2 1.
5
Recall that a definite integral such as ∫ dt is a real number, in this case
0 2
1+
t
approximately 2.165. If you change the upper limit of integration from 2 to 5, you get a
5 1.
5
different real number that is ∫ dt ≈ 3.
469. In fact, you could make that upper
0 2
1+
t
limit any real number and get a different value for the integral. If you let the limit of
x 1.
5
integration be the independent variable, you can define a function g( x)
= ∫ dt.
0 2
1+
t
x 1.
5
Notice that if x < 0, then g( x)
= ∫ dt is negative. One way to see why is by
0 2
1+
t
x 1. 5
0 1.
5
0 1.
5
recognizing that ∫ dt = −
0 ∫ dt. If x < 0, the value of
2 x
∫ dt is
2
x 2
1+
t
1+
t
1+
t
positive since the lower limit of integration is less than the upper limit, and the integrand
x 1.
5
is positive. Therefore ∫ dt is negative.
0 2
1+
t
In this activity, you will explore several functions of the form
are constants.
∫
x
a
k dt, where a and k
x
First, consider the function f0 ( x ) = ∫ 1.
5 dt. For x ≥ 0 , the function f measures the area
0
under the graph of y = 1.5 from 0 up to x. (The notation f0 ( x) is used to emphasize that
f, 0, and x are the key elements in this definition: f is the integrand, 0 is the lower limit,
and x is the upper limit.)
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AP® Calculus: 2006–2007 Workshop Materials