AP Calculus

Special Focus: The Fundamental
Theorem of Calculus
In this question we are given a function defined by a definite integral. One difficulty for
students in this type of question is computing g( − ) = − 1
1 f ( t)
dt because the upper limit
is less than the lower limit, and the integrand is described by a graph, not a symbolic
function. It is probably best to have the students use the properties of definite integrals to
switch the order of the limits to get the integral in “standard” form, then evaluate the new
definite integral using the geometry of the triangle:
−1
0 ⎛ 1 ⎞
g( − 1) = ∫ f ( t) dt = − ∫ f ( t)
dt = − ⋅ ⋅
⎝
⎜
⎠
⎟ =
−
2 1 3 − 3 0
1
2 .
The key to answering parts (b) and (c) is using the FTC to recognize that g′ ( x) = f ( x).
Therefore information about the behavior of the function g can be read from the
graph of f. For example, the function g is increasing on the interval − 1 < x < 1 because
g′ ( x) = f ( x) > 0 on this interval, and the graph of g is concave down on 0 < x < 2
because g′ ( x) = f ( x) is decreasing on this interval.
For similar questions, see 1995 AB6, 1995 BC6, 1997 AB5/BC5, 1999 AB5, 2002 (Form
B) AB4/BC4, 2003 (Form B) AB5/BC5, and 2004 AB5.
2003 AB4/BC4
∫
0
Let f be a function defined on the closed interval −3 ≤ x ≤ 4 with f ( 0) = 3.
The graph of
f ′, the derivative of f, consists of one line segment and a semicircle, as shown above.
(d) Find f(–3) and f(4). Show the work that leads to your answers.
Here the Fundamental Theorem can be used to compute values for f since we are given
an initial value and a rate of change via the graph of the derivative f ′. Hence
0 1
f ( 0) f ( 3) f ( x) dx ( )( ) ( )( )
2 1 1 1
2 2 2 3
− − = ∫ ′ = − = − , and so f f
−3
2 ( − 3) = ( 0)
+ 3
2
=
9
2 ;
80
AP® Calculus: 2006–2007 Workshop Materials