AP Calculus

Special Focus: The Fundamental
Theorem of Calculus
Students should be given many exercises like these four examples, and teachers may need
to supplement the students’ textbook to emphasize these ideas.
Domain Questions
Where are these rules valid? The rules are only valid on intervals on which f is continuous
x 1
and which contain both limits of integration. Thus if g( x)=
∫ dt , the domain
1 2
t − 4
of g is only the interval − 2 < x < 2; it is not the set ( −∞, − 2) ∪ ( −2, 2) ∪ ( 2, ∞)
. The
1
domain of g is different than the domain of the integrand. Thus g′ ( x)
= only
2
x − 4
for − 2 < x < 2.
Here varying the lower limit of integration can change the domain of the function; for
x 1
1
example, the domain of h( x) = ∫ dt is the set ( 2, ∞ ), and h′ ( x)
= only
5 2
2
t − 4
x − 4
for x > 2.
Integration by Substitution
The technique of using a generic antiderivative, as we did in example 2 and example 4
with h( t), can also be useful when dealing with integration by substitution. In terms
of antidifferentiation, if h′ = f , then ∫ f ( g( t)) ⋅ g′ ( t) dt = h( g( t))
. This last equation
is verified simply by differentiating h( g( t)) to see that it is an antiderivative of
f ( g( t)) ⋅ g′
( t)
. When we consider definite integrals, however, we must note that
a definite integral is a number and remember to change the bounds when doing
integration by substitution.
Example of a Common Error
3 2t
To evaluate ∫ dt , let u =
2
1+ t , so that
0 2
1+
t
3 2t
3 1
3
3
2
dt = 1 10
0 2
1 t
0 u du = u = +
+
t =
0
0
∫ ∫ ln ln( ) ln( ) − =
ln( 1) ln( 10 ).
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AP® Calculus: 2006–2007 Workshop Materials