AP Calculus

Special Focus: The Fundamental
Theorem of Calculus
Example 2
x 3
Let g( x) = ∫ 1+
t dt . Then what is g′
( x)?
1
Solution: We don’t know a formula for an antiderivative of 1+ t , but let’s assume that
h( t) is one. Then g( x) = h( t) x
= h( x) − h( 1 ). Thus g′ ( x) = h′ ( x) = 1+
x
3 . Note the
derivative of h( 1 ) is 0, since h( 1 ) is a number.
1
We repeat the pattern of these two examples to see how the chain rule works with
functions defined by integrals.
Example 3
3x
Let g( x) = ∫ 2 sin( t)
dt. What is g′
( x)?
0
Solution: As in example 1, we have
2
3x
3x
2
∫ 0
0
2
g( x) = sin( t) dt = − cos( t) = − cos( 3x
) + 1.
Now compute g′ ( x) = sin( 3x ) ⋅6x
.
This example demonstrates that
2
The derivative of g( x) is the value of the integrand at the upper limit of integration
times the derivative of the upper limit of integration.
Note that this procedure also works in example 1 and example 2, where the upper limit of
integration is simply x.
Example 4
sin( x)
3
Let g( x)
= ∫ 1+
t dt . What is g′
( x)?
1
Solution: Again using h(t) as we did in example 2,
sin( x)
sin( x )
1 1
3
g( x) = ∫ 1+ t dt = h( t) = h(sin( x))
− h( 1 ).
Thus g′ ( x) = h′( sin( x) )⋅ cos( x) = 1+ (sin( x)) ⋅cos( x)
. Note the careful use of the chain
rule in the process of evaluating h ′.
3
3
AP® Calculus: 2006–2007 Workshop Materials 59