290
5.3. INTEGRATION BY SUBSTITUTION
iv. v(x) = (2 − 7x) 3
v. w(x) = 3 4−11x
(c) Based on your experience in parts (a) and (b), conjecture an antiderivative for each
of the following functions. Test your conjectures by computing the derivative of
each proposed antiderivative.
i. a(x) = cos(πx)
ii. b(x) = (4x + 7) 11
iii. c(x) = xe x 2
⊲⊳
Reversing the Chain Rule: First Steps
In Preview Activity 5.3, we saw that it is usually straightforward to antidifferentiate a
function of the form
h(x) = f (u(x)),
whenever f is a familiar function whose antiderivative is known and u(x) is a linear
function. For example, if we consider
h(x) = (5x − 3)
6
,
in this context the outer function f is f (u) = u 6 , while the inner function is u(x) = 5x − 3.
Since the antiderivative of f is F(u) =
1
7 u 7 + C, we see that the antiderivative of h is
H(x) =
1
7
(5x − 3)
7 ·
1
5
+ C =
1
35
(5x − 3)
7 + C.
The inclusion of the constant
1
5 is essential precisely because the derivative of the inner
function is u ′ (x) = 5. Indeed, if we now compute H ′ (x), we find by the Chain Rule (and
Constant Multiple Rule) that
H
′ (x) =
1
35
· 7(5x − 3)
6 · 5 = (5x − 3)
6 = h(x),
and thus H is indeed the general antiderivative of h.
Hence, in the special case where the outer function is familiar and the inner function
5.3. INTEGRATION BY SUBSTITUTION
iv. v(x) = (2 − 7x) 3
v. w(x) = 3 4−11x
(c) Based on your experience in parts (a) and (b), conjecture an antiderivative for each
of the following functions. Test your conjectures by computing the derivative of
each proposed antiderivative.
i. a(x) = cos(πx)
ii. b(x) = (4x + 7) 11
iii. c(x) = xe x 2
⊲⊳
Reversing the Chain Rule: First Steps
In Preview Activity 5.3, we saw that it is usually straightforward to antidifferentiate a
function of the form
h(x) = f (u(x)),
whenever f is a familiar function whose antiderivative is known and u(x) is a linear
function. For example, if we consider
h(x) = (5x − 3)
6
,
in this context the outer function f is f (u) = u 6 , while the inner function is u(x) = 5x − 3.
Since the antiderivative of f is F(u) =
1
7 u 7 + C, we see that the antiderivative of h is
H(x) =
1
7
(5x − 3)
7 ·
1
5
+ C =
1
35
(5x − 3)
7 + C.
The inclusion of the constant
1
5 is essential precisely because the derivative of the inner
function is u ′ (x) = 5. Indeed, if we now compute H ′ (x), we find by the Chain Rule (and
Constant Multiple Rule) that
H
′ (x) =
1
35
· 7(5x − 3)
6 · 5 = (5x − 3)
6 = h(x),
and thus H is indeed the general antiderivative of h.
Hence, in the special case where the outer function is familiar and the inner function
