James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

396 Chapter 5 Integrals
Example 4 Find d
y x4
sec t dt.
dx 1
SOLUTION Here we have to be careful to use the Chain Rule in conjunction with FTC1.
Let u − x 4 . Then
d
y x4
sec t dt − d
dx 1
dx yu sec t dt
1
− d
Fy u
sec t dtG du
du 1 dx
− sec u du
dx
(by the Chain Rule)
(by FTC1)
− secsx 4 d ? 4x 3 n
In Section 5.2 we computed integrals from the definition as a limit of Riemann sums
and we saw that this procedure is sometimes long and difficult. The second part of
the Fun damental Theorem of Calculus, which follows easily from the first part, provides
us with a much simpler method for the evaluation of integrals.
The Fundamental Theorem of Calculus, Part 2 If f is continuous on fa, bg, then
We abbreviate this theorem as FTC2.
y b
f sxd dx − Fsbd 2 Fsad
a
where F is any antiderivative of f, that is, a function such that F9− f.
Proof Let tsxd − y x f std dt. We know from Part 1 that t9sxd − f sxd; that is, t is an
a
antiderivative of f. If F is any other antiderivative of f on fa, bg, then we know from
Corollary 4.2.7 that F and t differ by a constant:
6
Fsxd − tsxd 1 C
for a , x , b. But both F and t are continuous on fa, bg and so, by taking limits of
both sides of Equation 6 (as x l a 1 and x l b 2 ), we see that it also holds when x − a
and x − b. So Fsxd − tsxd 1 C for all x in fa, bg.
If we put x − a in the formula for tsxd, we get
tsad − y a
f std dt − 0
So, using Equation 6 with x − b and x − a, we have
Fsbd 2 Fsad − ftsbd 1 Cg 2 ftsad 1 Cg
a
− tsbd 2 tsad − tsbd − y b
f std dt
a
n
Part 2 of the Fundamental Theorem states that if we know an antiderivative F of f,
then we can evaluate y b a
f sxd dx simply by subtracting the values of F at the endpoints
of the interval fa, bg. It’s very surprising that y b a
f sxd dx, which was defined by a complicated
pro cedure involving all of the values of f sxd for a < x < b, can be found by
knowing the val ues of Fsxd at only two points, a and b.
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