FDWK_3ed_Ch05_pp262-319

Section 5.2 Definite Integrals 277
The notation that Leibniz introduced for the definite integral was equally inspired. In his
derivative notation, the Greek letters (“” for “difference”) switch to Roman letters (“d”
for “differential”) in the limit,
lim y dy
.
x→0 x
d x
In his definite integral notation, the Greek letters again become Roman letters in the
limit,
lim
n→∞ n
f c k x b
f x dx.
k1
a
Notice that the difference Dx has again tended to zero, becoming a differential dx. The
Greek “” has become an elongated Roman “S,” so that the integral can retain its identity
as a “sum.” The c k ’s have become so crowded together in the limit that we no longer think of
a choppy selection of x values between a and b, but rather of a continuous, unbroken sampling
of x values from a to b. It is as if we were summing all products of the form f x dx as
x goes from a to b, so we can abandon the k and the n used in the finite sum expression.
The symbol
b
f x dx
a
is read as “the integral from a to b of f of x dee x,” or sometimes as “the integral from a to
b of f of x with respect to x.” The component parts also have names:
Upper limit of integration
Integral sign
Lower limit of integration
b
a
f x dx
Integral of f from a to b
The function is the integrand.
x is the variable of integration.
When you find the value
of the integral, you have
evaluated the integral.
The value of the definite integral of a function over any particular interval depends on
the function and not on the letter we choose to represent its independent variable. If we
decide to use t or u instead of x, we simply write the integral as
b
a
f t dt or b
f u du instead of b
f x dx.
a
No matter how we represent the integral, it is the same number, defined as a limit of Riemann
sums. Since it does not matter what letter we use to run from a to b, the variable of integration
is called a dummy variable.
a
EXAMPLE 1 Using the Notation
The interval 1, 3 is partitioned into n subintervals of equal length Dx 4n. Let m k
denote the midpoint of the k th subinterval. Express the limit
lim
n→∞ n
3m k 2 2m k 5 x
k1
as an integral.
continued