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4.3. THE DEFINITE INTEGRAL
Since integrals arise from Riemann sums in which we add n values of a function, it
should not be surprising that evaluating an integral is something like averaging the output
values of a function. Consider, for instance, the right Riemann sum R n of a function f ,
which is given by
R n = f (x 1 )△x + f (x 2 )△x + · · · + f (x n )△x = ( f (x 1 ) + f (x 2 ) + · · · + f (x n ))△x.
Since △x =
b−a
n , we can thus write
R n = ( f (x 1 ) + f (x 2 ) + · · · + f (x n )) ·
b − a
n
= (b − a)
f (x 1 ) + f (x 2 ) + · · · + f (x n )
n
. (4.1)
Here, we see that the right Riemann sum with n subintervals is the length of the interval
(b − a) times the average of the n function values found at the right endpoints. And just as
with our efforts to compute area, we see that the larger the value of n we use, the more
accurate our average of the values of f will be. Indeed, we will define the average value of
f on [a, b] to be
f AVG[a, b] = lim
n→∞
f (x 1 ) + f (x 2 ) + · · · + f (x n )
n
.
But we also know that for any continuous function f on [a, b], taking the limit of a
Riemann sum leads precisely to the definite integral. That is, lim
n→∞
R n =
b
a
f (x) dx, and
thus taking the limit as n → ∞ in Equation (4.1), we have that
b
a
f (x) dx = (b − a) · f AVG[a, b] .
(4.2)
Solving Equation (4.2) for f AVG[a, b] , we have the following general principle.
Average value of a function: If f is a continuous function on [a, b], then its average
value on [a, b] is given by the formula
f AVG[a, b] =
1
b − a
·
b
a
f (x) dx.
Observe that Equation (4.2) tells us another way to interpret the definite integral:
the definite integral of a function f from a to b is the length of the interval (b − a)
times the average value of the function on the interval. In addition, Equation (4.2) has
a natural visual interpretation when the function f is nonnegative on [a, b]. Consider
Figure 4.28, where we see at left the shaded region whose area is
b
a
f (x) dx, at center
the shaded rectangle whose dimensions are (b − a) by f AVG[a, b] , and at right these two
figures superimposed. Specifically, note that in dark green we show the horizontal line
y = f AVG[a, b] . Thus, the area of the green rectangle is given by (b − a) · f AVG[a, b] , which
is precisely the value of
b
a
f (x) dx. Said differently, the area of the blue region in the
4.3. THE DEFINITE INTEGRAL
Since integrals arise from Riemann sums in which we add n values of a function, it
should not be surprising that evaluating an integral is something like averaging the output
values of a function. Consider, for instance, the right Riemann sum R n of a function f ,
which is given by
R n = f (x 1 )△x + f (x 2 )△x + · · · + f (x n )△x = ( f (x 1 ) + f (x 2 ) + · · · + f (x n ))△x.
Since △x =
b−a
n , we can thus write
R n = ( f (x 1 ) + f (x 2 ) + · · · + f (x n )) ·
b − a
n
= (b − a)
f (x 1 ) + f (x 2 ) + · · · + f (x n )
n
. (4.1)
Here, we see that the right Riemann sum with n subintervals is the length of the interval
(b − a) times the average of the n function values found at the right endpoints. And just as
with our efforts to compute area, we see that the larger the value of n we use, the more
accurate our average of the values of f will be. Indeed, we will define the average value of
f on [a, b] to be
f AVG[a, b] = lim
n→∞
f (x 1 ) + f (x 2 ) + · · · + f (x n )
n
.
But we also know that for any continuous function f on [a, b], taking the limit of a
Riemann sum leads precisely to the definite integral. That is, lim
n→∞
R n =
b
a
f (x) dx, and
thus taking the limit as n → ∞ in Equation (4.1), we have that
b
a
f (x) dx = (b − a) · f AVG[a, b] .
(4.2)
Solving Equation (4.2) for f AVG[a, b] , we have the following general principle.
Average value of a function: If f is a continuous function on [a, b], then its average
value on [a, b] is given by the formula
f AVG[a, b] =
1
b − a
·
b
a
f (x) dx.
Observe that Equation (4.2) tells us another way to interpret the definite integral:
the definite integral of a function f from a to b is the length of the interval (b − a)
times the average value of the function on the interval. In addition, Equation (4.2) has
a natural visual interpretation when the function f is nonnegative on [a, b]. Consider
Figure 4.28, where we see at left the shaded region whose area is
b
a
f (x) dx, at center
the shaded rectangle whose dimensions are (b − a) by f AVG[a, b] , and at right these two
figures superimposed. Specifically, note that in dark green we show the horizontal line
y = f AVG[a, b] . Thus, the area of the green rectangle is given by (b − a) · f AVG[a, b] , which
is precisely the value of
b
a
f (x) dx. Said differently, the area of the blue region in the