Average Value of a Function
Volume of a solid above the region D
It is easy to calculate the average
value of finitely many numbers
y1, y2 , . . . , yn :
yave
y1  y2    yn

n
However, how do we compute the
average temperature during a day if
infinitely many temperature readings
are possible?
In general, let’s try to compute
the average value of a function
y = f(x), a ≤ x ≤ b.
We start by dividing the interval [a, b]
into n equal subintervals, each with
length
x  (b  a) / n
Then, we choose points x1*, . . . , xn* in
successive subintervals and calculate the
average of the numbers f(x1*), . . . , f(xn*):
f ( xi *)      f ( xn *)
n
* For example, if f represents a temperature function
and n = 24, then we take temperature readings every
hour and average them.
Since ∆x = (b – a) / n, we can write n = (b – a) / ∆x and
the average value becomes:
f ( x1 *)      f ( xn *)
ba
x
1

 f ( x1 *)x      f ( xn *)x 
ba
n
1

f ( xi *)x

b  a i 1
* If we let n increase, we would be
computing the average value of a large
number of closely spaced values.
– For example, we would be averaging temperature
readings taken every minute or even every second.
• By the definition of a definite integral,
the limiting value is:
n
1
1 b
lim
f ( xi *)x 
f
(
x
)
dx


a
n  b  a
b

a
i 1
So, we define the average value of f
on the interval [a, b] as:
1 b
f ave 
f
(
x
)
dx

ba a
Illustration:
The average value of f on an interval [a,b] is
given as:
𝑏
1
𝑓
𝑏−𝑎 𝑎
𝑥 𝑑𝑥
Note: For a positive function, we can think of this definition as
saying area/width = average height
Average Value of a function of two variables
defined on Rectangle R
AVERAGE VALUE
• If f(x, y) ≥ 0, the equation
A( R)  f ave   f ( x, y) dA
• says that:
R
– The box with base R and height fave
has the same volume as the solid that
lies under the graph of f.
Illustration:
• Suppose z = f(x, y) describes a mountainous region and
you chop off the tops of
the mountains at height fave .
– Then, you can use them
to fill in the valleys so
that the region becomes
completely flat.
Volume of solid that lies under
the surface f(x,y) and above the
region D in the xy-plane
Example-1:
• Find the volume of the solid that lies
below the surface
Z = 16xy + 200
And lies above the region in the xy-plane
bounded by y = x2 and y = 8 – x2.
Sketch
The region in the xy-plane
Consider the following picture:
• How high would the water level be if the
waves all settled?