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In this section, we will learn about:

Applying integration to find out the average value of a function.

Applying integration to find out the average value of a function.

1.
6

APPLICATIONS OF INTEGRATION

APPLICATIONS OF INTEGRATION

2.
APPLICATIONS OF INTEGRATION

6.5

Average Value

of a Function

In this section, we will learn about:

Applying integration to find out

the average value of a function.

6.5

Average Value

of a Function

In this section, we will learn about:

Applying integration to find out

the average value of a function.

3.
AVERAGE VALUE OF A FUNCTION

It is easy to calculate the average

value of finitely many numbers

y1, y2 , . . . , yn :

y1 + y2 + ⋅⋅⋅+ yn

yave =

n

It is easy to calculate the average

value of finitely many numbers

y1, y2 , . . . , yn :

y1 + y2 + ⋅⋅⋅+ yn

yave =

n

4.
AVERAGE VALUE OF A FUNCTION

However, how do we compute the

average temperature during a day if

infinitely many temperature readings

are possible?

However, how do we compute the

average temperature during a day if

infinitely many temperature readings

are possible?

5.
AVERAGE VALUE OF A FUNCTION

This figure shows the graph of a

temperature function T(t), where:

t is measured in hours

T in °C

Tave , a guess at the average temperature

This figure shows the graph of a

temperature function T(t), where:

t is measured in hours

T in °C

Tave , a guess at the average temperature

6.
AVERAGE VALUE OF A FUNCTION

In general, let’s try to compute

the average value of a function

y = f(x), a ≤ x ≤ b.

In general, let’s try to compute

the average value of a function

y = f(x), a ≤ x ≤ b.

7.
AVERAGE VALUE OF A FUNCTION

We start by dividing the interval [a, b]

into n equal subintervals, each with

length Δx = (b −a ) / n .

We start by dividing the interval [a, b]

into n equal subintervals, each with

length Δx = (b −a ) / n .

8.
AVERAGE VALUE OF A FUNCTION

Then, we choose points x1*, . . . , xn* in

successive subintervals and calculate the

average of the numbers f(xi*), . . . , 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.

Then, we choose points x1*, . . . , xn* in

successive subintervals and calculate the

average of the numbers f(xi*), . . . , 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.

9.
AVERAGE VALUE OF A FUNCTION

Since ∆x = (b – a) / n, we can write

n = (b – a) / ∆x and the average value

f ( x1 *) + ⋅ ⋅ ⋅ + f ( xn *)

b−a

Δx

1

= [ f ( x1 *)Δx + ⋅ ⋅ ⋅ + f ( xn *)Δx ]

b−a

n

1

= ∑

b − a i =1

f ( xi *)Δx

Since ∆x = (b – a) / n, we can write

n = (b – a) / ∆x and the average value

f ( x1 *) + ⋅ ⋅ ⋅ + f ( xn *)

b−a

Δx

1

= [ f ( x1 *)Δx + ⋅ ⋅ ⋅ + f ( xn *)Δx ]

b−a

n

1

= ∑

b − a i =1

f ( xi *)Δx

10.
AVERAGE VALUE OF A FUNCTION

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.

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.

11.
AVERAGE VALUE OF A FUNCTION

By the definition of a definite integral,

the limiting value is:

1 n 1 b

lim ∑ f ( xi *)Δx = ∫ f ( x)dx

n → ∞ b −a b −a a

i =1

By the definition of a definite integral,

the limiting value is:

1 n 1 b

lim ∑ f ( xi *)Δx = ∫ f ( x)dx

n → ∞ b −a b −a a

i =1

12.
AVERAGE VALUE OF A FUNCTION

So, we define the average value of f

on the interval [a, b] as:

1 b

f ave = ∫ f ( x)dx

b −a a

So, we define the average value of f

on the interval [a, b] as:

1 b

f ave = ∫ f ( x)dx

b −a a

13.
AVERAGE VALUE Example 1

Find the average value of

the function f(x) = 1 + x2 on

the interval [-1, 2].

Find the average value of

the function f(x) = 1 + x2 on

the interval [-1, 2].

14.
AVERAGE VALUE Example 1

With a = -1 and b = 2,

we have:

1 b

f ave =

b−a a ∫ f ( x) dx

1 2

= ∫

2 − (−1) −1

(1 + x ) dx

2

2

1⎡ x ⎤ 3

= ⎢x + ⎥ = 2

3⎣ 3 ⎦−1

With a = -1 and b = 2,

we have:

1 b

f ave =

b−a a ∫ f ( x) dx

1 2

= ∫

2 − (−1) −1

(1 + x ) dx

2

2

1⎡ x ⎤ 3

= ⎢x + ⎥ = 2

3⎣ 3 ⎦−1

15.
AVERAGE VALUE

If T(t) is the temperature at time t,

we might wonder if there is a specific time

when the temperature is the same as

the average temperature.

If T(t) is the temperature at time t,

we might wonder if there is a specific time

when the temperature is the same as

the average temperature.

16.
AVERAGE VALUE

For the temperature function graphed here,

we see that there are two such times––just

before noon and just before midnight.

For the temperature function graphed here,

we see that there are two such times––just

before noon and just before midnight.

17.
AVERAGE VALUE

In general, is there a number c at which

the value of a function f is exactly equal

to the average value of the function—that is,

f(c) = fave?

In general, is there a number c at which

the value of a function f is exactly equal

to the average value of the function—that is,

f(c) = fave?

18.
AVERAGE VALUE

The mean value theorem for

integrals states that this is true

for continuous functions.

The mean value theorem for

integrals states that this is true

for continuous functions.

19.
MEAN VALUE THEOREM

If f is continuous on [a, b], then there exists

a number c in [a, b] such that

1 b

f (c) = f ave = ∫ f ( x) dx

b −a a

that is,

b

∫a

f ( x) dx = f (c )(b − a )

If f is continuous on [a, b], then there exists

a number c in [a, b] such that

1 b

f (c) = f ave = ∫ f ( x) dx

b −a a

that is,

b

∫a

f ( x) dx = f (c )(b − a )

20.
MEAN VALUE THEOREM

The Mean Value Theorem for Integrals is

a consequence of the Mean Value Theorem

for derivatives and the Fundamental Theorem

of Calculus.

The Mean Value Theorem for Integrals is

a consequence of the Mean Value Theorem

for derivatives and the Fundamental Theorem

of Calculus.

21.
MEAN VALUE THEOREM

The geometric interpretation of the Mean

Value Theorem for Integrals is as follows.

For ‘positive’ functions f, there is a number c such that

the rectangle with base [a, b] and height f(c) has

the same area as the region under the graph of f

from a to b.

The geometric interpretation of the Mean

Value Theorem for Integrals is as follows.

For ‘positive’ functions f, there is a number c such that

the rectangle with base [a, b] and height f(c) has

the same area as the region under the graph of f

from a to b.

22.
MEAN VALUE THEOREM Example 2

Since f(x) = 1 + x2 is continuous on the

interval [-1, 2], the Mean Value Theorem for

Integrals states there is a number c in [-1, 2]

such that:

2

∫ (1 + x ) dx =

2

f (c)[2 − (−1)]

−1

Since f(x) = 1 + x2 is continuous on the

interval [-1, 2], the Mean Value Theorem for

Integrals states there is a number c in [-1, 2]

such that:

2

∫ (1 + x ) dx =

2

f (c)[2 − (−1)]

−1

23.
MEAN VALUE THEOREM Example 2

In this particular case, we can

find c explicitly.

From Example 1, we know that fave = 2.

So, the value of c satisfies f(c) = fave = 2.

Therefore, 1 + c2 = 2.

Thus, c2 = 1.

In this particular case, we can

find c explicitly.

From Example 1, we know that fave = 2.

So, the value of c satisfies f(c) = fave = 2.

Therefore, 1 + c2 = 2.

Thus, c2 = 1.

24.
MEAN VALUE THEOREM Example 2

So, in this case, there happen to be

two numbers c = ±1 in the interval [1, 2]

that work in the Mean Value Theorem for

So, in this case, there happen to be

two numbers c = ±1 in the interval [1, 2]

that work in the Mean Value Theorem for

25.
MEAN VALUE THEOREM

Examples 1 and 2 are illustrated

Examples 1 and 2 are illustrated

26.
MEAN VALUE THEOREM Example 3

Show that the average velocity of a car

over a time interval [t1, t2] is the same

as the average of its velocities during

the trip.

Show that the average velocity of a car

over a time interval [t1, t2] is the same

as the average of its velocities during

the trip.

27.
MEAN VALUE THEOREM Example 3

If s(t) is the displacement of the car at time t,

then by definition, the average velocity of

the car over the interval is:

Δs s (t2 ) −s (t1 )

=

Δt t2 −t1

If s(t) is the displacement of the car at time t,

then by definition, the average velocity of

the car over the interval is:

Δs s (t2 ) −s (t1 )

=

Δt t2 −t1

28.
MEAN VALUE THEOREM Example 3

On the other hand, the average value of the

velocity function on the interval is:

1 t2 1 t2

vave =

t2 − t1 ∫

t1

v(t ) dt =

t2 − t1 ∫

t 1

s '(t ) dt

1

= [s(t2 ) − s(t1 )] (by the Net Change Theorem )

t2 − t1

s (t2 ) − s (t1 )

= = average velocity

t2 − t1

On the other hand, the average value of the

velocity function on the interval is:

1 t2 1 t2

vave =

t2 − t1 ∫

t1

v(t ) dt =

t2 − t1 ∫

t 1

s '(t ) dt

1

= [s(t2 ) − s(t1 )] (by the Net Change Theorem )

t2 − t1

s (t2 ) − s (t1 )

= = average velocity

t2 − t1