Calculus Of One Real Variable By Pheng Kim Ving
Chapter 12: Applications Of The Integral Section 12.1: The Mean-Value Theorem For Integrals

 

12.1
The Mean-Value Theorem For Integrals

 

 

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1. Average Values Of Functions

 

 

Fig. 1.1

 

Average value of a function is limit of average of n function
values as n approaches infinity.

 

Definition 1.1

 

 

 

 

Indeed it does.

 

Fig. 1.2

 

Average value of f(x) = y = 2x on [1, 5] is 6.

 

Example 1.1

 

Find the average value of f(x) = x2 on [1, 4]. Show that this average value is the height of a rectangle with base width
equal to 4 1 = 3 and area equal to the definite integral of f over [1, 4].

 

Solution

Fig. 1.3

 

 

Refer to Fig. 1.3. We have:

 

 

Consider the rectangle with its lower base being the interval [1, 4] and its height equal to 7. Its base width is 4 1 = 3, so
its area is (3)(7), which is equal to the definite integral of f over [1, 4]. Thus it's the required rectangle.

EOS

 

 

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2. The Mean-Value Theorem For Integrals

 

The average value of S1 = {1, 2, 3} is (1 + 2 + 3)/3 = 2, which belongs to S1, while that of S2 = {1, 2, 9} is (1 + 2 + 9)/3
= 4, which doesn't belong to S2. The average value of a collection of a finite number of quantities may or may not belong
to that collection. Now for a function f that's continuous on [a, b], it of course has an infinite number of values on [a, b].
We'll now see that the average value of f, which is the average value of the collection of all the values of f on [a, b], is
also a value of f, ie, belongs to that collection, provided f is continuous on [a, b]. This means that the average or mean
value of f continuous on [a, b] is attained at some point in [a, b]. This property constitutes the mean-value theorem for
integrals. Recall that the mean-value theorem for derivatives is the property that the average or mean rate of change of a
function continuous on [a, b] and differentiable on (a, b) is attained at some point in (a, b); see Section 5.1 Remarks 5.1.

 

Theorem 2.1 The Mean-Value Theorem For Integrals

 

If f is continuous on [a, b] then there exists c in [a, b] such that:

 

 

See Fig. 2.1.

 

 

Proof

 

A similar argument leads to the same conclusion if m > M.
EOP

 

Fig. 2.1

 

 

Remarks 2.1

 

i. The theorem says that the average value of f on [a, b] is attained on [a, b], ie, it's one of the values of f on [a, b].

 

 

Example 2.1

 

Let f(x) = 5x4. + 2. Determine c such that f(c) is the average value of f on [1, 2].

 

Solution

The average value of f on [1, 2] is:

 

EOS

 

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3. An Alternative Proof

 

Define the function F by:

 

 

for all x in [a, b]. By the fundamental theorem of calculus (see Section 9.4 Theorem 2.1), f is the derivative (or rate of
change) of F. So the average value of f on [a, b] is the average rate of change of F on [a, b], and the value of f at a
point in [a, b] is the instantaneous rate of change of F at that point. Using this observation we can prove the mean-value
theorem for integrals by applying the mean-value theorem for derivatives to F. This alternative proof is given in Problem
& Solution 5
.

 

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Problems & Solutions

 

1. Find the average value of f(x) = 2x on [0, 3].

 

Solution

 

 

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2. Calculate the average value of g(x) = 1 x2 on [2, 2].

 

Solution

 

 

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Solution

 

 

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4. Let f(x) = 3x2. Determine c such that f(c) is the average value of f on [0, 2].

 

Solution

 

The average value of f on [0, 2] is:

 

 

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Solution

 

By the fundamental theorem of calculus, F is differentiable on [a, b]. So, applying the mean-value theorem for
derivatives to F on [a, b], there's c in [a, b] such that:

 

 

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