Calculus Of One Real Variable –
By Pheng Kim Ving |
12.1 |
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1. Average Values Of Functions |
Fig. 1.1
Average value of a function is limit of average of n function |
|
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].
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
|
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].
Let f(x) = 5x4. + 2. Determine c such that f(c) is the average value of f on [–1, 2].
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.
Problems & Solutions |
1. Find the average value of f(x) = 2x on [0, 3].
Solution
2. Calculate the average value of g(x) = 1 – x2 on [–2, 2].
Solution
Solution
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:
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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