Page 285 Problem 15a
Prove the extended mean value theorem for integrals: If f and g are continuous on [a, b] and g(x) ≥ 0
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for all x ∈ [a, b], then there exists c ∈ [a, b] such that a (f g) = f (c) a g.
Solution
Let m = inf{f (x) : x ∈ [a, b]} and M = sup{f (x) : x ∈ [a, b]}. Thus, m ≤ f (x) ≤ M for all
x ∈ [a, b]. Since g(x) ≥ 0, the last inequality gives mg(x) ≤ f (x)g(x) ≤ M g(x) for all x ∈ [a, b], so
b
Z
b
Z
g≤
m
a
Z
b
fg ≤ M
a
g.
(1)
a
A continuous function on a closed set attains its maximum and minimum values, so there exists
x1 , x2 ∈ [a, b] such that f (x1 ) = m and f (x2 ) = M . Inequality (1) then becomes
Z
b
Z
g≤
f (x1 )
a
b
Z
f g ≤ f (x2 )
a
b
g.
(2)
a
If f (x1 ) = f (x2 ) then inf{f (x) : x ∈ [a, b]} = sup{f (x) : x ∈ [a, b]}, so f is constant: say f (x) = K
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for all x ∈ [a, b]. The result then immediately follows, for a (f g) = a Kg = K a g = f (c) a g,
where c is any point in [a, b].
If f (x1 ) < f (x2 ) suppose, without loss of generality, that x1 < x2 , and consider the continuous
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function h : [x1 , x2 ] → R defined by h(x) = f (x) a g. There are three cases that arise from
inequality (2):
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f g = f (x1 ) a g = h(x1 ).
In this case the result follows with c = x1 ∈ [a, b].
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2. a f g = f (x2 ) a g = h(x2 ).
In this case the result follows with c = x2 ∈ [a, b].
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3. h(x1 ) = f (x1 ) a g < a f g < f (x2 ) a g = h(x2 ).
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In this case the number a f g is between h(x1 ) and h(x2 ). By the intermediate value theorem
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there exists c ∈ (x1 , x2 ) ⊆ [a, b] such that a f g = h(c) = f (c) a g. Voilà!
1.
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a
Remark: We actually proved a stronger result because we did not need the function g to be
continuous, only that it be integrable, and that g(x) ≥ 0 for all x ∈ [a, b].