數學問題：關於微積分

If f and g are continuous functions defined on [a, b] so that g(x)
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[a, b] with


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g(x) dx
Proof
Define the numbers
m = inf{ f(x): x
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[a, b] }
M = sup{ f(x): x
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[a, b] }
Then we have m
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M and since g is non-negative we also have
m g(x)
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M g(x)
By the properties of the Riemann integral this implies that
m
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g(x) dx
Therefore there exists a number d between m and M such that
d
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f(x) g(x) dx
But since f is continuous on [a, b] and d is between m and M, we can apply the Intermediate Value Theorem to find a number c such that f(c) = d. Then
f(c)
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f(x) g(x) dx
which is what we wanted to prove.

Interesting question and good answers!! But be careful what is the definition of " 可積 " . There may be some traps here...................

2006-12-05 21:02:54 補充：
e.g. Let f(x)=x(x -1/2)/(x -1/2) and g(x)=1 on [0, 1] , then Integral[f(x)*g(x),0,1]=1/2 and Integral[g(x),0,1] = 1However, there is no c in [0, 1] such that f(c) =1/2 and henceIntegral[f(x)*g(x),0,1]=f(c)*Integral[g(x),0,1] has no solution for c.

m = inf{ f(x): x [a, b] }
M = sup{ f(x): x [a, b] }

Then we have m f(x) M and since g is non-negative we also have

m g(x) f(x) g(x) M g(x)

By the properties of the Riemann integral this implies that

m g(x) dx f(x) g(x) dx M g(x) dx

Therefore there exists a number d between m and M such that

d g(x) dx = f(x) g(x) dx

But since f is continuous on [a, b] and d is between m and M, we can apply the Intermediate Value Theorem to find a number c such that f(c) = d. Then

f(c) g(x) dx = f(x) g(x) dx

which is what we wanted to prove.

Interactive Real Analysis, ver. 1.9.4

2006-12-05 16:43:55 補充：
http://web01.shu.edu/projects/reals/integ/proofs/mvtint.html