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Let $f$ and $g$ be two continuous functions on $[ a,b ]$ and assume $g$ is positive. Prove that $$\int_{a}^{b}f(x)g(x)dx=f(\xi )\int_{a}^{b}g(x)dx$$ for some $\xi$ in $[ a,b ]$.

Here is my solution:

Since $f(x)$ and $g(x)$ are continuous, then $f(x) g(x)$ is continuous. Using the Mean value theorem, there exists a $\xi$ in $[ a,b ]$ such that $\int_{a}^{b}f(x)g(x)dx= f(\xi)g(\xi) (b-a) $ and using the Mean value theorem again, we can get $g(\xi) (b-a)=\int_{a}^{b}g(x)dx$ which yields the required equality.

Is my proof correct? If not, please let me know how to correct it.

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Are you sure it is the same $\xi$ in both steps? – Aleks Vlasev Nov 18 '11 at 3:05
The same $\xi \in [a,b]$ does not (necessarily) work for both $fg$ and $g$ by itself. Also, where have you used the hypothesis that $g > 0$? – cardinal Nov 18 '11 at 3:05
Note that there are counterexamples if you don't have "$g$ is positive." If $g$ changes sign and $f$ doesn't, the left side could be zero even though the right side doesn't take the value zero for any $\xi$. – Gerry Myerson Nov 18 '11 at 3:35
up vote 1 down vote accepted

The integrals on both sides of the problem are well defined since $f$ and $g$ are continuous, and $g$ is positive so $ \displaystyle \int^b_a g(x) dx > 0.$ Thus there exists some constant $K$ such that $$ \int^b_a f(x) g(x) dx = K\int^b_a g(x) dx . $$

If $\displaystyle K > \max_{x\in [a,b]} f(x) $ then the left side is smaller than the right.

If $\displaystyle K < \min_{x\in [a,b]} f(x) $ then the left side is larger than the right.

Thus $ K \in f( [a,b] ).$

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There is a typo in your post that invalidates the current version of the proof. Cheers. :) – cardinal Nov 18 '11 at 22:36


  1. For $y \in [a,b]$, let $\newcommand{\rd}{\,\mathrm{d}}h(y) = \int_a^b (f(y)-f(x))g(x) \rd x$. Then $h$ is continuous on $[a,b]$ since $f$ is continuous.
  2. The interval $[a,b]$ is compact and since $f$ is continuous then $f$ attains both a minimum $m$ and a maximum $M$ on $[a,b]$.
  3. What do you know about the value of $\int_a^b (M-f(x))g(x) \rd x$? How about $\int_a^b (m-f(x))g(x) \rd x$?
  4. Now, apply a very famous theorem with the initials IVT to $h(y)$ to conclude.
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"apply a very famous theorem with the initials IVT" - not much of a hint... :D but +1 of course. – J. M. Nov 18 '11 at 3:47
@J.M.: It's always a little tough to gauge what hints will click immediately in another's brain and which ones might not. Hopefully, I've still left enough work to the OP that they'll have to think a little. If you think I've left too much in the open, let me know and I will revise. I usually post hints as comments rather than answers. – cardinal Nov 18 '11 at 3:51
Hey, I upvoted... if I had any complaints I'd have said so already. :) – J. M. Nov 18 '11 at 3:53
@J.M.: I suppose that's true! Thanks. :) – cardinal Nov 18 '11 at 3:56

Actually, this therom is called the first mean value theorem for integration. There is a neat proof on Wiki:

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