$\begingroup$Problem:
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.
$\endgroup$33 Answers
$\begingroup$Hints:
- 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.
- The interval $[a,b]$ is compact and since $f$ is continuous then $f$ attains both a minimum $m$ and a maximum $M$ on $[a,b]$.
- What do you know about the value of $\int_a^b (M-f(x))g(x) \rd x$? How about $\int_a^b (m-f(x))g(x) \rd x$?
- Now, apply a very famous theorem with the initials IVT to $h(y)$ to conclude.
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] ).$
$\endgroup$1$\begingroup$Actually, this therom is called the first mean value theorem for integration. There is a neat proof on Wiki:
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