# Second Mean Value Theorem for Integrals, Variant

I want to prove the following theorem, which Wikipedia refers as 'Second Mean Value Theorem'

Suppose that $g(x)$ is a non-negative monotonically decreasing function on the interval $[a, b]$, and its derivative is continuous. For $f(x)$ continuous on $[a, b]$, prove that there exists $c \in [a, b]$ such that $$\int_a^bf(x)g(x)dx=g(a)\int_a^cf(x)dx$$

[My Work]

Since $f(x)$ is continuous, define $F(x)=\int_a^x f(t)dt$. Since $F(x)$ is differentiable, it is therefore continuous. Thus for $[a, b]$, $F(x)$ is bounded. Suppose $m \leq F(x) \leq M$.

So proving the given problem would be equivalent to proving that $$m\leq \frac{1}{g(a)}\int_a^b f(x)g(x)dx \leq M \ \ \ \ \cdots(1)$$ so that I can use the Intermediate Value Theorem to conclude that there exist $c$ in $(a, b)$ such that $\frac{1}{g(a)}\int_a^b f(x)g(x)dx=F(c)$.

I am stuck on (1). Can anyone give me hints to prove this? If there are other ways to prove the theorem, please let me know. Thanks.

• The far-left and far-right sides of (1) should be $m(b-a)$ and M(b-a).$– DanielWainfleet Jun 25 '16 at 1:00 • @user254665 If so, can you please post it as an answer? I actually made up that inequality so that I could prove the theorem. I'm not even sure if the inequality holds or not. – zxcvber Jun 25 '16 at 1:03 • On re-reading it: With$a=0$,$b=1$,$f(x)=1$for all$x$, and$g(x)=1-x$, we have$M=m=m(b-a)=1$and the middle term in the inequality is$1/2$, so whether or not we write$m$or$m(b-a)\$ on the far LHS, the left inequality is not valid. – DanielWainfleet Jun 25 '16 at 2:18