Question

Suppose 𝑓 and 𝑔 are continuous functions on [𝑎, 𝑏] and that

𝑔(𝑥) ≥ 0 for all 𝑥 ∈ [𝑎, 𝑏]. Prove that there exists 𝑥 in [𝑎, 𝑏] such that

𝑏 𝑏

∫ 𝑓(𝑡)𝑔(𝑡)𝑑𝑡 = 𝑓(𝑥) ∫ 𝑔(𝑡)𝑑𝑡.

𝑎 𝑎

Answer 1

Proof.

Since $f$ is continuous on the segment $[a,b]$ then $f$ is bounded. We denote $M=\max _{ [a,b] }{ f(x)\ ,\ m=\min _{ [a,b] }{ f(x) } }$ .Thus $m\le f(x)\le M$ . Multiplying by $g(x)$ , we obtain the inequality $mg(x)\le f(x)g(x)\le Mg(x)$ . All function in the inequality are continuous , hence integrable. We integrate the inequality:

$m\int _{ a }^{ b }{ g(x)dx\le \int _{ a }^{ b }{ f(x)g(x)dx\le M\int _{ a }^{ b }{ g(x)dx } } }$ (1)

Since, $g(x)\ge 0$ , then $\int _{ a }^{ b }{ g(x)dx } \ge 0$ . If $\int _{ a }^{ b }{ g(x)dx } =0$ , then from (1) it follows $\int _{ a }^{ b }{ f(x)g(x)dx } =0$ . In this case the equality

$\int _{ a }^{ b }{ f(x)g(x)dx } =f(x)\int _{ a }^{ b }{ \ g(x)dx }$ (2)

has the form $0=f(x)\cdot 0$ and is satisfied for any $x\in \left[ a,b \right]$ .

If $\int _{ a }^{ b }{ \ g(x)dx } \neq 0\quad$, then $\int _{ a }^{ b }{ \ g(x)dx } >0\quad$. Denote $c=\frac { \int _{ a }^{ b }{ f(x)g(x)dx } }{ \int _{ a }^{ b }{ \ g(x)dx } }$ .

Dividing all parts of inequality (1) by $\int _{ a }^{ b }{ \ g(x)dx }$ we have $m\le c\le M$ . By the Theorem on the intermediate value of a continuous function there is a point $x\in \left[ a,b \right]$ such that $f(x)=c.$ So there exists point $x\in \left[ a,b \right]$ such that equality (2) is satisfied.