Question

Exercise 25: Let f : [a, b] → R be continuous and ϕ : [a, b] → R be integrable and
non-negative, i.e. ϕ ≥ 0. Then there exists c ∈ (a, b) such that
the integral of f(x)ϕ(x)dx from a to b = f(c) the integral of ϕ(x)dx from a to b.

Answer 1

Let $\int_a^b \phi(x)=0.$ Since for all $x \in[a, b], \phi(x)\geq0,$ then it suffices that $\forall x \in[a, b];\phi(x)=0.$

Then, $\int_a^bf(x)\phi(x)dx=\int_a^bf(x).0dx=f(k).0=0$

Now suppose $\int_a^b\phi(x)dx>0$ since $f$ is continuous on $[a, b]$ Then $f$ is integrable on $[a, b]$. Then by Extreme value theorem, there exists $m, M$ such that

$f(m)=min f(x)\\ f(M)=max f(x)\\ x \in[a, b]$

Them we have the inequality

$f(m)\leq f(x) \leq f(M) ...........(1)$

Multiplying (1) by $\phi(x)$ we have

$f(m)\phi(x) \leq f(x) \phi(x) \leq f(M)\phi(x)$

Integrating both sides, we have;

$f(m)\int_a^b\phi(x)dx \leq f(x) \int_a^b \phi(x)dx \leq f(M)\int_a^b\phi(x)dx$

Dividing by $\int_a^b\phi(x)dx,$ we have;

$f(m) \leq \frac{\int_a^bf(x)\phi(x)dx}{\int_a^b\phi(x)dx} \leq f(M)$

Then by the intermediate value theorem there exists $c\in(a, b)$ such that

$\frac{\int_a^bf(x)\phi(x)dx}{\int_a^b\phi(x)dx} =f(c)$

$\implies \int_a^bf(x)\phi(x)dx=f(c)\int_a^b\phi(x)dx.$