Some notations. First, for every $x$ in $[a,b]$, let $F(x)=\displaystyle\int_a^x\phi(t)\mathrm{d}t$, hence $F$ is continuous and $F(a)=0$. Second, assume that $G$ is nondecreasing, the other case being similar. Thus there exists a nonnegative measure $\mu$ such that, for every $x$ in $[a,b]$, $G(x)=G(a)+\displaystyle\int_a^x\mathrm{d}\mu(t)$.
Now to the proof. Using an integration by parts, the integral $I$ of $G\phi$ over $[a,b]$ is
$$
I=[F(t)G(t)]_a^b-\int_a^bF(t)\mathrm{d}\mu(t)=F(b)G(b)-\int_a^bF(t)\mathrm{d}\mu(t).
$$
The hypothesis made on $G$ means that $\mu$ is a nonnegative measure hence the first mean value theorem for integration yields the existence of a point $x$ in $[a,b]$ such that
$$
\int_a^bF(t)\mathrm{d}\mu(t)=F(x)\int_a^b\mathrm{d}\mu(t)=F(x)(G(b)-G(a)).
$$
Coming back to $I$, one gets
$$
I=F(b)G(b)-F(x)(G(b)-G(a))=G(a)F(x)+G(b)(F(b)-F(x)),
$$
which, by definition of $F$, is the desired assertion.
(No continuity of $G$ is required.)
In case you are wondering, the first mean value theorem used above is not the usual one but it has the same proof than the usual one. Namely, $F$ being continuous on the compact set $[a,b]$ has a maximum $M$ and a minimum $m$ on $[a,b]$, thus
$$
m\int_a^b\mathrm{d}\mu(t)\le\int_a^bF(t)\mathrm{d}\mu(t)\le M\int_a^b\mathrm{d}\mu(t).
$$
(This step uses the hypothesis that $\mu$ has constant sign.) In other words, there exists $u$ in $[m,M]$ such that
$$
\int_a^bF(t)\mathrm{d}\mu(t)=u\int_a^b\mathrm{d}\mu(t).
$$
Now, the continuity of $F$ implies that there exists $x$ in $[a,b]$ such that $u=F(x)$ and you are done.
Previous version When $\phi$ has a constant sign, this is a consequence of the intermediate value theorem. (An earlier version of this post did not assume $\phi$ of constant sign, hence an argument was wrong. Thanks to the OP for having asked some explanations.)
Let $I$ denote the integral of $G\phi$ over $[a,b]$, $J$ the integral of $\phi$ over $[a,b]$, and $H(x)$ the RHS of the equality you want to prove. Assume that $\phi$ is nonnegative and $G$ is nondecreasing, the other cases being similar.
Since $G(a)\le G(t)\le G(b)$ and $\phi(t)\ge0$ for every $t$ in $[a,b]$, $G(a)\phi(t)\le G(t)\phi(t)\le G(b)\phi(t)$ for every $t$, hence $G(a)J\le I\le G(b)J$. The function $x\mapsto H(x)$ is continuous on the interval $[a,b]$, $H(a)=G(b)J\ge I$ and $H(b)=G(a)J\le I$, hence by the intermediate value theorem there exists $x$ in $[a,b]$ such that $H(x)=I$.
(No continuity of $G$ is required.)
Let f:[0,1]→R be differentiable with f(0)=0, and satisfying
$$|f'(x)|\le M|f(x)|, x\in[0,1] $$
for some $M>0$
a.) Use the Mean Value Theorem to show that for all $x \le x_0 \in[0,1], y\in[0,x_0]:$
$$ |f(x)|\le x_0 \sup |f'(y)|\le M x_0 \sup |f(y)|$$
b.) Use the previous part to show that f is the zero-function on [0,1].
Let's begin with part a. So fix an $x_0$ in $(0,1)$. Suppose that there is an $x$ in $[0,x_0]$ such that $|f(x)| > x_0 \sup|f'(y)|$. Then in particular, $\dfrac{|f(x) - f(0)|}{x-0} = \dfrac{|f(x)|}{x} > \dfrac{x_0}{x}\sup f'(y) \geq \sup f'(y)$ as $x_0 \geq x$.
But then by the mean value theorem, there must exist a $c$ such that $f'(c) = \dfrac{f(x) - f(0)}{x - 0}$. This is a contradiction, as then $f'(c) > \sup f'(y)$.
The second inequality is much easier, relying just on using $|f(x)|\le x_0 \sup |f'(y)|\le M x_0 \sup |f(y)|$ and interpreting sup.
And then you can follow the hint for part b, and the answer falls out as $|f(x)| < \sup |f(y)|$ is nonsense.
Best Answer
The mean value theorems you cite apply to integrals over bounded intervals, so some additional analysis is required.
Note that the integral $\int_0^\infty f(x) e^{-x} \, dx$ is convergent by the Weierstrass test.
Let $m = \inf_{x \in[0,\infty)} f(x)$ and $M = \sup_{x \in[0,\infty)} f(x)$. Since $me^{-x} \leqslant f(x)e^{-x} \leqslant Me^{-x}$, we have
$$m = \int_0^\infty me^{-x} \, dx \leqslant I = \int_0^{-\infty}f(x) e^{-x} \, dx \leqslant \int_0^\infty Me^{-x} \, dx = M,$$
and $m \leqslant I \leqslant M$.
If $m < I < M$ then there exist $a$ and $b$ such that $m \leqslant f(a) < I < f(b) \leqslant M$. (This is a basic property of the infimum and supremum.) By the familiar intermediate value theorem -- which applies to continuous functions on compact intervals -- there exists $c \in (a,b) \subset [0,\infty)$ such that
$$f(c) = I = \int_0^\infty f(x) e^{-x} \, dx.$$
See if you can handle the cases where $I = m$ and $I = M$.
Hint: If $I = m$ then $\int_0^\infty [f(x) - m]e^{-x} \, dx = 0$ where $f(x) - m \geqslant 0$.