Mean value theorem for integration Can anyone hint or give the outline of the proof. I am a bit of confusing how to find $x_0$. 

Let $\phi(x) \geq 0$ for $x \in [a,b]$, and $\phi$ decreasing on $[a,b]$, let $h : [a,b] \rightarrow \mathbb{R}$ be continuous on $[a,b]$. Prove that there is a point $x_0$ in $[a,b]$ such that $$\int_a^b \phi h = \phi(a)\int_a^{x_0}h.$$ 

 A: Let
$$H(x) = \int_a^xh(t)\,dt.$$
Since $h$ is continuous, $H$ is differentiable and
$$\int_a^b\phi(x)h(x) \,dx= \int_a^b\phi(x)H'(x) \,dx=\int_a^b\phi \, dH.$$
Integrating by parts,
$$\int_a^b\phi(x)h(x) \,dx= \phi(b)H(b)-\phi(a)H(a)-\int_a^bH \, d\phi.$$
Apply the MVT to the integral on the RHS. There is a point $x_0 \in [a,b]$ such that
$$\int_a^b\phi(x)h(x) \,dx= \phi(b)H(b)-\phi(a)H(a)-H(x_0)[\phi(b) - \phi(a)] \\=\phi(a)[H(x_0)-H(a)]+\phi(b)[H(b)-H(x_0)]\\=\phi(a)\int_a^{x_0}h(x) \, dx + \phi(b)\int_{x_0}^{b}h(x) \, dx.$$
Since $\phi$ is nonnegative and decreasing, we can repeat the above derivation for the nonnegative, decreasing function $\hat{\phi}$  such that $\hat{\phi}(x) = \phi(x)$ for $x \in [a,b)$ and $\hat{\phi}(b) = 0$, and find
$$\int_a^b\phi(x)h(x) \,dx = \int_a^b\hat{\phi}(x)h(x) \,dx=\phi(a)\int_a^{x_0}h(x) \, dx.$$
The integrals involving $\phi$ and $\hat{\phi}$ are equal since changing the value of a function at a finite number of points does not change the value of a Riemann integral.
