prove part (b) of this integration inequality
(a) Suppose that
)
is a continuous solution of
\leq f(t)+\int_{t_0}^t h(s)u(s)ds)
,

,
where

are continuous and
\geq 0)
for

. Show that
![u(t)\leq f(t)+\int_{t_0}^t f(s)h(s)\exp\left[\int_s^t h(\sigma)d\sigma\right]ds](http://latex.codecogs.com/png.latex?u(t)\leq f(t)+\int_{t_0}^t f(s)h(s)\exp\left[\int_s^t h(\sigma)d\sigma\right]ds)
,

.
This I have done! But then we have...
(b) Moreover, if
)
is increasing for

, show that
![u(t)\leq f(t)\exp\left[\int_{t_0}^t h(s)ds\right]](http://latex.codecogs.com/png.latex?u(t)\leq f(t)\exp\left[\int_{t_0}^t h(s)ds\right])
,

.
In order to do the exercise, we are given as a theorem the following formulation of the Gronwall inequality:
If
\leq M+\int_{t_0}^t h(s)u(s)ds)
,

,
where

,
\geq 0)
for all

and

is continuous, and
)
is a continuous solution, then
![u(t)\leq M\exp\left[\int_{t_0}^t h(s)ds\right]](http://latex.codecogs.com/png.latex?u(t)\leq M\exp\left[\int_{t_0}^t h(s)ds\right])
,

.
However, I believe this theorem may only be useful for part (a), which, as I said, I have already done. I provide it just in case.
Thanks in advance!