Prove $\int_a^b f(x)dx \leq \frac{e^{2L\beta}-1}{2L\alpha}\int_c^d f(x)dx$

I got something which is rather close to your result but couldn't get rid of an additional term. I'm hoping someone will find the development useful in order to give a complete answer. The question reminded me somehow of the proof of Gronwall's inequality and my answer is based on that.

Let $h(t)=\int_a^t f(s)ds$ taking the derivative with respect to $t$ yields $h'(t) = f(t)$. Rewriting $f(t)=f^2(t)/f(t)$, we can find an upper bound for $f^2(t)$ indeed, we have: $$f^2(t) = \int_a^t (f^2(s))'ds + f^2(a) = 2\int_a^t f(s)f'(s)ds +f^2(a)$$

but then, the Lipschitz continuity implies that $|f'(s)|\le L$ which yields

$$f^2(t) \le 2L \int_a^t f(s)ds + f^2(a)\quad \Longrightarrow \quad h'(t) \le \left(2Lh(t) +f^2(a)\right)\left({1\over f(t)}\right) $$ we can then introduce the additional positive term $\left(\int_c^df(t)dt\right)/\alpha$ with: $$ h'(t) \le \left(2L h(t)+f^2(a) + {\int_c^d f(s)ds\over \alpha}\right)\left({1\over f(t)}\right) $$ Remark: the additional term comes from the $f^2(a)$. We could get rid of it if $\left(\int_c^d f(s)ds\right)/\alpha > f^2(a)$ which I wasn't able to prove.

To simplify notations, let $a=f^2(a) + {\int_c^d f(s)ds\over \alpha},$ $b=2L$ and $g(t)=1/f(t)$ then. Then the previous inequality reads $$h'(t) \le g(t)(a+bh(t))$$ which can be rewritten as follows (here is where it starts to look like Gronwall's inequality):

$$ {(a+bh(t))'\over a+bh(t)} \le bg(t) $$

the left-hand side is the logarithmic derivative of $a+bh(t)$, integrating both sides from $a$ to $t$ yields

$$ a+bh(t) \le a\exp\left(b\int_a^t g(t) \right) $$

plugging the values of $a$,$b$ and using $\int_a^b g(t) = \beta$, we finally get:

$$\int_a^b f(s)ds = h(b) \le \color{green}{\left[{\exp(2L\beta)-1\over 2L\alpha}\right]\int_c^d f(s)ds} + \color{red}{{f^2(a)\over 2L}\left(\exp(2L\beta)-1\right)} $$