Do a specific inequality hold under integration? As we know that if  we have the inequality $f\leq g$, it does not imply $f'\leq g'$. 
Now Let  $f(x)\leq g(x)$ for each $x\in [a,b]$,
 where $0<a,b<\infty$. 
Is it possible to prove
$$\int_{a}^{b} f'(x)dx\leq \int_{a}^{b}g'(x)dx$$
,where $f$ denotes the derivative of $f$?
Thanking you in advance.
 A: The question you ask has an obvious simpliciation: by the fundamental theorem of calculus, you are asking:

Is it possible to prove
  $$ f(b) - f(a) \leq g(b) - g(a) $$

Including the constraints, we can simplify even further

Suppose $p \leq q$ and $r \leq s$. Can we prove $p-r \leq q-s$?

And the answer is no: you cannot subtract like inequalities. The problem is basically that you are flipping the sign, which reverses the inequality: i.e. $-r \geq -s$.

There is a theorem regarding integration and inequalities, however:

If $f \leq g$, $a \leq b$, and the following integrals are defined, then you have
  $$ \int_a^b f(x) \, \mathrm{d}x \leq \int_a^b g(x) \, \mathrm{d}x  $$

This does not hold with strict inequality unless you make additional assumptions. (e.g. $f,g$ continuous and $a < b$)
A: Take $f=0$ adn $g$ to be any non-negative strictly decreasing function. Then $\int_a^{b} g'(x) \, dx=g(b)-g(a) <0=\int_a^{b} f'(x) \, dx$. 
A: Take the functions $f(x)=0$ and $g(x)=-1/x$. We have $f(x)>g(x)$ on $(0,\infty)$, but also
$$f'(x)=0< \frac1{x^2} = g'(x).$$
But for $[a,b]\subseteq (0,\infty)$ it holds
$$f'(x)\le g'(x) \quad\implies\quad \int_a^b f'(x)\,\mathrm dx \le \int_a^b g'(x)\,\mathrm dx,$$
so we have a counter-example to your desired inequality. It cannot be proven.
