Simple inequality question of functions and their integrals

Question

Consider the following real valued function $f$ for which the following holds,

$$\frac{f(x_1)+c}{a}>\frac{f(x_2)}{b}$$

For all $x_1<x_2$, where $x_1$, $x_2$, $a$ and $b$ are positive real values. Does this imply that

$$\int_0^{x_1}\frac{f(x)+c}{a}>\int_0^{x_2}\frac{f(x)}{b}?$$

Intuitively, it should, as the expression on the LHS has a higher value for values of $x$. Hence, the area under the function should also be greater? However, $x_2>x_1$ and so the 'width' of the RHS function is greater.

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Some rearranging, might be useful: $$\frac{1}{a}\int_0^{x_1}(f(x)+c)>\frac{1}{b}\int_0^{x_2}f(x)$$

$$\iff \frac{1}{a}\int_0^{x_1}f(x)+\int_0^{x_1}c>\frac{1}{b}\int_0^{x_2}f(x)$$

Answer 1

NO. Suppose $1+c>a/b$ and $f(x)=1$ for all $x.$ Then the LHS integral is $x_1(1+c)/a$ but the RHS integral is $x_2/b,$ which is larger than the LHS if $x_2> x_1(1+c)b/a.$

Answer 2

If $c=0,a=1,b=1$ and $f$ is a positive decreasing function then the reverse inequality holds!.