We have two equations with well defined integrals:
$$\int_{a_1}^{b_1}f(x)dx = c, \ \ \int_{a_2}^{b_2}f(x)g(x)dx = c,$$
where $f(x)$ is continous, decreasing and positive while $g(x)$ is continuous, increasing and positive functions, $0\leq a_1,b_2,a_2,b_2 < \infty$ with $b_1>a_1$ and $b_2>a_2$, and $c>0$ is some constant.
If $g(x)>1$ for any $x$, can we conclude that
$$b_1 - a_1 > b_2-a_2?$$
Attempt: If we replace $g(x)$ with any constant greater than $1$, the inequality holds. I am not sure whether such line of argument suffices.
(A) No , we can not conclude that , in general.
(B) In case, we change it slightly , we can try something like that.
(A) Consider $f(x)=e^{-x}$ :
Integrating, we get $C=e^{-1.0}-e^{-1.1}$ between Initial Interval $1.0$ & $1.1$
Shown in the Picture, the RED Area is $C$ between Interval $1.0$ & $1.1$ , while the BLUE Area is also $C$ over larger Interval & GREEN Area is also $C$ with even larger Interval.
To get the same $C$ further in the curve, we require longer interval.
Scaling with $g(x)=1.1,10,100,1000,$ , we can make the area below $f(x)g(x)$ arbitrarily large & we can reduce the interval required to get $C$ when moving further away from the RED Area.
Scaling with very high value , we can eventually make the New Interval EQUAL to Initial Interval, yet having same Area.
Scaling with very even higher value , we can eventually make the New Interval SMALLER than Initial Interval, yet having same Area.
In Other words, given the Initial Interval of length 0.1, we can get the New Interval of length 100, 10, 1, 0.1, 0.01 or 0.001 , yet having Same Area.
(B) Changing the Condition to state the the Initial Interval (for $f(x)$) & the New Interval (for $f(x)g(x)$) have same Starting Point, with no restriction on Ending Point.
We can now claim that the Initial Interval is SMALLER than the New interval.
[[ Proof : All values at all Points in the New Interval have larger (Scaling $f(x)$ by $g(x)$) values.
Shown in Picture, the Initial Interval where RED Area & BLUE Area (below $f(x)$) have total Area $C$ , while the New Interval with Common Starting Point has RED Area & GREEN Area (below $f(x)g(x)$) have same Area $C$ .
With Starting Point same, we require shorter Interval to get same Area. ]]
To make things "uncomplicated", we can take $g(x)$ a function which Starts at $K-1$ and gradually moves to $K$ in the limit, where $K$ is some Constant.
Eg $g(x)=K-e^{-x}$
In our Discussions, we can then remove $g(x)$ & use Constant $K$