To find area under a curve you need to find definite integral of your function `y` over a given region `-48<x<0`.

`int_-48^0 11/(sqrt(25-2x))dx=|(t=25-2x),(dt=-2dx =>dx=-1/2dt),(t_1=25-2(-48)=121),(t_2=25-2cdot0=25)|=`

Above we use substitution `t=25-2x`, `t_1` and `t_2` are new limits of integration for variable `t`. In the following line we will use the fact that `int dx/sqrt x=2sqrt x`.

`-11/2 int_121^25 dt/sqrt(t)=-11/2cdot 2 sqrt(t)|_121^25=-11(sqrt(25)-sqrt(121))=-11(5-11)=66`

**So the area of region `R` is 66.**