Superscript of an integral of a marginal probability

Problem: $X$ is uniformly distributed over $(0,1)$. When $X=x$ is observed where $0<x<1$, $Y$ is uniformly distributed over $(x,1)$. Find $Y$'s marginal probability density function $f_Y(y)$.

Solution: From the problem statement, we know $$f_X(x)=\begin{cases}1,&0<x<1;\\0,&\rm{otherwise}.\end{cases}$$ $$f_{Y\mid X}(y\mid x)=\begin{cases}\frac{1}{1-x},&x<y<1;\\0,&\rm{otherwise}.\end{cases}$$ Therefore, $$f(x,y)=f_{Y\mid X}(y\mid x)\cdot f_X(x)=\begin{cases}\frac{1}{1-x},&0<x<y<1;\\0,&\rm{otherwise}.\end{cases}$$ Hence, $$f_Y(y)=\begin{cases}\int_0^yf(x,y)dx=\int_0^y\frac{1}{1-x}dx=-\ln(1-y),&0<y<1;\\0,&\rm{otherwise}.\end{cases}$$

My question is why the superscript of the last equation is $y$ and not $1$. Alongside the solution, there is also a picture, in which the area enclosed by the $y$-axis, $y=1$ and $y=x$ is shaded. I can't make much sense out of the picture, other than the fact that it's the support.

Your joint density is the following

$$f_{XY}(x,y)=\frac{1}{1-x}\cdot\mathbb{1}_{0<x<y<1}$$

the joint support is a triangle

$$\underbrace{0<x<y}_{\text{X - support}}<1$$

that is the reason why the superscript in the integral is $y$ and not 1

On the other hand, if you want to derive $f_X(x)$ you will integrate $f(x,y)$ in the following support

$$0<\underbrace{x<y<1}_{\text{Y - support}}$$

$$f_X(x)=\int_x^1\frac{1}{1-x}dy=\mathbb{1}_{(0;1)}(x)$$

that is exactly $X\sim U(0;1)$ as already known