Range of an integral operator

Question

I tried to determine the range of the integral operator $T\in\mathcal{L}(L^2(0,1))$ given by $$(Tx)(t) = \int^1_t x(s)ds \text{ for } x \in L^2(0,1), t\in[0,1]$$ but it doesn't sound very logical. This is my approach to the solution: \begin{align*} y(t) = Tx(t) &= \int^1_t x(s)ds\\ &= \alpha - \int^t_0 x(s)ds ~~~~~(\alpha = \int^1_0 x(s)ds<\infty) \\ \Rightarrow -y'&=x \end{align*} Then \begin{align*} (Tx)(0) = \alpha <\infty~~~~~ &, ~~~~~(Tx)(1) = 0\\ (Tx)(t) = -(Ty')(t) &= \int^t_1 y'(s)ds = y(t) - y(1) \end{align*} if the integral exist, i.e. $y'(s)$ exists, $y' \in L^2(0,1)$ and $y(1) = 0$. Therefore $$\mathcal{R}(T) = \lbrace y \in H^1(0,1)~ \vert ~y(1) = 0\rbrace$$