Consider the Banach Space $C[0,1]$. Find decomposition of spectrum of the indefinite integral operator. 
Consider the Banach Space $C[0,1]$ of real-valued continuous function on $[0,1]$ with the supremum norm. and the linear operator
  $$A: x(t)\mapsto\int\limits_0^tx(s)\,\mathrm{d}s.$$
  Find its eigenvalues, regular values and continuous spectrum.

I already proved that it has not any eigenvalues but I have problems finding the classification for the elements in $(-1,0)\cup(0,1)$.
By theorem, since $\|A\|=1$, then for $|\lambda|>1$ we have that $\lambda$ is a regular value.
Question: How do I know if $(Ax-\lambda x)^{-1}$ exists for $\lambda\in(-1,0)\cup(0,1)$.
 A: We will show that $\sigma(A) = \{0\}$, and that $0$ belongs to the residual spectrum. As you have shown, $0$ is not an eigenvalue, and as $\operatorname{im} A \subseteq \{x \in C[0,1]: x(0) = 0\}$, $A$ does not have dense image. Hence $0 \in \sigma_r(A)$. 
To see that $A - \lambda$ is invertible for $\lambda \ne 0$, let $y \in C[0,1]$ be given. We have to solve 
$ Ax - \lambda x = y $
for $x$, or equivalently, with $z := Ax$, 
$$ z - \lambda z' = y, \quad z(0) = 0  $$
We rewrite it as
$$ z' - \frac 1\lambda z =- \frac 1\lambda y, \quad z(0) = 0 $$
Variation of constants gives 
$$ z(t) =- \exp\frac t\lambda \cdot \frac 1\lambda \int_0^t\exp\left(-\frac s\lambda\right) y(s) \, ds $$
Hence
$$ x(t) = \frac 1\lambda \bigl(z(t) - y(t)\bigr)
   =- \frac 1{\lambda^2} \int_0^t \exp\left(\frac{t-s}\lambda\right) y(s)\, ds  - \frac 1\lambda y(t) = (A - \lambda)^{-1}y(t)  $$
So, $A-\lambda$ is invertible for $\lambda \ne 0$.
A: Here is an alternative proof of the fact that $r(T)=0$ ( I add it because it does not seem to appear in the related posts and I find it quite nice).
First show by induction that $|A^n x(t)|\leq \frac{t^n}{n!}\|x\|_\infty$ for all $t\in[0,1]$:
Basis $n=0$: $|x(t)|\leq \|x\|_\infty$ for all $t\in[0,1]$ by definition of the supremum norm.
Inductive step: For all $t\in[0,1]$ we have
$$|A^{n+1}x(t)|=\lvert\int_0^t A^n x(s)\,ds\rvert
\leq \int_0^t |A^n x(s)|\,ds\leq\|x\|_\infty\int_0^t \frac{s^n}{n!}\,ds=\frac{t^{n+1}}{(n+1)!}\|x\|_\infty.$$
It follows that $\|A^n\|=\frac{1}{n!}$ and consequently $r(T)=\lim_{n\to\infty}\|A^n\|^{\frac 1 n}=0$.
