we have the following integral operator:
$T:C^0[0,1] \rightarrow C^0[0,1]$, with $Tu(x)=e^x \int_{0}^{x}e^{-t}u(t)dt$.
This operator is not injective since I found $Tu(x)=0$ for $u(t)=e^t(2t-x)$. So $\ker(T) \ne \langle {\vec 0} \rangle$, thus it's not injective.
Am I wrong?
You are wrong ! Your function $u$ contains the parameter $x$. $u$ has to be independent of $x$.
$T$ is injective. To this end let $u \in C^0[0,1]$ and $Tu=0$. This means that
$e^x \int_{0}^{x}e^{-t}u(t)dt=0$ for all $x \in [0,1].$
Hence $\int_{0}^{x}e^{-t}u(t)dt=0$ for all $x \in [0,1]$ and therefore $\frac{d}{dx}\int_{0}^{x}e^{-t}u(t)dt=e^{-x}u(x)=0$ for all $x \in [0,1]$.
Conclusion: $u(x)=0$ for all $x \in [0,1]$.