1. The problem is to evaluate the function $x(t)$ given by the integral expression: $$x(t) = \int_0^1 x'(s) \, ds + x(0)$$ 2. Recall the Fundamental Theorem of Calculus, which states that if $x'(s)$ is the derivative of $x(s)$, then: $$\int_a^b x'(s) \, ds = x(b) - x(a)$$ 3. Applying this theorem to the integral from 0 to 1, we get: $$\int_0^1 x'(s) \, ds = x(1) - x(0)$$ 4. Substitute this back into the original expression: $$x(t) = x(1) - x(0) + x(0)$$ 5. Simplify the right-hand side: $$x(t) = x(1)$$ 6. Therefore, the value of $x(t)$ is equal to $x(1)$, which is the value of the function $x$ at 1, independent of $t$ in this expression. Final answer: $$x(t) = x(1)$$