Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $g(x) = \int^x_0 f(t) \, dt = \int^x_0 t^{1/3} \, dt = \frac{3}{4}x^{4/3}$, where $f(t) = t^{1/3}$.
Show that the function $h(x) = \int^x_0 g(t) \, dt$ is $C^2$ but not $C^3$ at $x = 0$.
Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $g(x) = \int^x_0 f(t) \, dt = \int^x_0 t^{1/3} \, dt = \frac{3}{4}x^{4/3}$, where $f(t) = t^{1/3}$.
Show that the function $h(x) = \int^x_0 g(t) \, dt$ is $C^2$ but not $C^3$ at $x = 0$.
$f(x)$ is continuous, but not differentiable at $0$; so $f$ is in $\mathcal{C}^0$, but not in $\mathcal{C}^1$, with the problem being at $x=0$.
Since $f(t)$ is continuous by the Second Fundamental theorem of Calculus we have that $g(x)$ is differentiable and: $\frac{d}{dx} g(x) = \frac{d}{dx}\int_0^x f(t)\,dt = f(x),$ so $g(x)$ has continuous derivative; but the derivative is not differentiable at $0$. That is, $g\in\mathcal{C}^1$, but not in $\mathcal{C}^2$, with the problem being at $x=0$.
Since $g(x)$ is continuous (it is differentiable), then again by the Second Fundamental Theorem of Calculus we know that $h(x)$ is differentiable, and $\frac{d}{dx}h(x) = \frac{d}{dx}\int_0^x g(t)\,dt = g(x).$ Therefore, \frac{d^2}{dx^2}h(x) = \frac{d}{dx}h'(x) = \frac{d}{dx}g(x) = f(x), so $h(x)$ has continuous second derivative; that is, $h\in\mathcal{C}^2$.
However, the second derivative is not differentiable everywhere (not differentiable at $0$), so $h(x)$ does not have a third derivative defined everywhere, and so cannot be $\mathcal{C}^3$, with the problem being at $x=0$.
More generally, if $f(x)$ is in $\mathcal{C}^k$ but not in $\mathcal{C}^{k+1}$, then $\mathcal{F}(x) = \int_0^x f(t)\,dt$ is in $\mathcal{C}^{k+1}$ but not in $\mathcal{C}^{k+2}$, by the Second Fundamental Theorem of Calculus.