continuity of a integral upper limit function As we know, if $f(x)$ is integrable on $[a,b]$, then the integral upper limit  function $\int_{a}^{x} f(t)dt$ is continuous.
Given a integrable piecewise function, why is the upper limit function discontinuous?
$$
f(x) = \begin{cases} x^2, & x \in[0,1) \\ x+1, & x \in[1,2] \end{cases}
$$
Obviously, the upper limit function is not continuous.
 A: You have the piecewise function
$$f(x) = \begin{cases} x^2, & x \in[0,1) \\ x+1, & x \in[1,2] \end{cases} \tag{1}\label{eq1A}$$
Define for $x \in [0,2]$ what you call the "upper limit" function of
$$g(x) = \int_{0}^{x}f(t)dt \tag{2}\label{eq2A}$$
As you stated in your comment, for $x \in [0,1)$, you have
$$g(x) = \int_{0}^{x}t^2 dt = \frac{x^3}{3} \tag{3}\label{eq3A}$$
For $x \ge 1$, though, the integration involves using the $x^2$ for $f(x)$ given in \eqref{eq1A}, namely the first part for up to $1$, and then the second case of $x + 1$ for any part greater than $1$. Thus, for $x \in [1,2)$, you have
$$\begin{equation}\begin{aligned}
g(x) & = \int_{0}^{1}t^2 dt + \int_{1}^{x}(t+1) dt \\
& = \left. \frac{t^3}{3} \right\rvert_{0}^{1} + \left. \left(\frac{t^2}{2} + t\right) \right\rvert_{1}^{x} \\
& = \frac{1}{3} + \left(\frac{x^2}{2} + x\right) - \left(\frac{1}{2} + 1\right) \\
& = \frac{x^2}{2} + x - \frac{7}{6}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
This matches the expression you gave in your comment. Note $g(x)$ is continuous on each sub-interval where it's defined. Also, from \eqref{eq3A}, you have
$$\lim_{x \to 1^{-1}}g(x) = \frac{1}{3} \tag{5}\label{eq5A}$$
In addition, from \eqref{eq4A}, you have
$$\lim_{x \to 1^{+}}g(x) = g(1) = \frac{1}{2} + 1 - \frac{7}{6} = \frac{1}{3} \tag{6}\label{eq6A}$$
This shows, by the definition of continuity, that $g(x)$ is continuous at $x = 1$ and, thus, for all $x \in [0,2]$.
In effect, $g(x)$ has become a piecewise function that is continuous at it's point of change, i.e., $x = 1$. If you were to graph it, you would see the $2$ parts would join at $x = 1$, but there would be a "kink" there. This is because for $x \lt 1$, you get $g'(x) = x^2$, so $\lim_{x \to 1^{-}}g'(x) = 1$, but for $x \gt 1$, you have $g'(x) = x + 1$, so $\lim_{x \to 1^{+}}g'(x) = 2$. Thus, the curve's slope makes a sudden change at $x = 1$. This shows that $g(x)$ is not differentiable at $x = 1$.
Another way to look at this is that, for $x \ge 1$, you are taking the cumulative "area" defined by the integral up to $x = 1$, and then adding to it the part which is greater than $1$, so there's a continuous change of the "area". In general, integration tends to have a "smoothing" effect on functions, such as this example shows where the function you are integrating is a discontinuous piecewise function but its integral is continuous.
