Is the indefinite integral of a piecewise continuous function a continuous function?

Question

I had looked around on the web and can't find much information related to the integration of piecewise continuous functions.

Let's say we have a simple function

$$f(x)= \begin{cases} 0 & x\leq 0 \\ x & 0\leq x\leq 1 \\ 0 & x>1 \end{cases}$$

and we are looking to find the integral $\int f(x) dx$

WolframAlpha gives me the following result:

$$\int f(x) dx= \begin{cases} c & x\leq 0 \\ \frac{x^2}{2}+c & 0\leq x\leq 1 \\ \frac{1}{2}+c & x>1 \end{cases}$$

The cases $x\leq 0$ and $0\leq x\leq 1$ are clear. But why do we need the $\frac{1}{2}$ for the $x>1$ case? I see that this makes the integral continuous but is it necessary, i.e. is it wrong to have simply $c$ for the $x>1$ case for the integral?

Answer 1

The short answer is no. An indefinite integral is not a continuous function, because an indefinite integral is not a function. However, once you fix $a\in \mathbb{R}$, $$\int_a^x f(t)dt$$ is a continuous function. Note that once we fix $a$, the constant of integration is completely determined.

Answer 2

Actually, the indefinite integral of a continuous function is a family of continuous functions. If you want to find exactly one of these functions, you have to fix an initial condition $F(x_0) = C_0$.

If you want to obtain the same function obtained by WolframAlpha, solve for the following equalities: lateral limits