Integrable function on $[0,2]$ and its antiderivative I got this question:
Let $f$ be the integrable function defined on the interval $[0,2]$ by the rule:
$f(x)=
\begin{cases} 
4x^3 & \text{if $0 \leq x \leq 1$} \\
x^2+2 & \text{if $1<x \leq 2$} \\
\end{cases}$
Does there exist real numbers $c1$ and $c2$ such that the function $F$ defined below is an antiderivative for $f$ (i.e. $F'=f$)?
$F(x)=
\begin{cases} 
x^4 + c_{1} & \text{if $0 \leq x \leq 1$} \\
\frac{x^3}{3}+2x+c_{2} & \text{if $1<x \leq 2$} \\
\end{cases}$
I got stuck since the function $f$ got step/jump discontinuity at $x=1$.
And I don't know how to proceed from there.
Thanks.
 A: If when you say $F' = f$ you mean $F'(x) = f(x)$ for every $x \in [0,2]$ so that $F$ is differentiable then note that $F$ is not differentiable at $x=1$. So then $\forall x \in [0,2] \setminus \{1\}$ we have $F'(x) = f(x)$ so in this sense the equality is found for any $c_1,c_2 \in \mathbb{R}$.
However there is a technical aspect that says these functions are not completely identical. Note that $F'$ is not defined at $x=1$ but $f$ is defined at $x=1$, and no $c_1, c_2$ can satisfy these needs.
A: Let's start with the original question as stated, and consider the requirement that $F' = f$ everywhere on $[0,2]$.  As we approach $1$, the left and right derivatives (left and right limits in the definition of the derivative) do not approach the same limit, so while $f(1)$ is defined, $F'(1)$ is not, so in this instance, one cannot choose $c_1$, $c_2$ to satisfy the hypotheses.

If instead of requiring $F' = f$ everywhere on $[0,2]$, we allow $F'=f$ almost everywhere on $[0,2]$ (i.e., on $[0,1)\cup (1,2]$), then we can note that since $f$ is absolutely integrable, $F(x) = \int_0^x f(t)\,\text{d}t$ is a continuous function of $x$ on $[0,2]$.
So to satisfy $F' = f$, for this definition of $F(x)$, on $[0,1)\cup (1,2]$, we could choose $c_1 = 0$, $c_2 = -4/3$; more generally, we could choose $c_1 \in \mathbb{R}$, $c_2 = c_1 - \frac{4}{3}$.  (These values would make the function $F(x)$ as defined immediately above continuous.)
However, if we simply require $F'=f$ on $[0,1)\cup(1,2]$ and do not ask $F$ to be continuous, then as the answer above states, we could choose any real numbers $c_1$, $c_2$, which would lead to (at worst) $F$ having a jump discontinuity at $x=1$.
