Let $f(x)$ be $$f(x) =\Biggr\{ \begin{matrix} x, \;\;\; x\in[0,1] \\ x+1, ;\;\; x\in(1,2]\\ \end{matrix} $$

We define $F$ with $F(0) = 0\;$ and

$$F(x) = \int_{0}^{x} f(t) dt, \;\;\; x \in (0,2]$$

Determine $F$ and prove that $F$ is continuous in the $[0,2]$ interval even if $f$ is not.

I did the integral of $f(x)$ and I obtained these results:

$$F(x) =\Biggr\{ \begin{matrix} \frac{x^2}{2}, \;\;\; x\in[0,1] \\ \frac{x^2}{2}+x+C, \;\;\; x\in(1,2]\\ \end{matrix} $$

I did the lateral limits as well, but $F(x)$ continued without being continuous. What am I doing wrong?

Thanks in advance.

  • $\begingroup$ With this definition of $F$ it is constant, because it doesn't depend on $x$. $\endgroup$
    – user289143
    Commented Jan 19, 2020 at 0:49
  • $\begingroup$ I think you meant x, not 2, as the limit of the integral in the first integral $\endgroup$ Commented Jan 19, 2020 at 0:51
  • $\begingroup$ Note that the continuity of $F(x)$ follows from the fundamental theorem of calculus (or at least a slight generalization of it); If $f(x)$ is integrable, then $F(x)$ is continuous. $f(x)$ is integrable because it has a single jump discontinuity. $\endgroup$
    – bjorn93
    Commented Jan 19, 2020 at 0:57
  • $\begingroup$ Yes, I have just corrected it. $\endgroup$
    – user9867
    Commented Jan 19, 2020 at 0:58
  • $\begingroup$ If C=-1 then F(x) is continuous ? $\endgroup$ Commented Jan 19, 2020 at 1:00

1 Answer 1


Perhaps you meant $F(x)=\int_0^x f(t)dt$ for $x \in (0,2]$

Now, for $x \in (0,1]$ we have $F(x)=\int_0^x tdt=\frac{x^2}{2}$

Instead, for $x \in (1,2]$ we have $F(x)=\int_0^1 tdt + \int_1^x (t+1) dt=\frac{1}{2}+\frac{x^2}{2}+x-\frac{1}{2}-1=\frac{x^2}{2}+x-1$

Clearly $F$ is continuous on $[0,1)$ and $(1,2]$ but taking the limits at $1$ and $F(1)$ we can see that they all coincide and they are equal to $\frac{1}{2}$. Therefore $F(x)$ is continuous on $[0,2]$.


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