# Prove F is continuous in the [0,2] interval even if f is not.

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?

• With this definition of $F$ it is constant, because it doesn't depend on $x$. Commented Jan 19, 2020 at 0:49
• I think you meant x, not 2, as the limit of the integral in the first integral Commented Jan 19, 2020 at 0:51
• 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. Commented Jan 19, 2020 at 0:57
• Yes, I have just corrected it. Commented Jan 19, 2020 at 0:58
• If C=-1 then F(x) is continuous ? Commented Jan 19, 2020 at 1:00

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 $$[0,2]$$.