MHB First fundamental theorem of Calculus

[evinda]

Hello! :)

I am looking at the theorem:
"$f:[a,b] \to \mathbb{R}$ integrable
We suppose the function $F:[a,b] \to \mathbb{R}$ with $F(x)=\int_a^x f$.If $x_0$ a point where $f$ is continuous $\Rightarrow F$ is integrable at $ x_0$ and $F'(x_0)=f(x_0)$".

There is a remark that the theorem stands only if $f$ is continuous and there is the following example:

$f(x)=\left\{\begin{matrix}
-1,-1 \leq x \leq 0\\
1,0 < x \leq 1
\end{matrix}\right.$
which is not continuous at $0$.
Then,according to the textbook, the function $F(x)$ is equal to $|x|-1$..But,why is it like that??And also why is $F$ not differentiable at $0$??

 

[Deveno]

Well on the intervals [-1,0] and (0,1] $f$ is constant, so it doesn't matter which points you pick for the Riemann sums you get:

[math]\int_{-1}^x f(t)\ dt = -x - 1[/math] for $x \in [-1,0]$

[math]\int_0^x f(t)\ dt = x[/math] for $x \in (0,1]$

(I urge you to actually calculate these Riemann sums).

From these integrals, we see that for $x \in [0,1]$:

[math]\int_{-1}^x f(t)\ dt = \int_{-1}^0 f(t)\ dt + \int_0^x f(t)\ dt = -1 + x[/math],

so that in all cases, $F(x) = |x| - 1$.

The problem with the differentiability of $F$ is in trying to determine $F'(0)$, we have:

[math]\lim_{h \to 0^-} \frac{F(0+h) - F(0)}{h}[/math]

[math] = \lim_{h \to 0^-} \frac{[-(0+h) - 1] - (0 - 1)}{h}[/math]

[math] = \lim_{h \to 0^-} \frac{-h}{h} = -1[/math]

whereas:

[math]\lim_{h \to 0^+} \frac{F(0+h) - F(0)}{h}[/math]

[math] = \lim_{h \to 0^+} \frac{[(0+h) - 1] - (0 - 1)}{h}[/math]

[math] = \lim_{h \to 0^+} \frac{h}{h} = 1[/math],

so this limit does not exist.

 

[evinda]

Well on the intervals [-1,0] and (0,1] $f$ is constant, so it doesn't matter which points you pick for the Riemann sums you get:

[math]\int_{-1}^x f(t)\ dt = -x - 1[/math] for $x \in [-1,0]$

[math]\int_0^x f(t)\ dt = x[/math] for $x \in (0,1]$

(I urge you to actually calculate these Riemann sums).

From these integrals, we see that for $x \in [0,1]$:

[math]\int_{-1}^x f(t)\ dt = \int_{-1}^0 f(t)\ dt + \int_0^x f(t)\ dt = -1 + x[/math],

so that in all cases, $F(x) = |x| - 1$.

The problem with the differentiability of $F$ is in trying to determine $F'(0)$, we have:

[math]\lim_{h \to 0^-} \frac{F(0+h) - F(0)}{h}[/math]

[math] = \lim_{h \to 0^-} \frac{[-(0+h) - 1] - (0 - 1)}{h}[/math]

[math] = \lim_{h \to 0^-} \frac{-h}{h} = -1[/math]

whereas:

[math]\lim_{h \to 0^+} \frac{F(0+h) - F(0)}{h}[/math]

[math] = \lim_{h \to 0^+} \frac{[(0+h) - 1] - (0 - 1)}{h}[/math]

[math] = \lim_{h \to 0^+} \frac{h}{h} = 1[/math],

so this limit does not exist.

I understand..Thanks a lot! (Nod)