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First fundamental theorem of Calculus!

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,723
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 $0$?? :confused:
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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:

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

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

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

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

\(\displaystyle \int_{-1}^x f(t)\ dt = \int_{-1}^0 f(t)\ dt + \int_0^x f(t)\ dt = -1 + x\),

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:

\(\displaystyle \lim_{h \to 0^-} \frac{F(0+h) - F(0)}{h}\)

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

\(\displaystyle = \lim_{h \to 0^-} \frac{-h}{h} = -1\)

whereas:

\(\displaystyle \lim_{h \to 0^+} \frac{F(0+h) - F(0)}{h}\)

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

\(\displaystyle = \lim_{h \to 0^+} \frac{h}{h} = 1\),

so this limit does not exist.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,723
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:

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

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

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

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

\(\displaystyle \int_{-1}^x f(t)\ dt = \int_{-1}^0 f(t)\ dt + \int_0^x f(t)\ dt = -1 + x\),

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:

\(\displaystyle \lim_{h \to 0^-} \frac{F(0+h) - F(0)}{h}\)

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

\(\displaystyle = \lim_{h \to 0^-} \frac{-h}{h} = -1\)

whereas:

\(\displaystyle \lim_{h \to 0^+} \frac{F(0+h) - F(0)}{h}\)

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

\(\displaystyle = \lim_{h \to 0^+} \frac{h}{h} = 1\),

so this limit does not exist.
I understand..Thanks a lot!!! (Nod)