Domanda su Teorema Fondamentale del Calcolo integrale
Ciao , per applicare l'equazione proposta dal Teorema Fondamentale del calcolo integrale ( prima forma) , oltre alla condizione che f sia integrabile secondo Riemann in [a,b] è necessario che :
-la primitiva F sia continua in [a,b]
-F'(x)=f(x) tranne al più in un insieme finito di punti.
Non riesco a capire intuitivamente perchè è necessario presupporre queste due condizioni . Cosa potrebbe accadere se non fossero soddisfatte ? Grazie
-la primitiva F sia continua in [a,b]
-F'(x)=f(x) tranne al più in un insieme finito di punti.
Non riesco a capire intuitivamente perchè è necessario presupporre queste due condizioni . Cosa potrebbe accadere se non fossero soddisfatte ? Grazie
Risposte
They are necessary conditions for $F$ to be a primitive of $f$. If it's not continuous, then it cannot be an indefinite integral. If $F'(x)$ is not $f(x)$ except for a finite set, then $F$ is not a primitive of $f$ by definition (actually, I think this condition can be relaxed, but I think that for you is fine, because I suppose that you don't know measure theory).
What happens if a point b ( which i use for calculate the integral ) is a vertical inflection of the primitive function ?
What do you mean by a vertical inflection?
In italian it's said "flesso a tangente verticale"
So, in mathematical terms, you mean a function $F$ such that $\lim_{x\to b^-}F'(x)=\infty$, right?
yes
Well, if this happens, then it has no sense to apply this theorem using Riemann integrals at least in the proper sense. I mean, if you want a function to Riemann integrable (in order to use the theorem) it has to be bounded. Then, these functions you want to consider are not included in the standard theory.
On the other hand, there exists something called Riemann improper integrals, which are integrals defined as limits of Riemann integrals. In this sense, you can consider as an example the function $f(x)=\frac{1}{(1-x)^{1/2}}$. This is the derivative of $F(x)=-2(1-x)^{1/2}+2$ in $(0,1)$, you have that $\lim_{x\to 1^-}f(x)=\infty$ and the formula works perfectly by taking limits. Indeed, as $f$ is Riemann integrable in $[0,t]$ for all $0\leq t<1$ we can apply the Fundamental Theorem obtaining:
$$\int_0^t \frac{1}{(1-x)^{1/2}}dx=\left.-2(1-x)^{1/2}+2\right|_0^t=-2(1-t)+2-(-2+2)=-2(1-t)^{1/2}+2=F(t), \quad t>0.$$
Now, if you take limits when $t\to 1^-$, then you get that the improper integral $\int_0^1 \frac{1}{(1-x)^{1/2}}dx$ is exactly $F(1)$, because we define it as the limit of the integrals in $[0,t]$:
$$\int_0^1 \frac{1}{(1-x)^{1/2}}dx=\lim_{t\to 1^-}\int_0^t \frac{1}{(1-x)^{1/2}}dx=\lim_{t\to 1^-}-2(1-t)^{1/2}+2=\lim_{t\to 1^-}F(t)=F(1),$$
because $F$ is continuous at $1$.
So, as you can see, the theorem works in this case. This works because $f$ is a Lebesgue integrable function in $[0,1]$, so you can apply a Lebesgue's version of the Fundamental Theorem (you can find this version in Rudin's book about Real and Complex analysis).
In general, it works (by using another method of integration) for all differentiable functions. Indeed, the following general theorem is true (by using the correct method of integration):
Theorem: Let $F:[a,b]\to\mathbb{R}$ a differentiable function in $[a,b]$. Then $F'$ is integrable in $[a,b]$ and, furthermore:
$$\int_a^x F'(t)dt=F(x)-F(a),\qquad x\in [a,b].$$
Moreover, we can weaken this hypothesis by asking $F$ to be differentiable except for a countable quantity of points and, if I'm not mistaken, this countable set can contain points where $F'$ is infinite.
Unfortunately, this theorem is not very known, so you will have to use Riemann's or Lebesgue's version of it, which ask the derivative to be integrable in order to apply the theorem (in particular, for Riemann's version, the derivative has to be bounded and, for the Lebesgue's version, you can find examples of not bounded but integrable derivatives, as the example I gave you).
On the other hand, there exists something called Riemann improper integrals, which are integrals defined as limits of Riemann integrals. In this sense, you can consider as an example the function $f(x)=\frac{1}{(1-x)^{1/2}}$. This is the derivative of $F(x)=-2(1-x)^{1/2}+2$ in $(0,1)$, you have that $\lim_{x\to 1^-}f(x)=\infty$ and the formula works perfectly by taking limits. Indeed, as $f$ is Riemann integrable in $[0,t]$ for all $0\leq t<1$ we can apply the Fundamental Theorem obtaining:
$$\int_0^t \frac{1}{(1-x)^{1/2}}dx=\left.-2(1-x)^{1/2}+2\right|_0^t=-2(1-t)+2-(-2+2)=-2(1-t)^{1/2}+2=F(t), \quad t>0.$$
Now, if you take limits when $t\to 1^-$, then you get that the improper integral $\int_0^1 \frac{1}{(1-x)^{1/2}}dx$ is exactly $F(1)$, because we define it as the limit of the integrals in $[0,t]$:
$$\int_0^1 \frac{1}{(1-x)^{1/2}}dx=\lim_{t\to 1^-}\int_0^t \frac{1}{(1-x)^{1/2}}dx=\lim_{t\to 1^-}-2(1-t)^{1/2}+2=\lim_{t\to 1^-}F(t)=F(1),$$
because $F$ is continuous at $1$.
So, as you can see, the theorem works in this case. This works because $f$ is a Lebesgue integrable function in $[0,1]$, so you can apply a Lebesgue's version of the Fundamental Theorem (you can find this version in Rudin's book about Real and Complex analysis).
In general, it works (by using another method of integration) for all differentiable functions. Indeed, the following general theorem is true (by using the correct method of integration):
Theorem: Let $F:[a,b]\to\mathbb{R}$ a differentiable function in $[a,b]$. Then $F'$ is integrable in $[a,b]$ and, furthermore:
$$\int_a^x F'(t)dt=F(x)-F(a),\qquad x\in [a,b].$$
Moreover, we can weaken this hypothesis by asking $F$ to be differentiable except for a countable quantity of points and, if I'm not mistaken, this countable set can contain points where $F'$ is infinite.
Unfortunately, this theorem is not very known, so you will have to use Riemann's or Lebesgue's version of it, which ask the derivative to be integrable in order to apply the theorem (in particular, for Riemann's version, the derivative has to be bounded and, for the Lebesgue's version, you can find examples of not bounded but integrable derivatives, as the example I gave you).
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