Vrftsjtryk

Let f be a continuous function on the interval $[a, b]$. The function F defined by

$$ mathcal F(x) = int_a^x f(t)dt $$

is continuous on $[a,b]$, differentiable on $(a,b)$ and has derivative

$$mathcal F'(x) = mathcal f(x)$$

My question is the following: What will happen in this case?

$$mathcal H(x) = int_0^x e^{t^2} dt$$

Would the derivative be:

$$mathcal H'(x) = mathcal e^{x^2}$$
or

$$mathcal H'(x) = mathcal e^{x^2}-1$$

If we use the idea in your comment (to the question) then $mathcal{F} '(x) $ should equal $f(x) - f(a) $ and not just $f(x) $ as mentioned in your question. I guess the confusion comes from mixing two parts of the Fundamental Theorem of Calculus (henceforth referred to as FTC).

Part 1 of FTC deals with an integral of the form $int_{a} ^{x} f(t) , dt$ where the lower limit of integral is a constant $a$ and upper limit $x$ is a variable. This then defines a new function, say $mathcal{F} :[a, b] tomathbb{R} $ via the relation $$mathcal {F} (x) =int_{a} ^{x} f(t) , dttag{1}$$ The goal of part 1 of FTC is to study the properties of this new function $mathcal{F} $ in terms of properties of $f$. And it says that $mathcal{F} $ is continuous on $[a, b] $ and if $f$ is continuous at some $cin[a, b] $ then $mathcal{F} $ is differentiable at $c$ and $mathcal{F} '(c) =f(c) $.

You should notice that the lower limit $a$ does not figure out in conclusion of the theorem. The value $mathcal{F} (x) $ depends on $f, a$ and $x$ but the value $mathcal{F}' (x) $ depends on $f$ and $x$ only.

Part 2 of FTC deals with the evaluation of $int_{a} ^{b} f(x) , dx$ under certain conditions. It assumes that $f$ is Riemann integrable on $[a, b] $ and possesses an anti-derivative $mathcal{F} $ so that $$mathcal{F} '(x) =f(x), forall xin[a, b] $$ and then says that $$int_{a} ^{b} f(x) , dx=mathcal{F} (b) - mathcal {F} (a) tag{2}$$ It is here that both the upper and lower limits of integration play key role and the integral is expressed as difference between the values of the anti-derivative.

Note that the $mathcal{F} $ in both parts of FTC are different and in particular the $mathcal{F} $ in part 1 is not necessarily an anti-derivative of $f$.

Note that $F(x+h)= F(x) + int_x^{x+h} f(t)dt$ and for small $h$ we have $int_x^{x+h} f(t)dt approx int_x^{x+h} f(x)dt = f(x) h$. Hence we expect $F'(x) = f(x)$.

It is straightforward to make this argument rigorous.

The simplest when applying a new formula is to identify each component:

Let f be a continuous function on the interval $[a, b]$. The function F defined by

$$F(x) = int_a^x f(t)dt $$

is continuous on $[a,b]$, differentiable on $(a,b)$ and has derivative

$$F'(x) = mathcal f(x)$$

For $H(x) = int_0^x e^{t^2} dt$ we have $a=0$ (lower limit of integration) and $f(t)=e^{t^2}$.

It is also important to check that all conditions of the theorem are satisfied: here, the functions $g(t)=e^{t}$ and $h(t)=t^2$ are continuous on $mathbb R$, so their composition $f=gcirc h$ is continuous on $mathbb R$, hence on $[a,b]$.

Now, we can safely apply the formula: $H'(x) = f(x)=mathcal e^{x^2}$.

Edit: to answer a comment

What would happen if $a$ is not $0$?

Note that the formula depends only on $f$ and its continuity and not really on $a$. For example, consider $H_2(x)=int_1^{x} f(t); dt= int_1^x e^{t^2}; dt$. Then, all of the above applies here and we have

$$H_2'(x) = f(x) = e^{x^2} $$

Wait! Why do the functions $H_2$ and $H_1(x)=int^x_0 f(t); dt$ have the same derivative $f(x)=e^{x^2}$? That's clear when you remark that

$$H_1(x)=int_0^x f(t); dt=int_0^1 f(t); dt + int_1^x f(t); dt= C + H_2(x)$$

where $C=int_0^1 f(t); dt$. Since $C$ is a constant, we have $H_1'(x)=H_2'(x)$.

Recall that in general by Leibniz integral rule the following holds

$$F(x)=int_{a(x)}^{b(x)}g(u) duimplies F'(x)=g(b(x))cdot b'(x)-g(a(x))cdot a'(x)$$

therefore

$$mathcal H(x) = int_0^x e^{t^2} dtimplies mathcal H'(x)=e^{x^2}$$