Trying to prove a local minimum, calculus 2 I'm asked to prove that the function
$$
F(x)=\int_{0.5}^{x} e^{t^2}\cdot(t^2-1) \,dt\
$$
has a local minimum at $x=1$.
I've tried defining a new function, $f$ such that $f(x)=e^{x^2}\cdot(x^2-1)$, and then applying $x=1$,
then claiming that $f(x)$ is derivative of $F(x)$,therefor $f(1)=0$ and showing that for some $x_0<1 \rightarrow f(x_0)<0$
and for another $x_1>1 \rightarrow f(x_1)>0$ which confirms that $x=1$ is local minimum for $F(x)$.
Only to realize that $t$ is substitute for something, and therefor I cannot claim $t=x$.
I'm stuck in my thoughts, any direction/idea of what I'm trying/needed to prove would be really helpful.
Cheers!
 A: Your solution is essentially correct. One small point to consider is that you say that "$f(x)$ is the derivative of $F(x)$ therefore $f(1)=0$". This should be phrased more carefully: you want to show that $x=1$ is a local minimum, it is not that which you already know. So in order to show that $x=1$ is a critical point, you have to show that $f(1)=0$, which is indeed true. To show that it is a local minimum you can differentiate again:
$$f'(x)=e^{x^2}\cdot 2x + e^{x^2}2x(x^2-1)$$
therefore  $F''(1)=f'(1)>0$ which shows that $x=1$ is a local minimum (alternatively, the way you did it by analysing $f(x)$ for $x>1$ and $x<1$ separately is just fine as well).
You never substituted $t=x$. Recall that the fundamental theorem of calculus says that $F'(x)=f(x)$. The $t$ inside the definition of $F(x)$ is a dummy variable, the real variable is $x$ in both $F(x)$ and $f(x)$. To expand on this point:
In the definition of $F(x)=\int_{0.5}^x e^{t^2}\cdot (t^2-1)\,dt$, which is the variable of the function $F(x)$, is it $x$ or $t$? Surely it is $x$, for $t$ is but an internal dummy variable. Consider this analog: let $a(n) = \sum_{k=1}^n b(k)$. Which is the variable, $n$ or $k$? Surely it is $n$, for $k$ is but the internal variable for summation; it does not exist outside the summation, it only exists to perform the summation. We could have just as well written the summation without the variable $k$, say $b(1)+b(2)+\ldots+b(n)$. In the very same way in $F(x)$ the variable is $x$; it denotes the upper bound of integration. The variable $t$ is an internal variable which does not exist outside the integral. For this reason the fundamental theorem of calculus says that $F'(x)=f(x)$, rather than $F'(x)=f(t)$, which would not make sense, as $t$ would be a free variable on the right hand side but it does not exist as a free variable on the left hand side (as it exists only inside the definition of the integral).
A: We may use libenitz rule for computing f'(x) and then we can calculate f"(x) very easily.
https://ibb.co/1n6B1fk
https://ibb.co/YZC71B8
Hope it helps !
