Values of a primitive $F$ is a primite of function $f:\mathbb R \to \mathbb R, \: f(x)=e^{x^2}$. $F(-1) = 0$. Then:
A) $F(1) < 0$
B) $F(1) = 0$
C) $F(1) > 2$
D) $F(1) = 2$
We know that $F'(x)=f(x) \: \forall x \in \mathbb R$. $F'(x) > 0 \: \forall x \in \mathbb R $ so F is strictly ascending. Now we know that $F(1) > 0$. But The answer is C. I don't know from where that $2$ could come from.
 A: Notice that
$$f(0)=1$$
$$f'(0)=0$$
$$f''(0)>0$$
Thus,
$$f(x)\ge1$$
And likewise,
$$F(1)=\int_{-1}^1f(x)\ dx\ge\int_{-1}^11\ dx=2$$
and with some effort, it should be obvious that $F(1)=2$ is not possible.
A: By the Mean Value Theorem, there exists $\xi \in (-1,1)$ so that $$\frac{F(1)-F(-1)}{2} = F'(\xi) = f(\xi) = e^{\xi^2} \ge e^0 \ge 1,$$ from which we deduce that $F(1) \ge F(-1) + 2 = 2$.

I overlooked option (D).  To show strict inequality, we need to apply Lagrange Remainder Theorem twice on $F(1)$ and $F(-1)$.
\begin{align}
F'(x) &= f(x) = e^{x^2} \\
F''(x) &= f'(x) = 2xe^{x^2}
\end{align}
So
\begin{align}
F'(0) &= 1 \\
F''(0) &= 0 \\
\end{align}
\begin{align}
F(-1) &= F(0) - \frac{F'(0)}{1} + \underbrace{\frac{F''(\xi_1)}{2}(-1)^2}_{<0} \text{ for some } \xi_1 \in (-1,0) \\
0 &< F(0) - \frac{1}{1} \\
F(0) &> 1 \\
F(1) &= F(0) + \frac{F'(0)}{1} + \underbrace{\frac{F''(\xi_2)}{2}}_{>0} \text{ for some } \xi_2 \in (0,1) \\
&> 1 + 1 + 0 = 2
\end{align}
In fact, we only need $F''(x) = f'(x) = 2xe^{x^2}$ to exclude (D).  No further derivatives are needed.
