max and min of $f(x,y)=\int_{x}^{y} e^{-t^2} dt$ I want to study max e min of this function on $C=\{(x,y): x^2+y^2\le1\}$
$\nabla f(x,y)=(-e^{-x^2},e^{-y^2})$ so there isn't max and min in C because the gradient isn't null.
But on the boundary of C?
 A: Either by using Lagrange multipliers or by parameterizing $f$ on the boundary by the angle $\theta$ reduces the problem to solving the following equation:
$$
\cos\theta e^{-sin^2\theta} = -\sin\theta e^{-cos^2\theta}$$
This can be rewritten as $$ -\tan\theta = e^{\cos2\theta}$$
We first note that there are no solutions with $\theta \in (0,\pi/2)$ since on this interval $-\tan\theta <0$.  On the interval $(-\pi/2, 0)$ we have that $e^{\cos2\theta}$ is increasing and $-\tan\theta$ is decreasing, so there is at most one point of intersection, which by inspection is at $\theta = -\pi/4$.  Since both equations are $\pi$-periodic, the solutions on $[-\pi,\pi]$ are given by $\theta = -\pi/4, 3\pi/4$.  These correspond to the points $(\sqrt{2}/2, -\sqrt{2}/2)$ and $(-\sqrt{2}/2, \sqrt{2}/2)$.
A: I formulate my approach as an answer.
You already stated that there are no inner solutions. Observe that $exp(-t^2)$ is always positive, and symmetric via $t=0$(which is also its maximum value). So given these facts we just have to maximize the integration domain $[x,y]$ subject to C. (Why?)
$max ~ y-x$ s.t. $x^2+y^2=1$
which solution is $x= -\sqrt{1/2}$, $y= \sqrt{1/2}$
for the minimum value you just have to flip the signs. 
Sorry, this is a bit short. I hope you can follow my steps.
Greetings.
