Level curves for "unsolvable" integral

Question

Problem: Sketch the level curves of g defined by $$g(x,y)=\int_x^y{e^{-t^2}dt}$$

Attempts at solution:

(1) Apparently we could take $y=x$, then for $z=0$ the level curves will be just that. However, what about other $z$?

(2) We could $\frac{d}{dt}$ the integral using the Fundamental theorem of calculus, and then draw slopes (direction lines) of the level curves. But in this case we don't even know for what $z$ each level curve would be. On the other hand, we may not need to know that and simply use direction fields to make the sketch.

I'm convinced that there's a more clever approach to this problem. Any hints would be much appreciated.

Answer 1

By finding the gradient of $g$, you can show that the level curves should satisfy the differential equation

$$ \dfrac{dy}{dx} = e^{y^2-x^2} $$

This will be $1$ along the lines $y = \pm x$, and always positive. Of the four regions into which these lines divide the plane, it will be greater than $1$ in two and less in the other two.

That should give you enough to do a reasonable sketch.

Answer 2

Just define the function and plot it, confirming that the symmetry across $y=x$ is preserved.

You can also plot the two-dimensional gradient vector atop the magnitude of the gradient.