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@user599310, I am going to attempt some pseudo math to show it: $$ I^2 = \int e^-x^2 dx \times \int e^-x^2 dx = Area \times Area = Area^2$$ We can replace one x, with a dummy variable, move the dummy copy into the first integral to get a double integral. $$ I^2 = \int \int e^ {-x^2-y^2} dA $$ In context, the integrand a function that returns. You will get the same answer because when you perform a change of variables, you change the limits of your integral as well (integrating in the complex plane requires defining a contour, of course, so you'll have to be careful about this). The integral which you describe has no closed form which is to say that it cannot be expressed in elementary functions

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For example, you can express $\int x^2 \mathrm {d}x$ in elementary functions such as $\frac {x^3} {3} +c$. Also, it makes sense logically if you recall the fact that the derivative of the function is the function's slope, because any function f (x)=c will have a slope of zero at point on the function. Answers to the question of the integral of $\frac {1} {x}$ are all based on an implicit assumption that the upper and lower limits of the integral are both positive real numbers.

What is the value of this?

If by integral you mean the cumulative distribution function $\phi (x)$ mentioned in the comments by the op, then your assertion is incorrect. But what else is there I've been learning the fundamental theorem of calculus So, i can intuitively grasp that the derivative of the integral of a given function brings you back to that function

Is this also the case. The integral of 0 is c, because the derivative of c is zero