The integral which you describe has no closed form which is to say that it cannot be expressed in elementary functions. For example, you can express $\int x^2 \mathrm {d}x$ in elementary functions such as $\frac {x^3} {3} +C$. However, the indefinite integral from $ (-\infty,\infty)$ does exist and it is $\sqrt {\pi}$ so explicitly: $$\int^ {\infty}_ {-\infty} e^ {-x^2} = \sqrt {\pi}$$ Note ...
A different type of integral, if you want to call it an integral, is a "path integral". These are actually defined by a "normal" integral (such as a Riemann integral), but path integrals do not seek to find the area under a curve. I think of them as finding a weighted, total displacement along a curve.
5 An integral domain is a ring with no zero divisors, i.e. $\rm\ xy = 0\ \Rightarrow\ x=0\ \ or\ \ y=0:.:$ Additionally it is a widespread convention to disallow as a domain the trivial one-element ring (or, equivalently, the ring with $: 1 = 0:$). It is the nonexistence of zero-divisors that is the important hypothesis in the definition.
The integral of 0 is C, because the derivative of C is zero. 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.
This is the integral that is sometimes described as "the area under the curve" (although I would consider that an application of the definite integral, not a definition).
@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 ...