The area under the impulse (Dirac delta) function

Question

Why is the area under the Dirac delta function equal to one and not zero? Shouldn't it be zero since the function is symmetric and the area under of each side cancels out the other?

Answer 1

The Dirac delta is no function, but a distribution.

The desire is to have $$ \int\limits_{-\infty}^{\infty} \delta(x) f(x) \, dx = f(0) \quad (*) $$ It is clear, that no function $\delta$ can achieve this.

Interpreted as linear functional $$ \langle \delta, f \rangle = f(0) $$ where $\langle . , . \rangle$ is a scalar product, typically involving an integration. This works in the theory of distributions. It can be approximated by sequences, as achile pointed out in the comments.

So it is kind of fishy to speculate about the symmetry of $\delta$, which has to vanish for any point but the origin, and can not have a finite value at the origin. One might look at the symmetry of the approximating functions, but that will not necessarily allow a transfer to the limit function I believe.

If you apply the definition $(*)$ to $f(x) = 1$, you end up with $$ \int\limits_{-\infty}^{\infty} \delta(x)\, dx = 1 $$