Basically, there are the following inclusions
$(\star)$ $\mathcal{D}(\mathbb{R}^n) \subset \mathcal{S}(\mathbb{R}^n) \subset \mathcal{E}(\mathbb{R}^n)$ and $\mathcal{E}'(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n) \subset \mathcal{D}'(\mathbb{R}^n)$.
Also $L^p(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n)$. Almost all of these inclusions are also continuing, i.e. $\mathcal{D}_K(\mathbb{R}^n) \hookrightarrow \mathcal{S}(\mathbb{R}^n)$, or also $L^p(\mathbb{R}^n) \hookrightarrow \mathcal{S}'(\mathbb{R}^n)$, this means that the inclusion operator $\iota : L^p(\mathbb{R}^n) \longrightarrow \mathcal{S}'(\mathbb{R}^n)$ is continuous with respect to topology defined in those spaces, which basically may be a topology induced by a separable seminorms family or by norm. Where
(a) $\mathcal{D}(\mathbb{R}^n)$ is the space test functions
(b) $\mathcal{S}(\mathbb{R}^n)$ is the Schartz space
(c) $\mathcal{E}(\mathbb{R}^n)$ is the space of the regular functions
(d) $\mathcal{E}'(\mathbb{R}^n)$ is the space of the distribution with compact support
(e) $\mathcal{S}'(\mathbb{R}^n)$ is the space of the tempered distributions
(f) $\mathcal{D}'(\mathbb{R}^n)$ is the space of the distributions
For example, we have the inclusions $\mathcal{E}'(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n) \subset \mathcal{D}'(\mathbb{R}^n)$ because the inclusions $\mathcal{D}(\mathbb{R}^n) \hookrightarrow \mathcal{S}(\mathbb{R}^n) \hookrightarrow \mathcal{E}(\mathbb{R}^n)$ are continuous and dense with respect to topology of these spaces and then, for example, the application $v \in \mathcal{E}'(\mathbb{R}^n) \longrightarrow v=u_{\mathcal{D}(\mathbb{R}^n)} \in \mathcal{D}'(\mathbb{R}^n)$ is linear and one-to-one. Therefore each distribution determines a continuous linear functional $v : \mathcal{E}(\mathbb{R}^n) \longrightarrow \mathbb{C}$ (with respect to convergence in $\mathcal{E}(\mathbb{R}^n))$ which it is a compact support distribution, likewise $v : \mathcal{S}(\mathbb{R}^n) \longrightarrow \mathbb{C}$ is a temperate distribution.
Look, here are a lot of theorems to prove, I'm doing you a summary, and basically $\mathcal{S}'(\mathbb{R}^n)$, that is the space of tempered distributions, it's a good space to define the Fourier transform for duality, because essentially the Fourier transform of Schwartz functions satisfies the very useful properties.
Most of these I have studied in several books, there is not just one. "Real Analysis: Modern Techniques and Their Applications" by Folland and "Linear Functinoal Analysis" by "J. Cerda" cover a large part of these topics.
The idea is that the "dirac delta function" at $0,$ denoted by $\delta_0,$ is not an ordinary function. But somehow we can still integrate with it. For every continuous $g$ we have
$$\int_{\mathbb R} g\,\delta_0 = g(0).$$
However, no integrable function could have this property. That is, if $f$ is integrable on $[-a,a],$ then
$$\int_{-a}^a g(x) f(x)\,dx = g(0)$$
will fail for some continuous $g.$
On the other hand, and staying with elementary means, we can find a sequence $f_n$ of continuous functions on $\mathbb R$ such that
$$\lim_{n\to \infty}\int_{\mathbb R} g(x) f_n(x)\,dx = g(0)$$
for every continuous $g$ on $\mathbb R .$
Proof: Define $f_n$ on $[-1/n,1/n]$ to have the triangular graph that joins the points $(-1/n,0), (0,n), (1/n,0);$ define $f_n=0$ everywhere else. You can see that the $f_n$'s live in ever smaller intervals centered at $0,$ but nevertheless $\int_{\mathbb R} f_n = 1$ for every $n.$
Let $g$ be continuous. Then for each $n$ there exists $c_n\in [-1/n,1/n]$ such that $|g(c_n)-g(0)|$ is the maximum of $|g(x)-g(0)|$ on the interval $[-1/n,1/n].$ Thus
$$|\int_{\mathbb R} g(x) f_n(x)\,dx - g(0)| = |\int_{\mathbb R} [g(x)-g(0)] f_n(x)\,dx|$$ $$ \le |g(c_n)-g(0)| \int_{\mathbb R} f_n = |g(c_n)-g(0)|\cdot 1.$$
As $n\to \infty,$ $c_n\to 0,$ so the last expression $\to 0$ by the continuity of $g$ at $0.$
Best Answer
As you say, the "value" of the delta "function" is infinite at $0$ and the "area under it" is $1$. I am not aware of any time we take the value to be $1$. The unit impulse refers to the total impulse delivered, which is the area under the force-time curve.