How do I interpret the definition of a distribution in this book by Joshi & Friedlander?

Question

I was wondering about the following definition in the book Introduction to the Theory of Distributions, by Joshi and Friedlander:

In the Appendix the bracket operation is defined as

To me this definition of a distribution doesn't make sense if we view the bracket operation as an inner product, e.g., an integral, since the domain of $u$ is $C_c^\infty(X)$ whereas the domain of $\phi$ is $X$. So then how do we make sense of $\int u(x)\phi(x) dx$?

I would understand if $u$ were, say, a continuous function from $X$ to $\mathbb{R}$. Then $u$ would determine a linear form through $\langle u, \phi \rangle$. Is that what the definition means? We take some object $u$, then $u$ determines a linear form through the pairing $\langle u, \phi \rangle$? And then we refer to $u$ as a linear form because somehow the linear form determines $u$ uniquely? Perhaps this bit from the book is relevant:

Answer 1

The notation $\langle u,\varphi\rangle$ is just a fancy way of writing $u(\varphi)$ for a linear from $u\colon C^\infty_c(X)\to\mathbb{C}$. Examples of linear forms which are not defined by functions are $\delta_0$ or $\partial^\alpha\delta_0$ which are defined as $$ \langle \delta_0, \varphi\rangle = \varphi(0) \qquad \text{and}\qquad \langle \partial^\alpha\delta_0, \varphi\rangle = (-1)^{|\alpha|}(\partial^\alpha\varphi)(0). $$ For these two it is easy to check that they satisfy the condition stated above. A slightly more complicated example might be $$ \langle \mathrm{vp}\frac{1}{x}, \varphi\rangle = \lim_{\varepsilon\to 0} \int_{\mathbb{R}\setminus(-\varepsilon,\varepsilon)} \frac{\varphi(x)}{x}\,\mathrm{d}x. $$ Note that none of these distributions can be represented by a continuous function.