I'm just starting with this book to get a better grasp of FEM from a mathematical perspective, so my multivariable calculus is a bit rusty. While I intuitively understand what is mean by the work of 2 forces being zero (Eqn 1.9), I'm having problems with the integration by parts.
The relevant equations state that for functions $u$ and $\delta u$, if the function $u(x)$ satisfies the differential equation (Eqn. 1.11):
\begin{equation} -EAu''(x) - p(x) = 0, 0 < x < l, \end{equation}
then holds \begin{equation} \int_0^l(-EAu'' - p)\delta u dx = 0, \end{equation}
or after integration by parts assuming that $\delta u(0) = \delta u(l) = 0$,
\begin{equation} \int_0^l\frac{N\delta N}{EA}dx = \int_0^lp\delta udx, \end{equation}
where $N = EAu'$ is the normal force
How does the integration by parts work to get the last equation? If the integration by parts formula is:
\begin{equation} \int_0^lf(x)g'(x)dx = \left[f(x)g(x)\right]_0^l - \int_0^lg(x)f'(x)dx \end{equation}
I get that the $f(x)g(x)$ component will be zero when evaluated (based on the assumptions for $\delta u(0)$ and $\delta u(l)$, but I'm missing which components are evaluated as f and g for the integrals.
Integrating by parts yields $$ \int_0^\ell \left( -EAu''(x) \right)\delta u\;\mathrm dx\\ =-EA\delta u(\ell)u'(\ell)+EA \delta u(0)u'(0)+EA\int_0^\ell u'(x)\delta u'(x)\;\mathrm dx $$ with your assumption that $\delta u(0)=\delta u(\ell)=0$ implying that the boundary terms vanish, and the integral is just $$ EA\int_0^\ell u'(x)\delta u'(x)\;\mathrm dx $$ renaming $N$ for purposes unclear to me in the above gives you back your expression (with $\delta$ being treated as constant it seems).
Next if $$ -EA\int_0^\ell u''(x)\delta u(x)\;\mathrm dx-\int_0^\ell p(x)\delta u(x)\mathrm dx=0 $$ as was assumed, then our work gives you that $$ EA\int_0^\ell u'(x)\delta u'(x)\;\mathrm dx=\int_0^\ell p(x)\delta u(x)\mathrm dx $$