Goldstein's derivation of the 'principle of least action' I want make an punctual question ands it's about The derivation of the expression
$$ \Delta\int_{t_1}^{t_2} Ldt=L(t_2)\Delta t_2-L(t_1)\Delta t_1 + \int_{t_1}^{t_2} \delta L dt. \tag{8.74}$$
You can find it from Goldstein's Classical Mechanics section 8-6.
Somehow the previous expression comes from
$$ \Delta\int_{t_1}^{t_2} Ldt= \int_{t_1+\Delta t_1}^{t_2+\Delta t_2} L(\alpha) dt -  \int_{t_1}^{t_2} L(0) dt \tag{8.73}$$
but I'm not completely sure how?
$$ L(\alpha) $$ means a varied path and
$$ L(0) $$ means the actual path.
 A: There are already several good answers showing the algebra. Here we will make some comments to the question (v4) concerning terminology and notation, which may clarify a thing or two. (In the following we refer to the $q$ position space as the vertical space and the $t$ time axis as the horizontal space.)


*

*Usually, the principle of least action refers to the principle of stationary action/Hamilton's principle 
$$ \delta \int_{t_i}^{t_f}\! dt~ L~=~0. \tag{2.2} $$
In this variational principle, the infinitesimal variation $\delta q^i$ is purely vertical $\delta t=0$, and the initial and final times $t_i$ and $t_f$ are kept fixed.

*Note that what Goldstein in Ref. 1 confusingly calls the principle of least action is usually called the principle of abbreviated action/Maupertuis' principle 
$$ \Delta \int_{t_i}^{t_f}\! dt~p_j \dot{q}^j~~=~0, \tag{8.80}$$
cf. e.g. Ref. 2.
In this variational principle, the infinitesimal variation
$$ \Delta q^j~=~ \delta q^j + \dot{q}^j \Delta t \tag{8.76} $$
is now composed of both vertical and horizontal variations. The total energy $E$ of all paths is kept fixed and the same; while the initial and final times $t_i$ and $t_f$ are free.

*For autonomous systems, the two above variational principles can be viewed as Legendre transforms of each other with respect to the dual Legendre variables $$E\quad\longleftrightarrow\quad \Delta t~:=~t_f-t_i.$$ In both variational principles, we usually keep the initial and final positions $q^j_i$ and $q^j_f$ fixed.
References:


*

*H. Goldstein, Classical Mechanics; Section 8.6.

*L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1, 1976; $\S 44$.
A: You can break $\int_{t_1+\Delta t_1}^{t_2+\Delta t_2} L(\alpha) dt$ into $\left( \int_{t_1 + \Delta t_1}^{t_1} +\int_{t_1}^{t_2} + \int_{t_2}^{t_2+\Delta t_2}\right)L(\alpha) dt$. Then of these three pieces, the $\int_{t_1}^{t_2}$ piece combines with the $-\int_{t_1}^{t_2} L(0) dt$ piece to give you the $\int_{t_1}^{t_2} \delta L dt$.
This means that $\left( \int_{t_1 + \Delta t_1}^{t_1} + \int_{t_2}^{t_2+\Delta t_2}\right)L(\alpha) dt$ must give you $L(t_2)\Delta t_2-L(t_1)\Delta t_1$. Let's see how that happens. In general, we have $\int_x^{x+h} f(x) dx = F(x+h)-F(x) \approx  F'(x)h = f(x)h$, where $F$ is an antiderivative of $f$. Applying this to $\int_{t_2}^{t_2+\Delta t_2} L(\alpha) dt$, we obtain $L(t_2)\Delta t_2$. Notice here that we did not specify whether $L$ in this expression is to be evaluated on the actual or varied path. This is because those paths are very close to each other, so it does not matter at the level of approximation we are doing. Anyway, evaluating the $t_1$ piece, we find  $\int_{t_1 + \Delta t_1}^{t_1}L(\alpha) dt=-\int_{t_1 }^{t_1+ \Delta t_1}L(\alpha) dt = -L(t_1)\Delta t_1$. 
Adding the two resulting pieces from the previous paragraph to the resulting piece from the first paragraph, we obtain $L(t_2)\Delta t_2-L(t_1)\Delta t_1 + \int_{t_1}^{t_2} \delta L dt$, which is what we wanted.
