classical mechanics - 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.

Answer

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.