I have begun reading Feynman & Hibbs Quantum Mechanics and Path Integrals. Knowing little about variational calculus or Lagrangians I found the following integration by parts opaque. I think if I saw it done once methodically it would clarify a lot. I have no problem with integration by parts in general.

On p. 27 he says that upon integration by parts the variation in $S$ becomes

$$ \delta S = \left[\delta x \frac{\partial L}{\partial \dot{x}} \right]_{t_a}^{t_b} - \int_{t_a}^{t_b}\delta x\left[\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{x}}\right)-\frac{\partial L}{\partial x}\right] dt. \tag{2-6} $$


$$ S = \int_{t_a}^{t_b}L(\dot{x},x,t) dt \tag{2-1}$$

in which $L$ is the Lagrangian

$$ L = \frac{m}{2}\dot{x}^2 - V(x,t)\tag{2-2}$$

and he says that to a first order

$$\delta S = S[\bar{x}+ \delta x] - S[x] = 0. \tag{2-4}$$

He shows $S[x+\delta x]$ explicitly in (2-5) and from this derives (2-6).

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    $\begingroup$ You should not try to learn quantum mechanics, either in the path integral or in the canoncial quantiztation approach without having learned the Lagrangian and Hamiltonian formalisms of classical mechanics. You will only grow more and more confused if you try it. $\endgroup$
    – ACuriousMind
    Jul 5, 2014 at 15:52
  • $\begingroup$ I do agree with ACuriousMind above and strongly encourage you to learn about Lagrangian and Hamiltonian formalism. However, (2-6) must be thougt as the variation of S if you integrate L along a trajectory x(t) + dx(t), expanding L in Taylor series with respect to such dx(t) is the way to obtain (2-6). Tell me if you need more detail and I'll post this more accurately $\endgroup$ Jul 5, 2014 at 16:15
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    $\begingroup$ Do you know what $\delta L$ is, given that $L= L[x,\dot x] $? Hint: Think of differentials in ordinary calculus. $\endgroup$
    – Danu
    Jul 5, 2014 at 16:22
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    $\begingroup$ I believe it's $\delta \dot{x}\frac{\partial L}{\partial \dot{x}}+ \delta x\frac{\partial L}{\partial x}~ ?$ $\endgroup$
    – daniel
    Jul 5, 2014 at 16:25
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    $\begingroup$ Right. Now, apply partial integration to the first term. This should yield the desired result $\endgroup$
    – Danu
    Jul 5, 2014 at 16:34