variation of functional A little confused about finding the variation of the functional 
J = $\int_{t0}^{tf}(e^{x_1(t)+x_2(t)})dt$
When I perturb and find the increment, I get:
$\Delta J = \int_{t0}^{tf} (e^{x_1(t) + \delta x_1(t) + x_2(t) + \delta x_2(t)} - e^{x_1(t) + x_2(t)}$)dt
To find the variation, I must eliminate any terms that are non-linear in $\delta x$, which pretty much eliminates the left term due to the exponential:
$\delta J = \int_{t0}^{tf} (-e^{x_1(t) + x_2(t)})dt$
I'm not sure where to go from here. I tried integration by parts, but I got stuck in an infinite loop. Am I missing something?
 A: At first order:
$${{\rm e}^{x_{{1}} \left( t \right) +x_{{2}} \left( t \right) +\delta\,
x_{{1}} \left( t \right) +\delta\,x_{{2}} \left( t \right) }}={{\rm e}
^{x_{{1}} \left( t \right) +x_{{2}} \left( t \right) }}+{{\rm e}^{x_{{
1}} \left( t \right) +x_{{2}} \left( t \right) }} \left( \delta\,x_{{1
}} \left( t \right) +\delta\,x_{{2}} \left( t \right)  \right) 
$$
Then
$${{\rm e}^{x_{{1}} \left( t \right) +x_{{2}} \left( t \right) +\delta\,
x_{{1}} \left( t \right) +\delta\,x_{{2}} \left( t \right) }}-{{\rm e}
^{x_{{1}} \left( t \right) +x_{{2}} \left( t \right) }}={{\rm e}^{x_{{
1}} \left( t \right) +x_{{2}} \left( t \right) }} \left( \delta\,x_{{1
}} \left( t \right) +\delta\,x_{{2}} \left( t \right)  \right) 
$$
and the variation is
$$\delta J =\int _{t_{{0}}}^{t_{{f}}}\!{{\rm e}^{x_{{1}}
 \left( t \right) +x_{{2}} \left( t \right) }} \left( \delta\,x_{{1}}
 \left( t \right) +\delta\,x_{{2}} \left( t \right)  \right) {dt}
$$
Then the corresponding Euler-Lagrange equation is
$${{\rm e}^{x_{{1}} \left( t \right) +x_{{2}} \left( t \right) }}=0$$
Do you agree?
