So I was solving this problem and I wanted to know if I did everything right.
The problem is:
I have to calculate the first variation of the functional
$F:V\rightarrow \mathbb{R}\quad$ defined by
$\displaystyle\;y\rightarrow\!\int_{a}^{b} \!y(x)\,\mathrm dx\;$ with $\;V=C^{0}([a,b])\,.$
So the formula I used is this: $\;\delta F[y]=\dfrac{\partial F}{\partial y}\cdot v$
First, we need to calculate the functional derivative $\partial F/\partial y$
$\begin{align} \dfrac{\partial F}{\partial y}&=\lim\limits_{\varepsilon\rightarrow 0} \dfrac{(F[y+\varepsilon v]-F[y])}{\varepsilon}\\ &=\lim\limits_{\varepsilon\rightarrow 0}\dfrac{\int_{a}^{b}[y(x)+\varepsilon v(x)]\,\mathrm dx-\int_{a}^{b}y(x)\,\mathrm dx}{\varepsilon}\\ &=\lim\limits_{\varepsilon\rightarrow 0}\dfrac{\int_{a}^{b}\varepsilon v(x)\,\mathrm dx}{\varepsilon}=\int_{a}^{b}v(x)\,\mathrm dx\,. \end{align}$
So at the end, with $\;\delta F[y]=\partial F/\partial y\cdot v\,,\,$ I have :
$\displaystyle\delta F[y] =\int_{a}^{b}v(x)\,\mathrm dx\cdot v$
Am I right or am I missing something?