$ \underline{\text{Given:}}\ \ \ f(x) = 1 - f(2-x) \qquad \to (1) $
$ \underline{\text{To find:}}\ \ \int_{0}^{2} f(x) \,dx $
$ \underline{\large \text{Method 1:}} $
Notice that the continuous function $f(x) = 1/2$ from $\mathbb{R}$ to $\mathbb{R}$ satisfies eqn. $(1)$
So, let $f(x) = 1/2$.
Now,
$
\begin{align*}
\large \int_{0}^{2} f(x) \,dx &= \large \int_{0}^{2} \tfrac{1}{2} \cdot \,dx = \large \tfrac{x}{2}\Big|_0^2 = 1 \\
\end{align*}
$
$\bf \therefore Ans = B.$
$ \underline{\large \text{A more formal method:}} $
Let $u=a-x$, we have $du=-dx$, then
$
\begin{align*}
\large \int_{0}^{a}f(a-x)dx &= \large \int_{u=a}^{u=0}-f(u)du = \large -\int_a^0f(u)du = \large \int_0^af(u)du
\end{align*}
$
So, $ \large \int_{0}^{a} f(a-x) \,dx = \int_{0}^{a} f(u) \,dx = \int_{0}^{a} f(x) \,dx. \qquad \small \text{(} \because \text{variable of integration is a dummy variable)}$
Now,
$
\begin{align*}
\large \int_{0}^{2} f(x) \,dx &= \large \int_{0}^{2} f(2-x) \,dx = \large \int_{0}^{2} 1-f(x) \,dx = \large 2 - \int_{0}^{2} f(x) \,dx \\
\large 2 \cdot \int_{0}^{2} f(x) \,dx &= \large 2 \\
\large \int_{0}^{2} f(x) \,dx &= \large 1 \\
\blacksquare \\
\end{align*}
$