Integrate by parts with
[tex]u = f(\phi) \implies du = f'(\phi) \, d\phi = (2\phi - 1) \, d\phi[/tex]
[tex]dv = d\phi \implies v = \phi[/tex]
to evaluate the integral in the definition of [tex]f[/tex].
[tex]\displaystyle \int_0^1 f(\phi) \, d\phi = \phi\,f(\phi) \bigg|_0^1 - \int_0^1 \phi(2\phi-1) \, d\phi = f(1) - \frac16[/tex]
Now if [tex]\phi=1[/tex], we have
[tex]f(1) = 1^2 - 1 - \left(f(1) - \dfrac16\right) \implies 2f(1) = \dfrac16 \implies f(1) = \dfrac1{12}[/tex]
and so
[tex]\displaystyle \frac1{12} = 1^2 - 1 - \int_0^1 f(\phi) \, d\phi \implies \int_0^1 f(\phi) \, d\phi = -\frac1{12}[/tex]
It follows that
[tex]\displaystyle \int_0^2 f(\phi) \, d\phi = \int_0^2 \left(\phi^2 - \phi + \frac1{12}\right) \, d\phi \\\\ ~~~~~~~~~~~~~~~ = \left(\frac{\phi^3}3 - \frac{\phi^2}2 + \frac\phi{12}\right) \bigg|_0^2 \\\\ ~~~~~~~~~~~~~~~ = \frac{2^3}3 - \frac{2^2}2 + \frac2{12} \\\\ ~~~~~~~~~~~~~~~ = \boxed{\frac56}[/tex]
