I'm trying to read Kress' Linear integral equations, and I'm stuck at the first example. There must be something obvious I'm missing, and to that end, should I read something before this text?
$f(x)=\phi(x)-\int_a^b K(x,y)\phi(y)dy,\quad x\in[a,b]$
For the kernel $K(x,y)=c\neq0\quad\forall x,y\in[a,b]$
The solution is $\phi=f+\frac{c}{1-c(b-a)}\int_a^bf(y)dy$
On
You can always substitute $\phi$ in the equation to verify that it is a solution. But I believe you want to know how to obtain the solution. Here is one way:
Rewrite the equation as $$ \phi(x) = f(x) + c\int_a^b \phi(y) dy. $$ Now substitute this expression for $\phi$ inside the integral, that is $$ \phi(x) = f(x) + c \int_a^b \left( f(y) + c\int_a^b \phi(z) dz \right) dy, $$ which is the same as $$ \phi(x) = f(x) + c\int_a^b f(y) dy + c^2(b-a) \int_a^b \phi(y) dy. $$ Now we can replace $c\int_a^b \phi(y) dy$ by $\phi(x)-f(x)$, and hence we obtain $$ \phi(x) = f(x) + c\int_a^b f(y) dy + c(b-a)(\phi(x)-f(x)), $$ and after some simple manipulations this gives $$ \phi(x) = f(x) + \frac{c}{1-c(b-a)}\int_a^b f(y) dy. $$
For the other question, that depends on your background, but I'm not an specialist in integral equations, and I haven't read the book either.
If $K(x,y)=c$, then $$ f(x)=\phi(x)-cI, \tag{1} $$ where $I:=\int_a^b\phi(y)\,dy$. Integrating both sides of $(1)$ yields $$ \int_a^bf(x)\,dx=\int_a^b[\phi(x)-cI]\,dx=I-cI(b-a). \tag{2} $$ Solving $(2)$ for $I$ and plugging the result in $(1)$, one finally obtains $$ \phi(x)=f(x)+\frac{c}{1-c(b-a)}\int_a^bf(y)\,dy. \tag{3} $$