Is the integral as an operator linear? Consider the following operator:
$L[f(x)]=\int_0^x f(s) ds$.
Is this operator linear?
I think in order to answer this question, I need to consider all possible functions, $f(x)$. The book says an operator is linear if you get a constant * the operator back:
$L[f(x)]=cf(x)$.
I cannot integrate $f(s)ds$ without knowing what $f(s)$ is. So, suppose it is just $s$. Then the integral is $0.5s^2$ (evaluated from $0$ to $x$), or $0.5x^2$. This is not equal to $x$, so therefore the integral operator is not linear?
 A: You seem to be confusing what it means to be linear. Let $f(s)$ and $g(s)$ be functions. Then $L$ is linear if for all constants $A$, we have
$$ L(A\cdot f(s) + g(s)) = A\cdot L(f(s)) + L(g(s)).$$
This is really two different things. Let first ignore $A$ (i.e., let $A = 1$). Then, the definition of linear above becomes
$$ L(f(s) + g(s)) = L(f(s)) + L(g(s)).$$
I.e., we can either apply $L$ to the sum $f(s) + g(s)$, or we can apply $L$ to both $f(s)$ and $g(s)$ and then add the answers together.
Now, lets ignore $g(s)$ (which we can do without losing information since the above "breaking up" the sum of functions property shows we only need to figure out what to do with a single function!). Here, the definition of a linear operator becomes
$$ L(A\cdot f(s)) = A\cdot L(f(s)).$$
Again, we can either apply $L$ to the entire function (multiplying by $A$ before applying $L$), or we can apply it to $f(s)$, and then multiply by $A$ after having applied $L$. The notion of linear can be thought (loosely) of being able to "break things up" before/after and get the same result.
So in your question, we need to consider $L$ as integration: the question becomes, is
$$ \int_0^x A\cdot f(s) + g(s)ds = A\cdot\int_0^x f(s)ds + \int_0^x g(s)ds $$
for all functions $f(s),g(s)$ and constants $A$? Now you won't need to know anything about $f(s)$ or $g(s)$ (except that they are actually integrable, as well as their sum, of course) to see whether $L$ is or isn't a linear operator!
