The function f(u) can be interpreted as calculating area under the curve u(x) over the specified interval. We can visualize that sum of areas under two curves would be the same as the area under the curve that results by summing the two curves. Similar is the case with scaling a curve.

keenan

Right: all you need to do is check that it satisfies the two basic properties of a linear function: additivity and homogeneity, i.e., $f(u+v) = f(u) + f(v)$ and $f(au) = af(u)$ for all vectors $u,v$ and scalars $a$. In this case we specifically have

Of course, both of these statements depend on believing that these two properties hold for integration! Intuitively, why should those be true (if you remember the basic starting point for simple integrals)?

yes, it is.

Oh yeah? Why? :-)

The function f(u) can be interpreted as calculating area under the curve u(x) over the specified interval. We can visualize that sum of areas under two curves would be the same as the area under the curve that results by summing the two curves. Similar is the case with scaling a curve.

Right: all you need to do is check that it satisfies the two basic properties of a linear function: additivity and homogeneity, i.e., $f(u+v) = f(u) + f(v)$ and $f(au) = af(u)$ for all vectors $u,v$ and scalars $a$. In this case we specifically have

$$ f(u+v) = \int_0^1 u(x) + v(x) dx = \int_0^1 u(x) dx + \int_0^1 v(x) dx = f(u) + f(v)$$

and

$$ f(au) = \int_0^1 au(x) dx = a\int_0^1 u(x) dx = af(x). $$

Of course, both of these statements depend on believing that these two properties hold for integration! Intuitively, why should those be true (if you remember the basic starting point for simple integrals)?