Thinking of integration as an area is certainly a useful technique when it works, but the problem is that there are some integrable functions where that doesn't make sense.

One example would be

f(x) = {1, if x is rational.
          {0, otherwise.

The integral of that function on any interval is 0.
My lecturer only mentioned it in passing though, saying the proof was far beyond the scope of the course.

That said, it's kind of understandable since the rationals are countable and the irrationals are not.

My lecturer only mentioned it in passing though, saying the proof was far beyond the scope of the course.

That integral does not exist as a Riemann integral.  We can take the limit as Δx goes to zero by choosing all rationals for Δx, in which case each term in the sequence is 1, or we can take the limit by choosing Δx to be irrationals, in which case each term in the sequence is 0.  This prove the limit does not exist.

Now that is using the calculus definition of the integral, rather than the analysis definition.  The analysis definition will also not exist.

The integral is however zero when you consider it as a Lebesgue integral.  Lebesgue integration works like this: come up with a lower approximation for the integral using functions you can compute the integral of, and then call the highest one of those the value of the integral.  We make these approximations using simple functions which are functions which take on finitely many values.  The integral of a simple function f is

Where a_i is a value the simple function takes, E_i = f^{-1}(a_i), and m(E_i) is the measure of E_i.  This uses a bit of a fancy definition, but in this case you're helped a bit by the theorem that the measure of any countable set is 0, and m(A U B) = m(A) + m(B) when A and B are disjoint.  This means the measure of the rationals in [0,1] is 0, and so your function has integral 0.