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I've just realized I was being a little bit slow. I had already found on the internet that $n^{-2}\sum_{k=1}^nφ(k)$ is roughly $3/π^2$ and stupidly didn't notice that I could "differentiate" this to get exactly what I want. That is, $\sum_1^N φ(k)$ is about $3N^2/π^2$, so the difference between the sum to $N+M$ and the sum to $N$ is around $6NM/π^2$, from which it follows that the average value near $N$ is around $6N/π^2$, which is entirely consistent with the well-known fact that the probability that two random integers are coprime is $6/π^2$.

I'm adding this paragraph after Greg's comment. To argue that the probability that two random integers are coprime, you observe that the probability that they do not have p as a common factor is (1-1/p^2). If you take the product of that over all p then you've got the reciprocal of the Euler product formula for ζ(2), or 1^{-2}+2^{-2}+... = π^2/6. It's not that hard to turn these formal arguments into a rigorous proof, since everything converges nicely.

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I've just realized I was being a little bit slow. I had already found on the internet that $n^{-2}\sum_{k=1}^nφ(k)$ is roughly $3/π^2$ and stupidly didn't notice that I could "differentiate" this to get exactly what I want. That is, $\sum_1^N φ(k)$ is about $3N^2/π^2$, so the difference between the sum to $N+M$ and the sum to $N$ is around $6NM/π^2$, from which it follows that the average value near $N$ is around $6N/π^2$, which is entirely consistent with the well-known fact that the probability that two random integers are coprime is $6/π^2$.
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