TITLE: Find $f(1729)$ if $n^2\int_{x}^{x+\frac 1 n} f(t)\;\text{d}t=nf(x)+0.5$
QUESTION [9 upvotes]: I was asked this question-

Let $f$ be a real continuous function satisfying $f(0)=0$ and for each natural number $n$
$$n^2\int_{x}^{x+\frac 1 n} f(t)\;\text{d}t=nf(x)+0.5$$
Then find the value of $f(1729)$

I couldn't make much of a progress. But I tried using
$$F(x)=\int_0^x f(x)\;\text{d}x$$
so that
$$\int_{x}^{x+\frac 1 n} f(t)\;\text{d}t=F\left(x+\frac 1 n\right)-F(x)$$
and then maybe letting $n$ go to infinity such that we can think about the behaviour of
$$\lim_{n\to \infty} n^2\cdot \left(F\left(x+\frac 1 n\right)-F(x)\right)$$
But, soon I realised that we don't even know whether this limit exists, since we don't have any idea of whether the limit in the RHS exists, i.e., we don't know whether
$$\lim_{n\to \infty} nf(x)$$
exists. However, if this would somehow exist, we could have argued that the limit of the LHS also exists, and thus,
$$\left(F\left(x+\frac 1 n\right)-F(x)\right)=\frac {h(x)}{n^2}$$
But still, there's no progress.
Another approach that I tried was to sum up the equation from $1$ to $k$ to get
$$\sum_{n=1}^k n^2\int_{x}^{x+\frac 1 n} f(t)\;\text{d}t = \frac {k(k+1)}2 f(x) + \frac k 2$$
But how to calculate the sum on the LHS? I tried writing it as
$$\sum_{n=1}^k n^2\cdot \left(F\left(x+\frac 1 n\right)-F(x)\right)=\sum_{n=1}^k n^2\cdot F\left(x+\frac 1 n\right) - \sum_{n=1}^k n^2\cdot F(x)$$
where the second term maybe easy to calculate, but what about the first term?
Any help would be appreciated. Also, I'm not really sure about the tags- feel free to edit them.

REPLY [1 votes]: Plugging in $0$ on both sides, we obtain
$$n^2\int_0^{\frac{1}{n}} f(t) dt = nf(0)+\frac{1}{2} \implies \int_0^{\frac{1}{n}} f(t)dt = \frac{1}{2n^2}$$
which means $f(t)=t$ and $f(1729)=1729$
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