can you help me prove this equality?

I tried to use Riemann sums but I haven't succeeded to find something useful.

$$\lim_{n\to\infty} \sum_{k=1}^n f\left(\frac{k}{n}\right)\frac{1}{n} = \int_0^1f(x)dx $$

Thank you very much.


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    $\begingroup$ What did you get with Riemann sums? Also, summing from $n=1$ to $n$ doesn't make sense $\endgroup$ – J. W. Tanner Apr 2 at 17:02
  • $\begingroup$ This is an immediate consequence of the definition of Riemann integral. Here the partition is uniform with points $x_k=k/n$ and tags $t_k$ are choosen the same as $x_k$ so the Riemann sum becomes $\sum_{k=1}^{n}f(t_k)(x_k-x_{k-1})$. $\endgroup$ – Paramanand Singh Apr 3 at 17:24

Your approach using Riemann sums is probably correct. Try to think it this way, if you were given an integral of f(x) from 0 to 1, using Riemann sums you would break the interval in n equal delta intervals (where each delta interval is of length 1/n). And the value of the function f(x) at each subsequent interval would be f(1/n), f(2/n), ...., f(n/n). Summing the f(i/n).1/n for i=1 to n will give you the area under the curve f(x) from 0 to 1. Hence taking the limit n tending to infinity will give this summation an integral approximation.


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