Let f be an increasing function on (0,1) and let the improper integral $\int_{0}^{1}f(x)dx$ exist. Then, show that $\int_{0}^{1-\frac{1}{n}}f(x)dx \leq \frac{f(\frac{1}{n})+f(\frac{2}{n})... +f(\frac{n-1}{n})}{n} \leq \int_{\frac{1}{n}}^{1}f(x)dx$
Here's a problem I've been asked to solve.
What I've come up with so far:
- Looks like we've made a partition of the interval (0,1), creating n equally large sub-intervals.
- The leftmost integral represents area under the graph of f(x) from 0 to $\frac{1}{n}$ (similarly for the rightmost integral)
- Since f is increasing, the sum in the middle is basically the average of supremums of the sub-intervals of the partition we're working with
How do I relate these, so as to establish the required inequality? Please help, thanks!
$\int_{0}^{1-\frac{1}{n}}f(x)dx=\sum_{k=1}^{n-1}\int_{\frac{k-1}{n}}^{\frac{k}{n}}f(x)dx \leq \sum_{k=1}^{n-1}\int_{\frac{k-1}{n}}^{\frac{k}{n}}f(\frac{k}{n})dx =\frac{f(\frac{1}{n})+...+f(\frac{n-1}{n})}{n} \\ =\sum_{k=1}^{n-1}\int_{\frac{k}{n}}^{\frac{k+1}{n}}f(\frac{k}{n})dx\le \sum_{k=1}^{n-1}\int_{\frac{k}{n}}^{\frac{k+1}{n}}f(x)dx=\int_{\frac{1}{n}}^{1}f(x)dx$