Answer
$\lim\limits_{n \to \infty}\frac{1}{n}[(\frac{1}{n})^9+(\frac{2}{n})^9+(\frac{3}{n})^9+...+(\frac{n}{n})^9] = \frac{1}{10}$
Work Step by Step
An integral can be expressed as the limit of a Riemann sum:
$\int_{a}^{b}f(x)~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i^*)\Delta x$
Consider the interval $[0,1]$:
$\Delta x = \frac{b-a}{n} = \frac{1-0}{n} = \frac{1}{n}$
$x_i^* = a+i~\Delta x = 0+\frac{i}{n} = \frac{i}{n}$
Note that $x_i^*$ is the right endpoint of each subinterval.
$\lim\limits_{n \to \infty}\frac{1}{n}[(\frac{1}{n})^9+(\frac{2}{n})^9+(\frac{3}{n})^9+...+(\frac{n}{n})^9]$
$= \lim\limits_{n \to \infty}\sum_{i=1}^{n}(\frac{i}{n})^9\cdot \frac{1}{n}$
$= \int_{0}^{1}x^9~dx$
$= \frac{x^{10}}{10}~\vert_{0}^{1}$
$= \frac{(1)^{10}}{10}- \frac{(0)^{10}}{10}$
$= \frac{1}{10}$
