Answer
$\lim\limits_{n \to \infty}\sum_{i=1}^{n}(\frac{i^4}{n^5}+\frac{i}{n^2}) = \frac{7}{10}$
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.
$\lim\limits_{n \to \infty}\sum_{i=1}^{n}(\frac{i^4}{n^5}+\frac{i}{n^2})$
$= \lim\limits_{n \to \infty}\sum_{i=1}^{n}(\frac{i^4}{n^4}+\frac{i}{n})\cdot \frac{1}{n}$
$= \lim\limits_{n \to \infty}\sum_{i=1}^{n}[(\frac{i}{n})^4+(\frac{i}{n})]\cdot \frac{1}{n}$
$= \int_{0}^{1}(x^4+x)~dx$
$= (\frac{x^5}{5}+\frac{x^2}{2})~\vert_{0}^{1}$
$= (\frac{1^5}{5}+ \frac{1^2}{2}) + (\frac{0^5}{5}+ \frac{0^2}{2}) $
$= \frac{1}{5}+ \frac{1}{2}$
$= \frac{7}{10}$
