## Calculus: Early Transcendentals 8th Edition

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