# Question #786af

Jan 21, 2018

$\frac{1}{4} \left(4 {a}^{3} + 6 {a}^{3} + 4 {a}^{2} + a + 1\right)$

#### Explanation:

I'm interpreting the question as:
${\lim}_{h \to 0} \left(h \cdot \left({\left(a + 0 h\right)}^{3} + {\left(a + 1 h\right)}^{3} + \cdots + {\left(a + \left(n - 1\right) h\right)}^{3}\right)\right)$

${\lim}_{h \to 0} {\sum}_{k = 0}^{n - 1} \left(h \cdot f \left(a + k \cdot h\right)\right)$ where $f \left(x\right) = {x}^{3}$.

This is a left Riemann sum for the function $f \left(x\right) = {x}^{3}$ on the interval from $a$ to $a + 1$. In this case $h = \frac{a + 1 - a}{n} = \frac{1}{n}$.

We can evaluate the limit by evaluating the equivalent definite integral:

${\int}_{a}^{a + 1} {x}^{3} \mathrm{dx} = {\left[\frac{1}{4} {x}^{4}\right]}_{a}^{a + 1} = \frac{1}{4} \left({\left(a + 1\right)}^{4} - {a}^{4}\right)$

$= \frac{1}{4} \left({a}^{4} + 4 {a}^{3} + 6 {a}^{3} + 4 {a}^{2} + a + 1 - {a}^{4}\right)$

$= \frac{1}{4} \left(4 {a}^{3} + 6 {a}^{3} + 4 {a}^{2} + a + 1\right)$