Approximate \(\sum_{k = 1}^n k^3\) with an integral.
As \(k^3\) is a monotonically increasing function, we can write:
\[\begin {align} \int_0^n x^3 & \le \sum_{k = 1}^n k^3 \le \int_1^{n + 1} x^3 \\ \left[\frac {x^4} 4\right]_0^n & \le \sum_{k = 1}^n k^3 \le \left[\frac {x^4} 4\right]_1^{n + 1} \\ \left[\frac {n^4} 4 - 0\right] & \le \sum_{k = 1}^n k^3 \le \left[\frac {(n + 1)^4} 4 - \frac 1 4\right] \\ \frac {n^4} 4 & \le \sum_{k = 1}^n k^3 \le \frac {(n + 1)^4 - 1} 4 \end {align}\]