13 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = \sqrt [ 3 ] { 27 - 8 x ^ { 3 } }\). Jenny uses her scientific calculator to create a table of values for \(\mathrm { f } ( x )\) and \(\mathrm { f } ^ { \prime } ( x )\).
| \(x\) | \(f ( x )\) | \(f ^ { \prime } ( x )\) |
| 0 | 3 | 0 |
| 0.25 | 2.9954 | - 0.056 |
| 0.5 | 2.9625 | - 0.228 |
| 0.75 | 2.8694 | - 0.547 |
| 1 | 2.6684 | - 1.124 |
| 1.25 | 2.2490 | - 1.977 |
| 1.5 | 0 | ERROR |
- Use calculus to find an expression for \(\mathrm { f } ^ { \prime } ( x )\) and hence explain why the calculator gives an error for \(\mathrm { f } ^ { \prime } ( 1.5 )\).
- Find the first three terms of the binomial expansion of \(\mathrm { f } ( x )\).
- Jenny integrates the first three terms of the binomial expansion of \(\mathrm { f } ( x )\) to estimate the value of \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { 27 - 8 x ^ { 3 } } \mathrm {~d} x\). Explain why Jenny's method is valid in this case. (You do not need to evaluate Jenny's approximation.)
- Use the trapezium rule with 4 strips to obtain an estimate for \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { 27 - 8 x ^ { 3 } } \mathrm {~d} x\).
The calculator gives 2.92117438 for \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { 27 - 8 x ^ { 3 } } \mathrm {~d} x\). The graph of \(y = \mathrm { f } ( x )\) is shown in Fig. 13.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{904025c9-6d68-4344-bd41-8c0fccfcf92f-08_490_906_1505_568}
\captionsetup{labelformat=empty}
\caption{Fig. 13}
\end{figure} - Explain why the trapezium rule gives an underestimate.