Question 9
Values of a function $f$ are shown in the table.
| $t$ | 0 | 2.5 | 5 | 7.5 | 10 | 12.5 | 15 | 17.5 | 20 |
|---|---|---|---|---|---|---|---|---|---|
| $f(t)$ | 12.0 | 13.1 | 14.6 | 16.4 | 18.9 | 22.0 | 25.9 | 30.4 | 35.6 |
Use Simpson’s rule with all the values in the table to determine an approximation for $\displaystyle\int_{0}^{20}f(t)\,dt$.
Not quite. You swapped the weights, putting $2$ on the odd ordinates and $4$ on the even ones.
That’s it. Step through the hints below to see why.
Not quite. You used the trapezoidal rule, which gives $412.75$.
Not quite. You multiplied by $w$ instead of $\dfrac{w}{3}$.
Apply Simpson’s rule with $w=2.5$ (weights $1,4,2,\dots,2,4,1$).
$$\begin{aligned}&\frac{2.5}{3}\Big[12.0+4(13.1+16.4+22.0+30.4)\\&\quad+2(14.6+18.9+25.9)+35.6\Big]\\&=411.667\end{aligned}$$
$$\begin{aligned}&\frac{2.5}{3}\Big[12.0+4(13.1+16.4+22.0+30.4)\\&\quad+2(14.6+18.9+25.9)+35.6\Big]\\&=411.667\end{aligned}$$
The approximation is $411.7$, option (B).
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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