Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Discrete mathematics, continuous mathematics and probability / 6 / c

Two strips give $h=\pi /4$, with cosine values $1,1/\sqrt{2},0$. The trapezoidal rule therefore givesThe exact integral is $1$, so the estimate is low, consistent with the concavity of cosine on this interval. The table can be used as $T_2=\frac{1}{2}(0.7854)+0.5554$ before rounding.

With the same three cosine values, Simpson's rule givesUsing the table's $n=12$ row gives $0.2618+4(0.1851)=1.0022$, with the same final rounding. That table row refers to the prefactor $\pi /12$; it does not change the specified number of strips.