Numerical Integration Questions and Answers

Answer: D

The exact rule is the composite trapezoidal rule by dividing [-1, 2] into 3 equal subintervals.

\(f(x) = |x| + |x - 1| + |x - 2|\) is piecewise linear on [-1, 2], with corners only at \(x = 0\) and \(x = 1\). Dividing [-1, 2] into three equal parts puts the nodes at -1, 0, 1, 2, so on each subinterval \(f\) is linear and the trapezoidal rule is exact.

Check: \(f(-1) = 6, f(0) = 3, f(1) = 2, f(2) = 3\), so the composite rule gives \(\frac{1}{2}(6 + 3) + \frac{1}{2}(3 + 2) + \frac{1}{2}(2 + 3) = 9.5\), the exact integral. The single trapezoid and single Simpson's rule ignore the corners, and the four-subinterval Simpson's rule places nodes at -0.25 and 1.25 with panels straddling the corners, where a quadratic cannot match a kink.

Answer: A

The area is 11.5 sq. unit.

The table gives 9 equally spaced ordinates with \(h = 1\), that is 8 intervals, an even number, so Simpson's one-third rule applies:

\[ \begin{aligned} A &= \tfrac{h}{3}[(y_{1} + y_{9}) + 4(y_{2} + y_{4} + y_{6} + y_{8}) \\ &\quad + 2(y_{3} + y_{5} + y_{7})] \\ &= \tfrac{1}{3}[2.5 + 4(5.8) + 2(4.4)] \\ &= \tfrac{1}{3}(34.5) = 11.5 \end{aligned} \]

The trapezoidal rule gives \(1.25 + 10.2 = 11.45\), which also rounds to 11.5, so either standard rule leads to option A. 12.0, 10.5 and 10.2 are not produced by any rule; 10.2 is just the sum of the interior ordinates.

Answer: B

The trapezoidal rule gives \(\frac{h}{2}[(y_{0} + y_{n}) + 2(y_{1} + y_{2} + \cdots + y_{n-1})]\).

Each strip \([x_{i}, x_{i+1}]\) is replaced by a trapezium of area \(\frac{h}{2}(y_{i} + y_{i+1})\). Adding the \(n\) strips, the end ordinates \(y_{0}\) and \(y_{n}\) appear once, and every interior ordinate \(y_{1}, \ldots, y_{n-1}\) appears twice, once in each neighbouring strip.

The other options double the wrong ordinates: the first two or the end values instead of the interior ones.