§5.6 Numerical integration (continued)

Overall observations regarding

Summary

Three functions $f$, $g$, $h$ all satisfy value $2$ at $0$ and value $1$ at $1$. Their one panel sums $L_1$ and $R_1$ are

For the test functions, $S_2 = \frac{5}{3}$ for $f$ and $S_2 = \frac{7}{4}$ for $g$, matching the exact integrals $\frac{5}{3}$ and $\frac{7}{4}$, while the third function's exact value $\frac{9}{5}$ exceeds its $S_2 = \frac{43}{24}$. The explanation is

For an unknown monotone, fixed concavity function on $[3,6]$, we have $L_4 = 7.23 \gt R_4 = 6.75$, with midpoint $M_4 = 7.05$ above trapezoid $T_4 = 6.99$. The function is

The integral $\int_0^1 x\tan(x)\,dx$ cannot be evaluated exactly with the chapter's tools because

Flow rates $2000, 2100, 2400, 3000, 3900, 5100, 6500$ CFS are recorded every ten seconds. With $M_3 = 204000$, $T_3 = 211000$, and $S_6 = \frac{619000}{3}$, the best volume estimate is

With $M_1 = \frac{7}{4}$ and $T_1 = \frac{3}{2}$ for the test function $f$, compute Simpson's $S_2 = \frac{2M_1 + T_1}{3}$ as a fraction.

For the unknown function with $L_4 = 7.23$, $R_4 = 6.75$, $M_4 = 7.05$, and $T_4 = 6.99$, Simpson's estimate $S_8 = \frac{2M_4 + T_4}{3}$ is the best available. Compute $S_8$, rounded to two decimals.

The test function $g$ satisfies $g\left(\frac{1}{2}\right) = \frac{15}{8}$. Compute the one panel midpoint estimate $M_1$ for $\int_0^1 g(x)\,dx$ as a decimal.

Using midpoint samples at $t = 10, 30, 50$ seconds with panel width $20$, compute $M_3$ for the dam flow integral in cubic feet.

Simpson's rule integrates linear functions and quadratic functions exactly.