微积分I（标准路径） · free preview

§5.6 Numerical integration (continued)

当被积函数不存在初等代数原函数，因而无法使用微积分第一基本定理时，我们如何准确计算诸如 $\int_0^1 e^{-x^2} \, dx$ 这样的定积分？是否有一些方法可以在黎曼和中不使用极大的 $n$ 值的情况下得到精确的估计？

1. Overall observations regarding

Overall observations regarding

2. Summary

Summary

3. Endpoint sums depend only on endpoints

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

4. Simpson lands exactly on quadratics

For one family of test functions, $S_2 = \frac{5}{3}$ for $f$ and $S_2 = \frac{7}{4}$ for $g$ , while the exact integrals are $\frac{5}{3}$ and $\frac{7}{4}$ . The explanation is

5. Classifying a mystery function

For an unknown monotone, fixed concavity function on $[3,6]$ , we have $L_4 = 7.23 \gt R_4 = 6.75$ , with midpoint $7.20$ above trapezoid $6.86$ . The function is

6. Simpson value for the first test function

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.

7. Best estimate for the mystery integral

For the unknown function with $L_4 = 7.23$ , $R_4 = 6.75$ , $M_4 = 7.20$ , and $T_4 = 6.86$ , Simpson's estimate $S_8 = \frac{2M_4 + T_4}{3}$ is the best available. Compute $S_8$ , rounded to two decimals.

8. Exactness class of Simpson's rule

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

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