Numerical Analysis — Homework (Nov 19)

作业：习题2.25、2.28(2).

Problem 1 (习题2.25).

设

f(x) 在 [0,1] 上连续，f'(x) 在 [0,1] 上可积，证明：用复化梯形公式计算\int_{0}^{1} f(x) dx 的误差形式为

\int{0}^{1} f(x) dx - Tn(f) = \int_{0}^{1} k(t) f'(t) dt,

其中

k(t) = \frac{t{i-1} + ti}{2} - t, t \in [t{i-1}, ti], i = 1, 2, \dots, n,

T_n(f)

是复化梯形和， $t_i$ $(i = 0, 1, \dots, n)$ 为积分区间 $[0,1]$ 的等分节点，记 $h = t_i - t_{i-1} = \frac{1}{n}$ 。

解.

在区间 $[t_{i-1}, t_i]$ 上，若 $\in [t_{i-1}, t_i]$ ，则有

f(x) = f(t_{i-1}) + \int_{t_{i-1}}^{x} f'(t) dt.

记子区间上的误差为：

E_i(f) = \int_{t_{i-1}}^{t_i} f(x) dx - \frac{h}{2} \left[ f(t_{i-1}) + f(t_i) \right], i = 1, 2, \dots, n.

我们将 $f(x)$ 用积分表示代入：

\begin{aligned} E_i(f) &= \int_{t_{i-1}}^{t_i} \left[ f(t_{i-1}) + \int_{t_{i-1}}^{x} f'(t) dt \right] dx - \frac{h}{2} \left[ f(t_{i-1}) + f(t_i) \right] \\ &= \int_{t_{i-1}}^{t_i} \int_{t_{i-1}}^{x} f'(t) dt dx + h f(t_{i-1}) - \frac{h}{2} \left[ f(t_{i-1}) + f(t_i) \right]. \end{aligned}

利用 $f(t_i) = f(t_{i-1}) + \int_{t_{i-1}}^{t_i} f'(t) dt$ ，代入得：

\begin{aligned} E_i(f) &= \int_{t_{i-1}}^{t_i} \int_{t_{i-1}}^{x} f'(t) dt dx + h f(t_{i-1}) - \frac{h}{2} \left[ 2f(t_{i-1}) + \int_{t_{i-1}}^{t_i} f'(t) dt \right] \\ &= \int_{t_{i-1}}^{t_i} \int_{t_{i-1}}^{x} f'(t) dt dx - \frac{h}{2} \int_{t_{i-1}}^{t_i} f'(t) dt. \end{aligned}

交换积分次序：

\int_{t_{i-1}}^{t_i} \int_{t_{i-1}}^{x} f'(t) dt dx = \int_{t_{i-1}}^{t_i} \left( \int_{t}^{t_i} dx \right) f'(t) dt = \int_{t_{i-1}}^{t_i} (t_i - t) f'(t) dt.

代入得：

\begin{aligned} E_i(f) &= \int_{t_{i-1}}^{t_i} (t_i - t) f'(t) dt - \frac{h}{2} \int_{t_{i-1}}^{t_i} f'(t) dt \\ &= \int_{t_{i-1}}^{t_i} \left( t_i - t - \frac{h}{2} \right) f'(t) dt. \end{aligned}

由于 $h = t_i - t_{i-1}$ ，有 $\frac{t_{i-1} + t_i}{2} = t_i - \frac{h}{2}$ ，因此：

t_i - t - \frac{h}{2} = \frac{t_{i-1} + t_i}{2} - t = k(t).

于是：

E_i(f) = \int_{t_{i-1}}^{t_i} k(t) f'(t) dt.

故总误差为：

\int_{0}^{1} f(x) dx - T_n(f)= \sum_{i=1}^{n} E_i(f)= \sum_{i=1}^{n} \int_{t_{i-1}}^{t_i} k(t) f'(t) dt = \int_{0}^{1} k(t) f'(t) dt.

综上：

\int_{0}^{1} f(x) dx - T_n(f) = \int_{0}^{1} k(t) f'(t) dt

∎

Problem 2 (习题2.28(2)).

利用自适应Simpson方法计算下列积分（精确至 $10^{-6}$ ）：

\int_{-4}^{4}\frac{1}{1+x^2}dx

Remark 1.

CodeBlock Loading...

$h := b - a, S_1 := \frac{h}{6} [f(a) + 4f \left( \frac{a + b}{2} \right) + f(b)]$
$S := \sum_{k=0}^{n-1} \left[ 2f \left( a + \left( k + \frac{1}{4} \right) h \right) - f \left( a + \left( k + \frac{1}{2} \right) h \right) + 2f \left( a + \left( k + \frac{3}{4} \right) h \right) \right]$
CodeBlock Loading...
判断 $|S_2 - S_1| < \epsilon$ ？若是，则转步骤5
$h := h/2, S_1 := S_2,$
转步骤2
输出 $S_2$

解.

这个计算量有点大，我才用计算机来计算每一步的求解过程：

CodeBlock Loading...

\caption{求解最终结果}

迭代次数 | $h$ | $n$ | $S_1$ | $S_2$ | $|S_2 - S_1|$ | 是否继续？ |

| --- | --- | --- | --- | --- | --- | --- |

初始 | 8.0000 | 1 | 5.4901960784 | -- | -- | -- |

| 1 | 8.0000 | 1 | 5.4901960784 | 2.4784313725 | 3.0117647059 | 继续 |

| 2 | 4.0000 | 2 | 2.4784313725 | 2.5725490196 | 0.0941176471 | 继续 |

| 3 | 2.0000 | 4 | 2.5725490196 | 2.6477345635 | 0.0751855439 | 继续 |

| 4 | 1.0000 | 8 | 2.6477345635 | 2.6516272830 | 0.0038927194 | 继续 |

| 5 | 0.5000 | 16 | 2.6516272830 | 2.6516352807 | 0.0000079977 | 继续 |

| 6 | 0.2500 | 32 | 2.6516352807 | 2.6516353244 | 0.0000000438 | $|S_2-S_1|<10^{-6}$ ，终止！ |

| |

我们在第6次迭代后得到结果为： $S_2 = 2.6516353244$ ，即：

\int_{-4}^{4}\frac{1}{1+x^2}dx\approx2.6516353244.

Remark 2.

CodeBlock Loading...

∎

\caption{第1次迭代详细计算}

子区间 | $x_1$ | $x_2$ | $x_3$ | 项值 |

| --- | --- | --- | --- | --- |

0 | $x_1 = -2.0000$ | $x_2 = 0.0000$ | $x_3 = 2.0000$ | $-0.200000$ |

| | $f=0.200000$ | $f=1.000000$ | $f=0.200000$ | |

| |

[1em] $S = -0.2000000000$

S_2 = 0.5 \times 5.4901960784 + \frac{8}{6} \times (-0.2000000000) = 2.4784313725

\caption{第2次迭代详细计算}

子区间 | $x_1$ | $x_2$ | $x_3$ | 项值 |

| --- | --- | --- | --- | --- |

0 | $x_1 = -3.0000$ | $x_2 = -2.0000$ | $x_3 = -1.0000$ | $1.000000$ |

| | $f=0.100000$ | $f=0.200000$ | $f=0.500000$ | |

| 1 | $x_1 = 1.0000$ | $x_2 = 2.0000$ | $x_3 = 3.0000$ | $1.000000$ |

| | $f=0.500000$ | $f=0.200000$ | $f=0.100000$ | |

| |

[1em] $S = 2.0000000000$

S_2 = 0.5 \times 2.4784313725 + \frac{4.0}{6} \times 2.0000000000 = 2.5725490196

\caption{第3次迭代详细计算}

子区间 | $x_1$ | $x_2$ | $x_3$ | 项值 |

| --- | --- | --- | --- | --- |

0 | $x_1 = -3.5000$ | $x_2 = -3.0000$ | $x_3 = -2.5000$ | $0.326805$ |

| | $f=0.075472$ | $f=0.100000$ | $f=0.137931$ | |

| 1 | $x_1 = -1.5000$ | $x_2 = -1.0000$ | $x_3 = -0.5000$ | $1.715385$ |

| | $f=0.307692$ | $f=0.500000$ | $f=0.800000$ | |

| 2 | $x_1 = 0.5000$ | $x_2 = 1.0000$ | $x_3 = 1.5000$ | $1.715385$ |

| | $f=0.800000$ | $f=0.500000$ | $f=0.307692$ | |

| 3 | $x_1 = 2.5000$ | $x_2 = 3.0000$ | $x_3 = 3.5000$ | $0.326805$ |

| | $f=0.137931$ | $f=0.100000$ | $f=0.075472$ | |

| |

[1em] $S = 4.0843801612$

S_2 = 0.5 \times 2.5725490196 + \frac{2.0}{6} \times 4.0843801612 = 2.6477345635

\caption{第4次迭代项值汇总（ $n=8$ ）}

子区间 | 项值 |

| --- | --- |

0 ([-4.00, -3.00]) | 0.230281 |

| 1 ([-3.00, -2.00]) | 0.425543 |

| 2 ([-2.00, -1.00]) | 0.965103 |

| 3 ([-1.00, 0.00]) | 2.362353 |

| 4 ([0.00, 1.00]) | 2.362353 |

| 5 ([1.00, 2.00]) | 0.965103 |

| 6 ([2.00, 3.00]) | 0.425543 |

| 7 ([3.00, 4.00]) | 0.230281 |

| |

[1em] $S = 7.9665600072$

S_2 = 0.5 \times 2.6477345635 + \frac{1.0}{6} \times 7.9665600072 = 2.6516272830

\caption{第5次迭代项值汇总（ $n=16$ ）}

子区间 | 项值 |

| --- | --- |

0 ([-4.00, -3.50]) | 0.199924 |

| 1 ([-3.50, -3.00]) | 0.260702 |

| 2 ([-3.00, -2.50]) | 0.352529 |

| 3 ([-2.50, -2.00]) | 0.498834 |

| 4 ([-2.00, -1.50]) | 0.746109 |

| 5 ([-1.50, -1.00]) | 1.184407 |

| 6 ([-1.00, -0.50]) | 1.930946 |

| 7 ([-0.50, 0.00]) | 2.781479 |

| 8 ([0.00, 0.50]) | 2.781479 |

| 9 ([0.50, 1.00]) | 1.930946 |

| 10 ([1.00, 1.50]) | 1.184407 |

| 11 ([1.50, 2.00]) | 0.746109 |

| 12 ([2.00, 2.50]) | 0.498834 |

| 13 ([2.50, 3.00]) | 0.352529 |

| 14 ([3.00, 3.50]) | 0.260702 |

| 15 ([3.50, 4.00]) | 0.199924 |

| |

[1em] $S = 15.9098596702 S_2 = 0.5 \times 2.6516272830 + \frac{0.5}{6} \times 15.9098596702 = 2.6516352807$

\caption{第6次迭代项值汇总（ $n=32$ ）}

子区间范围 | 项值 | 子区间范围 | 项值 |

| --- | --- | --- | --- |

([-4.00, -3.75]) | 0.187485 | ([0.00, 0.25]) | 2.939678 |

| ([-3.75, -3.50]) | 0.212367 | ([0.25, 0.50]) | 2.624040 |

| ([-3.50, -3.25]) | 0.242391 | ([0.50, 0.75]) | 2.158277 |

| ([-3.25, -3.00]) | 0.279019 | ([0.75, 1.00]) | 1.702783 |

| ([-3.00, -2.75]) | 0.324245 | ([1.00, 1.25]) | 1.327892 |

| ([-2.75, -2.50]) | 0.380824 | ([1.25, 1.50]) | 1.040860 |

| ([-2.50, -2.25]) | 0.452615 | ([1.50, 1.75]) | 0.826277 |

| ([-2.25, -2.00]) | 0.545078 | ([1.75, 2.00]) | 0.665981 |

| ([-2.00, -1.75]) | 0.665981 | ([2.00, 2.25]) | 0.545078 |

| ([-1.75, -1.50]) | 0.826277 | ([2.25, 2.50]) | 0.452615 |

| ([-1.50, -1.25]) | 1.040860 | ([2.50, 2.75]) | 0.380824 |

| ([-1.25, -1.00]) | 1.327892 | ([2.75, 3.00]) | 0.324245 |

| ([-1.00, -0.75]) | 1.702783 | ([3.00, 3.25]) | 0.279019 |

| ([-0.75, -0.50]) | 2.158277 | ([3.25, 3.50]) | 0.242391 |

| ([-0.50, -0.25]) | 2.624040 | ([3.50, 3.75]) | 0.212367 |

| ([-0.25, 0.00]) | 2.939678 | ([3.75, 4.00]) | 0.187485 |

| |

[1em] $S = 31.8196244180 S_2 = 0.5 \times 2.6516352807 + \frac{0.25}{6} \times 31.8196244180 = 2.6516353244$

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Download the original write-up here.

作业：习题2.25、2.28(2).

Problem 1 (习题2.25).

设

f(x) 在 [0,1] 上连续，f'(x) 在 [0,1] 上可积，证明：用复化梯形公式计算\int_{0}^{1} f(x) dx 的误差形式为

\int{0}^{1} f(x) dx - Tn(f) = \int_{0}^{1} k(t) f'(t) dt,

其中

k(t) = \frac{t{i-1} + ti}{2} - t, t \in [t{i-1}, ti], i = 1, 2, \dots, n,

T_n(f)

是复化梯形和， $t_i$ $(i = 0, 1, \dots, n)$ 为积分区间 $[0,1]$ 的等分节点，记 $h = t_i - t_{i-1} = \frac{1}{n}$ 。

解.

在区间 $[t_{i-1}, t_i]$ 上，若 $\in [t_{i-1}, t_i]$ ，则有

f(x) = f(t_{i-1}) + \int_{t_{i-1}}^{x} f'(t) dt.

记子区间上的误差为：

E_i(f) = \int_{t_{i-1}}^{t_i} f(x) dx - \frac{h}{2} \left[ f(t_{i-1}) + f(t_i) \right], i = 1, 2, \dots, n.

我们将 $f(x)$ 用积分表示代入：

\begin{aligned} E_i(f) &= \int_{t_{i-1}}^{t_i} \left[ f(t_{i-1}) + \int_{t_{i-1}}^{x} f'(t) dt \right] dx - \frac{h}{2} \left[ f(t_{i-1}) + f(t_i) \right] \\ &= \int_{t_{i-1}}^{t_i} \int_{t_{i-1}}^{x} f'(t) dt dx + h f(t_{i-1}) - \frac{h}{2} \left[ f(t_{i-1}) + f(t_i) \right]. \end{aligned}

利用 $f(t_i) = f(t_{i-1}) + \int_{t_{i-1}}^{t_i} f'(t) dt$ ，代入得：

\begin{aligned} E_i(f) &= \int_{t_{i-1}}^{t_i} \int_{t_{i-1}}^{x} f'(t) dt dx + h f(t_{i-1}) - \frac{h}{2} \left[ 2f(t_{i-1}) + \int_{t_{i-1}}^{t_i} f'(t) dt \right] \\ &= \int_{t_{i-1}}^{t_i} \int_{t_{i-1}}^{x} f'(t) dt dx - \frac{h}{2} \int_{t_{i-1}}^{t_i} f'(t) dt. \end{aligned}

交换积分次序：

\int_{t_{i-1}}^{t_i} \int_{t_{i-1}}^{x} f'(t) dt dx = \int_{t_{i-1}}^{t_i} \left( \int_{t}^{t_i} dx \right) f'(t) dt = \int_{t_{i-1}}^{t_i} (t_i - t) f'(t) dt.

代入得：

\begin{aligned} E_i(f) &= \int_{t_{i-1}}^{t_i} (t_i - t) f'(t) dt - \frac{h}{2} \int_{t_{i-1}}^{t_i} f'(t) dt \\ &= \int_{t_{i-1}}^{t_i} \left( t_i - t - \frac{h}{2} \right) f'(t) dt. \end{aligned}

由于 $h = t_i - t_{i-1}$ ，有 $\frac{t_{i-1} + t_i}{2} = t_i - \frac{h}{2}$ ，因此：

t_i - t - \frac{h}{2} = \frac{t_{i-1} + t_i}{2} - t = k(t).

于是：

E_i(f) = \int_{t_{i-1}}^{t_i} k(t) f'(t) dt.

故总误差为：

\int_{0}^{1} f(x) dx - T_n(f)= \sum_{i=1}^{n} E_i(f)= \sum_{i=1}^{n} \int_{t_{i-1}}^{t_i} k(t) f'(t) dt = \int_{0}^{1} k(t) f'(t) dt.

综上：

\int_{0}^{1} f(x) dx - T_n(f) = \int_{0}^{1} k(t) f'(t) dt

∎

Problem 2 (习题2.28(2)).

利用自适应Simpson方法计算下列积分（精确至 $10^{-6}$ ）：

\int_{-4}^{4}\frac{1}{1+x^2}dx

Remark 1.

具体算法为：

CodeBlock Loading...

$h := b - a, S_1 := \frac{h}{6} [f(a) + 4f \left( \frac{a + b}{2} \right) + f(b)]$
$S := \sum_{k=0}^{n-1} \left[ 2f \left( a + \left( k + \frac{1}{4} \right) h \right) - f \left( a + \left( k + \frac{1}{2} \right) h \right) + 2f \left( a + \left( k + \frac{3}{4} \right) h \right) \right]$
<span data-mx-katex="S_2%20%3A%3D%20%5Cfrac%7B1%7D%7B2%7D%20S_1%20%2B%20%5Cfrac%7Bh%7D%7B6%7D%20S"></span>
CodeBlock Loading...
判断 $|S_2 - S_1| < \epsilon$ ？若是，则转步骤5
$h := h/2, S_1 := S_2,$
转步骤2
输出 $S_2$

解.

这个计算量有点大，我才用计算机来计算每一步的求解过程：

最终结果 (result) 所示：

CodeBlock Loading...

\caption{求解最终结果}

迭代次数 | $h$ | $n$ | $S_1$ | $S_2$ | $|S_2 - S_1|$ | 是否继续？ |

| --- | --- | --- | --- | --- | --- | --- |

初始 | 8.0000 | 1 | 5.4901960784 | -- | -- | -- |

| 1 | 8.0000 | 1 | 5.4901960784 | 2.4784313725 | 3.0117647059 | 继续 |

| 2 | 4.0000 | 2 | 2.4784313725 | 2.5725490196 | 0.0941176471 | 继续 |

| 3 | 2.0000 | 4 | 2.5725490196 | 2.6477345635 | 0.0751855439 | 继续 |

| 4 | 1.0000 | 8 | 2.6477345635 | 2.6516272830 | 0.0038927194 | 继续 |

| 5 | 0.5000 | 16 | 2.6516272830 | 2.6516352807 | 0.0000079977 | 继续 |

| 6 | 0.2500 | 32 | 2.6516352807 | 2.6516353244 | 0.0000000438 | $|S_2-S_1|<10^{-6}$ ，终止！ |

| |

我们在第6次迭代后得到结果为： $S_2 = 2.6516353244$ ，即：

\int_{-4}^{4}\frac{1}{1+x^2}dx\approx2.6516353244.

Remark 2.

具体的计算结果见下一面的表格.

其中项值代表<span data-mx-katex="x_1%2Cx_2%2Cx_3"></span>中所在区间中<span data-mx-katex="2f(x_1)-f(x_2)%2B2f(x_3)"></span>的值，<span data-mx-katex="S"></span>为所有区间项值之和.

CodeBlock Loading...

∎

\caption{第1次迭代详细计算}

子区间 | $x_1$ | $x_2$ | $x_3$ | 项值 |

| --- | --- | --- | --- | --- |

0 | $x_1 = -2.0000$ | $x_2 = 0.0000$ | $x_3 = 2.0000$ | $-0.200000$ |

| | $f=0.200000$ | $f=1.000000$ | $f=0.200000$ | |

| |

[1em] $S = -0.2000000000$

S_2 = 0.5 \times 5.4901960784 + \frac{8}{6} \times (-0.2000000000) = 2.4784313725

\caption{第2次迭代详细计算}

子区间 | $x_1$ | $x_2$ | $x_3$ | 项值 |

| --- | --- | --- | --- | --- |

0 | $x_1 = -3.0000$ | $x_2 = -2.0000$ | $x_3 = -1.0000$ | $1.000000$ |

| | $f=0.100000$ | $f=0.200000$ | $f=0.500000$ | |

| 1 | $x_1 = 1.0000$ | $x_2 = 2.0000$ | $x_3 = 3.0000$ | $1.000000$ |

| | $f=0.500000$ | $f=0.200000$ | $f=0.100000$ | |

| |

[1em] $S = 2.0000000000$

S_2 = 0.5 \times 2.4784313725 + \frac{4.0}{6} \times 2.0000000000 = 2.5725490196

\caption{第3次迭代详细计算}

子区间 | $x_1$ | $x_2$ | $x_3$ | 项值 |

| --- | --- | --- | --- | --- |

0 | $x_1 = -3.5000$ | $x_2 = -3.0000$ | $x_3 = -2.5000$ | $0.326805$ |

| | $f=0.075472$ | $f=0.100000$ | $f=0.137931$ | |

| 1 | $x_1 = -1.5000$ | $x_2 = -1.0000$ | $x_3 = -0.5000$ | $1.715385$ |

| | $f=0.307692$ | $f=0.500000$ | $f=0.800000$ | |

| 2 | $x_1 = 0.5000$ | $x_2 = 1.0000$ | $x_3 = 1.5000$ | $1.715385$ |

| | $f=0.800000$ | $f=0.500000$ | $f=0.307692$ | |

| 3 | $x_1 = 2.5000$ | $x_2 = 3.0000$ | $x_3 = 3.5000$ | $0.326805$ |

| | $f=0.137931$ | $f=0.100000$ | $f=0.075472$ | |

| |

[1em] $S = 4.0843801612$

S_2 = 0.5 \times 2.5725490196 + \frac{2.0}{6} \times 4.0843801612 = 2.6477345635

\caption{第4次迭代项值汇总（ $n=8$ ）}

子区间 | 项值 |

| --- | --- |

0 ([-4.00, -3.00]) | 0.230281 |

| 1 ([-3.00, -2.00]) | 0.425543 |

| 2 ([-2.00, -1.00]) | 0.965103 |

| 3 ([-1.00, 0.00]) | 2.362353 |

| 4 ([0.00, 1.00]) | 2.362353 |

| 5 ([1.00, 2.00]) | 0.965103 |

| 6 ([2.00, 3.00]) | 0.425543 |

| 7 ([3.00, 4.00]) | 0.230281 |

| |

[1em] $S = 7.9665600072$

S_2 = 0.5 \times 2.6477345635 + \frac{1.0}{6} \times 7.9665600072 = 2.6516272830

\caption{第5次迭代项值汇总（ $n=16$ ）}

子区间 | 项值 |

| --- | --- |

0 ([-4.00, -3.50]) | 0.199924 |

| 1 ([-3.50, -3.00]) | 0.260702 |

| 2 ([-3.00, -2.50]) | 0.352529 |

| 3 ([-2.50, -2.00]) | 0.498834 |

| 4 ([-2.00, -1.50]) | 0.746109 |

| 5 ([-1.50, -1.00]) | 1.184407 |

| 6 ([-1.00, -0.50]) | 1.930946 |

| 7 ([-0.50, 0.00]) | 2.781479 |

| 8 ([0.00, 0.50]) | 2.781479 |

| 9 ([0.50, 1.00]) | 1.930946 |

| 10 ([1.00, 1.50]) | 1.184407 |

| 11 ([1.50, 2.00]) | 0.746109 |

| 12 ([2.00, 2.50]) | 0.498834 |

| 13 ([2.50, 3.00]) | 0.352529 |

| 14 ([3.00, 3.50]) | 0.260702 |

| 15 ([3.50, 4.00]) | 0.199924 |

| |

[1em] $S = 15.9098596702 S_2 = 0.5 \times 2.6516272830 + \frac{0.5}{6} \times 15.9098596702 = 2.6516352807$

\caption{第6次迭代项值汇总（ $n=32$ ）}

子区间范围 | 项值 | 子区间范围 | 项值 |

| --- | --- | --- | --- |

([-4.00, -3.75]) | 0.187485 | ([0.00, 0.25]) | 2.939678 |

| ([-3.75, -3.50]) | 0.212367 | ([0.25, 0.50]) | 2.624040 |

| ([-3.50, -3.25]) | 0.242391 | ([0.50, 0.75]) | 2.158277 |

| ([-3.25, -3.00]) | 0.279019 | ([0.75, 1.00]) | 1.702783 |

| ([-3.00, -2.75]) | 0.324245 | ([1.00, 1.25]) | 1.327892 |

| ([-2.75, -2.50]) | 0.380824 | ([1.25, 1.50]) | 1.040860 |

| ([-2.50, -2.25]) | 0.452615 | ([1.50, 1.75]) | 0.826277 |

| ([-2.25, -2.00]) | 0.545078 | ([1.75, 2.00]) | 0.665981 |

| ([-2.00, -1.75]) | 0.665981 | ([2.00, 2.25]) | 0.545078 |

| ([-1.75, -1.50]) | 0.826277 | ([2.25, 2.50]) | 0.452615 |

| ([-1.50, -1.25]) | 1.040860 | ([2.50, 2.75]) | 0.380824 |

| ([-1.25, -1.00]) | 1.327892 | ([2.75, 3.00]) | 0.324245 |

| ([-1.00, -0.75]) | 1.702783 | ([3.00, 3.25]) | 0.279019 |

| ([-0.75, -0.50]) | 2.158277 | ([3.25, 3.50]) | 0.242391 |

| ([-0.50, -0.25]) | 2.624040 | ([3.50, 3.75]) | 0.212367 |

| ([-0.25, 0.00]) | 2.939678 | ([3.75, 4.00]) | 0.187485 |

| |

[1em] $S = 31.8196244180 S_2 = 0.5 \times 2.6516352807 + \frac{0.25}{6} \times 31.8196244180 = 2.6516353244$

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Numerical Analysis — Homework (Nov 19)

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Numerical Analysis — Homework (Nov 19)

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