Simpson's Rule

Use Simpson's rule to estimate the value of $\int_0^1 f(x)\, dx$ given the following table:

x	0	0.25	0.5	0.75	1
f(x)	1	0.8	1.3	1.1	1.6

Solution:

$a = 0$, $b = 1$, $n = 4$, $h = \frac{1 - 0}{4} = 0.25$

$\int_0^1 f(x)\, dx \approx \frac{0.25}{3}[1 + 4(0.8 + 1.1) + 2(1.3) + 1.6]$

$\approx \frac{0.25}{3}[1 + 4(1.9) + 2.6 + 1.6]$

$\approx \frac{0.25}{3}[1 + 7.6 + 2.6 + 1.6]$

$\approx \frac{0.25}{3}[12.8]$

$\approx 1.0667$

Use Simpson's rule to approximate the value of $\int_0^2 \sqrt{x^2 + 1}\, dx$ given that $n = 4$ subintervals.

Solution:

$f(x) = \sqrt{x^2 + 1}$, $a = 0$, $b = 2$, $n = 4$, $h = \frac{2 - 0}{4} = 0.5$

x	f(x)
0	1
0.5	1.1180
1	1.4142
1.5	1.8028
2	2.2361

$\int_0^2 \sqrt{x^2 + 1}\, dx \approx \frac{0.5}{3}[1 + 4(1.1180 + 1.8028) + 2(1.4142) + 2.2361]$

$\approx \frac{0.5}{3}[1 + 4(2.9208) + 2.8284 + 2.2361]$

$\approx \frac{0.5}{3}[1 + 11.6832 + 2.8284 + 2.2361]$

$\approx \frac{0.5}{3}[17.7477]$

$\approx 2.95795 \text{ or } 2.9580$

Evaluate $\int_1^2 \frac{1}{x}\, dx$ using Simpson's rule with $n = 10$ subintervals.

Solution:

Given $\int_1^2 \frac{1}{x}\, dx$, then $y = \frac{1}{x}$. Using Simpson's rule:

$\int_a^b y\, dx \approx \frac{h}{3}[(y_0 + y_n) + 4(y_1 + y_3 + \ldots + y_{n-1}) + 2(y_2 + y_4 + \ldots + y_{n-2})]$

$a = 1$, $b = 2$, $n = 10$, so $h = \frac{2 - 1}{10} = 0.1$

Table of values:

i	$x_i$	$y_i$
0	1.0	1.0000
1	1.1	0.9091
2	1.2	0.8333
3	1.3	0.7692
4	1.4	0.7143
5	1.5	0.6667
6	1.6	0.6250
7	1.7	0.5882
8	1.8	0.5556
9	1.9	0.5263
10	2.0	0.5000

Let:

$S_0 = y_0 + y_{10} = 1.0000 + 0.5000 = 1.5000$

$S_1 = y_1 + y_3 + y_5 + y_7 + y_9 = 0.9091 + 0.7692 + 0.6667 + 0.5882 + 0.5263 = 3.4595$

$S_2 = y_2 + y_4 + y_6 + y_8 = 0.8333 + 0.7143 + 0.6250 + 0.5556 = 2.7282$

$\int_1^2 \frac{1}{x}\, dx \approx \frac{0.1}{3}[1.5000 + 4(3.4595) + 2(2.7282)]$

$\approx \frac{0.1}{3}[1.5 + 13.838 + 5.4564]$

$\approx \frac{0.1}{3}[20.7944]$

$\approx 0.6931$

The coordinates of a curve $y = f(x)$ are given in the following table:

x	0	1	2	3	4	5	6
y	0	3.2	3.7	3.5	3	2.4	2.1

Use Simpson's rule to estimate the volume generated by revolving the area bounded by the curve between $x = 0$ and $x = 6$ about the x-axis.

Solution:

Volume, $V = \pi \int_a^b y^2\, dx$

$a = 0$, $b = 6$, $n = 6$, $h = \frac{6 - 0}{6} = 1$

Table of $y^2$ values:

x	y	$y^2$
0	0	0
1	3.2	10.24
2	3.7	13.69
3	3.5	12.25
4	3	9
5	2.4	5.76
6	2.1	4.41

$V \approx \pi \cdot \frac{1}{3}[0 + 4(10.24 + 12.25 + 5.76) + 2(13.69 + 9) + 4.41]$

$\approx \frac{\pi}{3}[0 + 4(28.25) + 2(22.69) + 4.41]$

$\approx \frac{\pi}{3}[113 + 45.38 + 4.41]$

$\approx \frac{\pi}{3}[162.79]$

$\approx 170.47 \text{ cubic units}$