Numerical Integration

The trapezoidal rule approximates the area under a curve by dividing it into trapezoids.

Divide the region under $y = f(x)$ from $a$ to $b$ into $n$ strips of equal width $h$:

$a = x_0 < x_1 < x_2 < ... < x_{n-1} < x_n = b$

$h = \frac{b - a}{n}$

Area of each trapezium:

Trapezium 1: $A_1 = \frac{h}{2}[f(x_0) + f(x_1)]$

Trapezium 2: $A_2 = \frac{h}{2}[f(x_1) + f(x_2)]$

...

Trapezium n: $A_n = \frac{h}{2}[f(x_{n-1}) + f(x_n)]$

The total area under the curve is approximately the sum of the trapezium areas:

$\int_a^b f(x) dx \approx \frac{h}{2}[f(x_0) + f(x_1)] + \frac{h}{2}[f(x_1) + f(x_2)] + ... + \frac{h}{2}[f(x_{n-1}) + f(x_n)]$

Simplifying:

$\int_a^b f(x) dx \approx \frac{h}{2}[f(a) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(b)]$

Or:

$A = \frac{h}{2}[(\text{first ordinate} + \text{last ordinate}) + 2(\text{sum of remaining ordinates})]$

Or:

$A = \frac{h}{2}[y_0 + 2(y_1 + y_2 + y_3 + ... + y_{n-1}) + y_n]$

Note: $n = \text{number of ordinates} - 1$

Use the trapezoidal rule with five strips to evaluate $\int_0^1 (e^{-x} - x^3 + 3x^2) dx$, correct to four decimal places.

Solution:

$f(x) = e^{-x} - x^3 + 3x^2$, $a = 0$, $b = 1$, $n = 5$

$h = \frac{1 - 0}{5} = 0.2$

Table of values:

n	x	y = f(x)
0	0	1
1	0.2	1.5187
2	0.4	1.9051
3	0.6	2.1161
4	0.8	2.1064
5	1	2.3679

$\int_0^1 f(x) dx \approx \frac{0.2}{2}[1 + 2(1.5187 + 1.9051 + 2.1161 + 2.1064) + 2.3679]$

$\approx 0.1[1 + 2(7.6463) + 2.3679]$

$\approx 0.1[1 + 15.2926 + 2.3679]$

$\approx 0.1[18.6605]$

$\approx 1.8661$

Estimate the value of $\int_1^2 \frac{1}{x} dx$ using the trapezoidal rule with six ordinates (5 strips).

Solution:

$f(x) = \frac{1}{x}$, $a = 1$, $b = 2$, $n = 5$

$h = \frac{2 - 1}{5} = 0.2$

n	x	f(x)
0	1	1
1	1.2	0.8333
2	1.4	0.7143
3	1.6	0.6250
4	1.8	0.5556
5	2	0.5

$\int_1^2 \frac{1}{x} dx \approx \frac{0.2}{2}[1 + 2(0.8333 + 0.7143 + 0.6250 + 0.5556) + 0.5]$

$\approx 0.1[1 + 2(2.7282) + 0.5]$

$\approx 0.1[1 + 5.4564 + 0.5]$

$\approx 0.6956$

Approximate the value of $\int_0^1 (1 + \cos(\sqrt{3x^2})) dx$ with $n = 10$ subintervals using the trapezoidal rule.

Solution:

Given $\int_0^1 (1 + \cos(\sqrt{3x^2})) dx$, then $f(x) = 1 + \cos(\sqrt{3}x)$. Note: the original text had a typo in the function under the cosine. It should be $\sqrt{3}x$ not $\sqrt{3x^2}$ if we are to get the values presented in the table. If it is indeed supposed to be $\sqrt{3x^2}$ then the y values in the table will be different.

Using the trapezoidal rule:

$\int_a^b f(x) dx \approx \frac{h}{2}[y_0 + 2(y_1 + y_2 + ... + y_{n-1}) + y_n]$

Here, $a = 0$, $b = 1$, $n = 10$, so $h = \frac{1 - 0}{10} = 0.1$.

Table of values (assuming $f(x) = 1 + \cos(\sqrt{3}x)$):

n	$x_n$	$y_n = f(x_n)$	$2y_n$ (for n=1 to 9)
0	0	2
1	0.1	1.173	2.346
2	0.2	1.691	3.382
3	0.3	1.955	3.910
4	0.4	1.995	3.990
5	0.5	1.802	3.604
6	0.6	1.401	2.802
7	0.7	0.838	1.676
8	0.8	0.165	0.330
9	0.9	-0.536	-1.072
10	1	-0.190
Sum of $y_0$ and $y_{10}$			1.81
Sum of $2y_1$ to $2y_9$			20.978

$\int_0^1 (1 + \cos(\sqrt{3}x)) dx \approx \frac{0.1}{2}[2 + 2(1.173 + 1.691 + 1.955 + 1.995+ 1.802 + 1.401 + 0.838 + 0.165 + -0.536) + -0.190]$

$\approx 0.05[2 + 2(10.484) - 0.190]$

$\approx 0.05[2 + 20.968 - 0.190]$

$\approx 0.05[22.778]$

$\approx 1.139$

If however the function is $f(x) = 1 + \cos(\sqrt{3x^2})$ or $f(x) = 1 + \cos(x\sqrt{3})$

n	$x_n$	$y_n = f(x_n)$	$2y_n$ (for n=1 to 9)
0	0	2
1	0.1	1.995	3.99
2	0.2	1.98	3.96
3	0.3	1.955	3.91
4	0.4	1.92	3.84
5	0.5	1.875	3.75
6	0.6	1.82	3.64
7	0.7	1.755	3.51
8	0.8	1.68	3.36
9	0.9	1.595	3.19
10	1	1.5
Sum of $y_0$ and $y_{10}$			3.5
Sum of $2y_1$ to $2y_9$			33.15

$\int_0^1 (1 + \cos(\sqrt{3x^2})) dx \approx \frac{0.1}{2}[2 + 2(1.995 + 1.98 + 1.955 + 1.92+ 1.875 + 1.82 + 1.755 + 1.68 + 1.595) + 1.5]$

$\approx 0.05[2 + 33.15 + 1.5]$

$\approx 0.05[36.65]$

$\approx 1.8325 \text{ or } 1.833$