How do you use the Trapezoidal Rule to approximate integral #int(2/x) dx# for n=4 from [1,3]?

Question

Elijah Abram · Accepted Answer

# int_1^3 \ 2/x \ dx ~~ 2.233333#

We have:

# y = 2/x #

We want to estimate #int \ y \ dx# over the interval #[1,3]# with #4# strips; thus:

# Deltax = (3-1)/4 = 0.5#

The values of the function are tabulated as follows;

Trapezium Rule

# A = int_a^b \ y \ dx #
# \ \ \ ~~ h/2{y_0+y_n+2(y_1+...+y_(n-1)) } #
# \ \ \ = 0.5/2 * { 2 + 0.666667 + 2*(1.333333 + 1 + 0.8) } #
# \ \ \ = 0.25 * { 2.666667 + 2*(3.133333) }#
# \ \ \ = 0.25 * { 2.666667 + 6.266667 }#
# \ \ \ = 0.25 * 8.933333#
# \ \ \ = 2.233333#

Actual Value

For comparison of accuracy:

# A = int_1^3 \ 2/x \ dx #
# \ \ \ = [2lnx]_1^3 #
# \ \ \ = 2ln3-2ln1 #
# \ \ \ = 2ln3 #
# \ \ \ ~~ 2.1972 #

Cameron Alexander · Answer

undefined To approximate the integral $ \int_{1}^{3} \frac{2}{x} \, dx $ using the Trapezoidal Rule with $ n = 4 $, follow these steps:

1. Divide the interval $[1, 3]$ into $ n $ subintervals of equal width. Since $ n = 4 $, each subinterval will have width $ \Delta x = \frac{3 - 1}{4} = 0.5 $.

2. Determine the function values at the endpoints of each subinterval. For $ n = 4 $, these points are $ x_0 = 1 $, $ x_1 = 1.5 $, $ x_2 = 2 $, $ x_3 = 2.5 $, and $ x_4 = 3 $.

3. Use the Trapezoidal Rule formula:

$$ \int_{a}^{b} f(x) \, dx \approx \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] $$

Substitute the values:

$$ \int_{1}^{3} \frac{2}{x} \, dx \approx \frac{0.5}{2} \left[ \frac{2}{1} + 2\left(\frac{2}{1.5}\right) + 2\left(\frac{2}{2}\right) + 2\left(\frac{2}{2.5}\right) + \frac{2}{3} \right] $$

4. Calculate the approximate value:

$$ \int_{1}^{3} \frac{2}{x} \, dx \approx \frac{1}{4} \left[ 2 + \frac{8}{1.5} + 4 + \frac{8}{2.5} + \frac{2}{3} \right] $$

$$ = \frac{1}{4} \left[ 2 + \frac{16}{3} + 4 + \frac{16}{5} + \frac{2}{3} \right] $$

$$ = \frac{1}{4} \left[ 2 + \frac{32}{15} + \frac{20}{5} + \frac{2}{3} \right] $$

$$ = \frac{1}{4} \left[ 2 + \frac{32}{15} + 4 + \frac{2}{3} \right] $$

$$ = \frac{1}{4} \left[ 6 + \frac{32}{15} + \frac{2}{3} \right] $$

$$ = \frac{1}{4} \left[ \frac{90 + 32 + 10}{15} \right] $$

$$ = \frac{1}{4} \left[ \frac{132}{15} \right] $$

$$ = \frac{33}{15} $$

$$ = 2.2 $$

So, the approximate value of $ \int_{1}^{3} \frac{2}{x} \, dx $ using the Trapezoidal Rule with $ n = 4 $ is $ 2.2 $.