How do you do the trapezoidal rule to compute #int logxdx# from [1,2]?

Question

Evelyn Agard · Accepted Answer

#int_1^2logx dx≈log2/2≈0.34657#

Using the trapezoidal rule, the expression #int_1^2logx dx# becomes #(2-1)[(log1+log2)/2]=log2/2≈0.34657#.

It's important to notice that it's only an approximate value (n=1) and it contain a percentage of error : the correct value would be around 0.38629, but with this formula we found approximately 0.34657.

Charles Anctil · Answer

undefined To use the trapezoidal rule to approximate the integral of $ \log(x) $ from 1 to 2, follow these steps:

1. Divide the interval $[1, 2]$ into $ n $ subintervals of equal width. Let $ n $ be the desired number of subintervals.
2. Calculate the width of each subinterval, $ \Delta x $, using the formula: $ \Delta x = \frac{b - a}{n} $, where $ a = 1 $ and $ b = 2 $.
3. Evaluate the function $ \log(x) $ at the endpoints of each subinterval and sum the values.
4. Multiply the sum by $ \Delta x $.
5. Adjust the result by subtracting half the first term and half the last term.

The formula for the trapezoidal rule is:

$$ \text{Trapezoidal Rule} = \Delta x \left( \frac{1}{2}f(a) + f(x_1) + f(x_2) + \cdots + f(x_{n-1}) + \frac{1}{2}f(b) \right) $$

Where $ x_1, x_2, \ldots, x_{n-1} $ are the interior points of the subintervals.

Substitute $ a = 1 $, $ b = 2 $, and $ f(x) = \log(x) $ into the formula, along with the calculated $ \Delta x $, and compute the result.