Question

For the following exercises, approximate the integrals using the midpoint rule, trapezoidal rule, and Simpson's rule using four subintervals, rounding to three decimals.[I] \(\int_{1}^{4} \frac{\ln (1 / x)}{x} d x\)

Short answer

Midpoint: -2.108, Trapezoidal: -1.897, Simpson's: -1.915.

Step-by-step solution

Calculate the Width of Each Subinterval

The integral is computed from 1 to 4, which is a total interval length of 3. By dividing this interval into four subintervals, the width \( h \) of each subinterval is calculated as follows:\[ h = \frac{b-a}{n} = \frac{4-1}{4} = 0.75 \]

Apply the Midpoint Rule

The Midpoint Rule formula is:\[ M_n = h \sum_{i=1}^{n} f(m_i) \]where \( m_i \) is the midpoint of each subinterval. The midpoints are: \( 1.375, 2.125, 2.875, \) and \( 3.625 \). Calculate \( M_4 \):\[ M_4 = 0.75 \left( f(1.375) + f(2.125) + f(2.875) + f(3.625) \right) \]After calculating \( f(x) = \frac{\ln(1/x)}{x} \) at each midpoint, approximate the integral. The sum is \(-2.811\), so:\[ M_4 = 0.75 \times (-2.811) \approx -2.108 \]

Apply the Trapezoidal Rule

The Trapezoidal Rule formula is:\[ T_n = \frac{h}{2} \left( f(a) + 2 \sum_{i=1}^{n-1} f(x_i) + f(b) \right) \]Calculate \( T_4 \) using sample points: 1, 1.75, 2.5, 3.25, and 4.\[ T_4 = \frac{0.75}{2} \left( f(1) + 2 \times (f(1.75) + f(2.5) + f(3.25)) + f(4) \right) \]Substituting the values gives \(-2.53\). Thus, \[ T_4 \approx -1.897 \]

Apply Simpson's Rule

The Simpson's Rule formula is:\[ S_n = \frac{h}{3} \left( f(a) + 4 \sum_{i=1,3,5,...}^{n} f(x_i) + 2 \sum_{i=2,4,6,...}^{n-1} f(x_i) + f(b) \right) \]Calculate \( S_4 \) using points: 1, 1.75, 2.5, 3.25, and 4:\[ S_4 = \frac{0.75}{3} \left( f(1) + 4 \times (f(1.75) + f(3.25)) + 2 \times f(2.5) + f(4) \right) \]Substituting the function values, the approximation is \(-2.44\). Thus,\[ S_4 \approx -1.915 \]

Key concepts

Midpoint Rule

The Midpoint Rule is a technique used in numerical integration for approximating the value of an integral. It is particularly useful when you want a quick and relatively simple estimation of the integral area. The central idea involves calculating the sum of the areas of rectangles, where each rectangle's height is determined by the function value at the midpoint of its base.

For our exercise, we divide the interval from 1 to 4 into four subintervals, each with a width of 0.75. These intervals help us find the midpoints: 1.375, 2.125, 2.875, and 3.625. We then evaluate the function \( f(x) = \frac{\ln(1/x)}{x} \) at each of these midpoints. These values are summed up and multiplied by the width, giving us \( M_4 \approx -2.108 \).

• Choose midpoints to simplify calculations.
• Ensure equal subinterval widths for consistency.
• Useful for providing a quick approximate of non-linear functions.

Trapezoidal Rule

The Trapezoidal Rule offers another method of numerical integration, involving the use of trapezoids instead of rectangles. Imagine a trapezoid under each curve of the function; the method improves precision by "trapping" the curve more accurately than rectangles.

In using the Trapezoidal Rule for approximating the integral from 1 to 4, we form trapezoidal sections by using sample points of 1, 1.75, 2.5, 3.25, and 4. The formula combines the endpoints with twice the value of the function at each internal point, creating an approximation of \( T_4 \approx -1.897 \).

• Incorporates both ends of intervals for better accuracy.
• Great for smooth, continuous functions.
• Balances additional complexity with improved precision over the Midpoint Rule.

Simpson's Rule

Simpson's Rule is often considered the most accurate among basic numerical integration techniques. It uses parabolic arcs instead of straight lines to approximate the integral, giving an excellent fit especially for functions that are well-behaved over the interval.

For Simpson’s Rule, calculated over the interval from 1 to 4, we use the points 1, 1.75, 2.5, 3.25, and 4. This technique involves a weighted sum: the points at odd positions are multiplied by four, and even positions by two, with the endpoints contributing uniquely. These are then multiplied by one-third of the width. This results in an estimation of \( S_4 \approx -1.915 \).

• Offers superior accuracy for polynomial-like functions.
• Uses a combination of both endpoints and midpoints.
• Suitable for functions with consistent curvature over the interval.