Problem 19 - Calculus: Early Transcendental Functions

Question

Approximate the given value using (a) Midpoint Rule, (b) Trapezoidal Rule and (c) Simpson's Rule with \(n=4\). $$\ln 4=\int_{1}^{4} \frac{1}{x} d x$$

Step-by-Step Solution

Answer

The approximated integral using each technique is as follows: \n(a) Midpoint Rule: \(M_4 = 0.75[f(1.375)+ f(2.125) + f(2.875) + f(3.625)]\), (b) Trapezoidal Rule: \(T_4 = 0.375[f(1) + 2f(1.375) + 2f(2.125) + 2f(2.875) + 2f(3.625) + f(4)]\), (c) Simpson's Rule: \(S_4 = 0.25[f(1) + 4f(1.375) + 2f(2.125) + 4f(2.875) + f(4)]\). Plug appropriate values in all the formulas to get the approximated values.

Step 1: Calculate the width of the subintervals.

Since we are using \( n=4 \), we will divide our interval \([1, 4]\) into 4 subintervals. First, we need to calculate the width \( h \) of the subintervals which is given by \( h = \frac{b - a}{n} = \frac{4 - 1}{4} = 0.75 \). So, the four subintervals are: [1, 1.75], [1.75, 2.5], [2.5, 3.25], and [3.25, 4].

Step 2: Approximate using the Midpoint Rule.

In the Midpoint Rule we evaluate our function at the midpoints of each subinterval. So we have: \[ M_4 = \frac{h}{2}[f(1.375) + f(2.125) + f(2.875) + f(3.625)] = 0.75[f(1.375) + f(2.125) + f(2.875) + f(3.625)]. \] The function we're integrating is \( \frac{1}{x} \), so substitute these values into the function and then sum them.

Step 3: Approximate using the Trapezoidal Rule.

In the Trapezoidal Rule, we evaluate our function at the endpoints of each subinterval. \[ T_{4} = \frac{h}{2} \left[ f \left(1 \right) + 2f \left( 1.375 \right) + 2f \left( 2.125 \right) + 2f \left( 2.875 \right) + 2f \left( 3.625 \right) + f \left( 4 \right) \right] . \] The function we're integrating is \( \frac{1}{x} \), so substitute these values into the function and then sum them.

Step 4: Approximate using Simpson’s Rule.

Simpson's Rule is a bit more involved. We will need both the midpoints and the endpoints of the subintervals. \[ S_4 = \frac{h}{3} \left[ f \left( a \right) + 4f \left( x_1 \right) + 2f \left( x_2 \right) + 4f \left( x_3 \right) + f \left( b \right) \right] = \frac{0.75}{3} \left[ f \left( 1 \right) + 4f \left( 1.375 \right) + 2f \left( 2.125 \right) + 4f \left( 2.875 \right) + f \left( 4 \right) \right] . \] The function we're integrating is \( \frac{1}{x} \), so substitute these values into the function and then sum them.

Key Concepts

Midpoint Rule

The Midpoint Rule is a numerical technique for estimating the value of a definite integral. This rule is straightforward: you evaluate the function at the midpoint of each subinterval, multiply by the width of the intervals, and sum the results.
For example, with a function like \( \frac{1}{x} \) over the interval \([1, 4]\) and \(n=4\), we first find the subinterval width, \(h = 0.75\).
Next, calculate the midpoints: \(1.375, 2.125, 2.875,\) and \(3.625\). Use these midpoints to evaluate the function and apply the formula:
\[ M_4 = 0.75[f(1.375) + f(2.125) + f(2.875) + f(3.625)]. \]
This technique is particularly useful for functions that are well-behaved between points, and it's often simpler than other methods.

Trapezoidal Rule

The Trapezoidal Rule provides another way to approximate the value of a definite integral. This approach uses trapezoids to estimate the area under the curve, summing the areas of each trapezoid as determined by the endpoints of each subinterval.
For our function \( \frac{1}{x} \) on \([1, 4]\), and \(n=4\), the subinterval width is once again \(h = 0.75\).
Using the formula:
\[ T_{4} = \frac{0.75}{2} \, [ f(1) + 2f(1.75) + 2f(2.5) + 2f(3.25) + f(4)]. \]
This method often provides good approximations and gives insight into the function's behavior over the interval, especially helpful for linear approximations.

Simpson's Rule

Simpson's Rule is a more refined method for numerical integration that combines aspects of the Midpoint and the Trapezoidal rules. It offers higher accuracy by using parabolic arcs instead of straight lines or single points.
For the function \( \frac{1}{x} \) between \(1\) and \(4\) with \(n=4\), the width \(h = 0.75\) is vital for calculation.
Using the formula:
\[ S_4 = \frac{0.75}{3} \, [ f(1) + 4f(1.375) + 2f(2.125) + 4f(2.875) + f(4)]. \]
The combination of multiplying certain terms by 4 and others by 2 mimics the arc of the parabola, which makes Simpson's rule more accurate for many functions.
This approach is often preferred for its precision, especially when dealing with smoother curves.

Definite Integrals

Definite integrals provide the total accumulation or area under a curve over a specified interval. They're an essential concept in calculus, useful for solving various real-world problems.
Given a function \( f(x) \) over an interval \([a, b]\), the definite integral \( \int_{a}^{b} f(x) \, dx \) measures the net area. This net area can be positive, negative, or zero, reflecting changes in the function over the interval.
For the integral \( \ln 4 = \int_{1}^{4} \frac{1}{x} \, dx \), the goal is to find the exact area under \( \frac{1}{x} \) from 1 to 4.
Numerical methods like the Midpoint, Trapezoidal, and Simpson's Rules help estimate these integrals when an anti-derivative is complex or cumbersome.

Riemann Sums

Riemann Sums form the foundational concept behind calculating definite integrals. They represent the summation of areas of rectangles under a curve, which can vary in height depending on particular criteria.
By partitioning an interval \([a, b]\) into smaller subintervals, we can choose various points to evaluate the function and calculate the area.
Different types of Riemann Sums arise from where you select your sample points—whether at the left endpoint, right endpoint, or midpoint, leading to Left, Right, and Midpoint Riemann Sums.
This approach is vital in approximations when direct integration is challenging, making Riemann Sums an essential tool in understanding the behavior of functions across intervals.