Question Statement

Evaluate $\int_{0}^{1} \frac{d x}{1 + x}$ using the following numerical methods with $n = 8$ subintervals:

Trapezoidal Rule
Simpson's $\frac{1}{3}$ Rule

Then compare both approximate results with the exact value of the integral.

Background and Explanation

This problem demonstrates two fundamental numerical integration techniques for approximating definite integrals when analytical solutions are difficult or when working with discrete data. The Trapezoidal Rule approximates the area under the curve using linear segments (trapezoids), while Simpson's $\frac{1}{3}$ Rule fits quadratic parabolas to successive pairs of intervals, typically yielding greater accuracy.

Solution

Setup and Parameters

First, we identify the given parameters by comparing $\int_{0}^{1} \frac{d x}{1 + x}$ with the standard form $\int_{a}^{b} f (x) d x$ :

$a = 0, b = 1, n = 8, f (x) = \frac{1}{1 + x}$

Calculate the step size $h$ : $h = \frac{b - a}{n} = \frac{1 - 0}{8} = \frac{1}{8} = 0.125$

Calculating Partition Points ( $x_{i}$ )

We divide the interval $[0, 1]$ into 8 equal subintervals: $x_{0} x_{1} x_{2} x_{3} x_{4} x_{5} x_{6} x_{7} x_{8} = 0 = x_{0} + h = 0 + 0.125 = 0.125 = x_{1} + h = 0.125 + 0.125 = 0.25 = x_{2} + h = 0.25 + 0.125 = 0.375 = x_{3} + h = 0.375 + 0.125 = 0.5 = x_{4} + h = 0.5 + 0.125 = 0.625 = x_{5} + h = 0.625 + 0.125 = 0.75 = x_{6} + h = 0.75 + 0.125 = 0.875 = x_{7} + h = 0.875 + 0.125 = 1$

Evaluating Function Values ( $f (x_{i})$ )

Calculate $f (x) = \frac{1}{1 + x}$ at each partition point:

$i$	$x_{i}$	$f (x_{i}) = \frac{1}{1 + x _{i}}$
0	0	$f (0) = \frac{1}{1 + 0} = 1$
1	0.125	$f (0.125) = \frac{1}{1.125} \approx 0.8889$
2	0.25	$f (0.25) = \frac{1}{1.25} = 0.8$
3	0.375	$f (0.375) = \frac{1}{1.375} \approx 0.7273$
4	0.5	$f (0.5) = \frac{1}{1.5} \approx 0.6667$
5	0.625	$f (0.625) = \frac{1}{1.625} \approx 0.6154$
6	0.75	$f (0.75) = \frac{1}{1.75} \approx 0.5714$
7	0.875	$f (0.875) = \frac{1}{1.875} \approx 0.5333$
8	1	$f (1) = \frac{1}{2} = 0.5$

Part (a): Trapezoidal Rule

The Trapezoidal Rule formula for $n$ subintervals is: $\int_{a}^{b} f (x) d x \approx \frac{h}{2} [f (x_{0}) + 2 \sum_{i = 1}^{n - 1} f (x_{i}) + f (x_{n})]$

Expanding for $n = 8$ : $\int_{0}^{1} \frac{d x}{1 + x} \approx \frac{h}{2} [f (x_{0}) + 2 (f (x_{1}) + f (x_{2}) + f (x_{3}) + f (x_{4}) + f (x_{5}) + f (x_{6}) + f (x_{7})) + f (x_{8})]$

Substituting the values: $\approx \frac{0.125}{2} [1 + 2 (0.8889 + 0.8 + 0.7273 + 0.6667 + 0.6154 + 0.5714 + 0.5333) + 0.5] \approx 0.0625 [1 + 2 (4.803) + 0.5] \approx 0.0625 [1 + 9.606 + 0.5] \approx 0.0625 (11.106) \approx 0.6941$

Result: $\approx 0.6941$

Part (b): Simpson's $\frac{1}{3}$ Rule

Simpson's $\frac{1}{3}$ Rule requires an even number of intervals and uses the pattern $1, 4, 2, 4, 2, \dots, 4, 1$ : $\int_{a}^{b} f (x) d x \approx \frac{h}{3} [f (x_{0}) + 4 (f (x_{1}) + f (x_{3}) + f (x_{5}) + f (x_{7})) + 2 (f (x_{2}) + f (x_{4}) + f (x_{6})) + f (x_{8})]$

Substituting the values (note: odd indices get coefficient 4, even indices get coefficient 2, except endpoints): $\approx \frac{0.125}{3} [1 + 4 (0.8889 + 0.7273 + 0.6154 + 0.5333) + 2 (0.8 + 0.6667 + 0.5714) + 0.5] \approx \frac{0.125}{3} [1 + 4 (2.7649) + 2 (2.0381) + 0.5] \approx \frac{0.125}{3} [1 + 11.0596 + 4.0762 + 0.5] \approx \frac{0.125}{3} (16.6358) \approx 0.6930$

(Note: Using the rounded values from the table: $4 (2.764) + 2 (2.0381) = 11.056 + 4.0762 = 15.1322$ , plus $1.5$ gives $16.6322$ , yielding $\frac{0.125}{3} \times 16.6322 \approx 0.6930$ )

Result: $\approx 0.6930$

Part (c): Exact Value and Comparison

Exact Integration: $\int_{0}^{1} \frac{d x}{1 + x} = [ln (1 + x)]_{0}^{1} = ln (2) - ln (1) = 0.6931 - 0 = 0.6931$

Error Analysis:

Trapezoidal Rule Error: $Error = ∣ Exact - Approximate ∣ = ∣0.6931 - 0.6941∣ = 0.001$

Simpson's $\frac{1}{3}$ Rule Error: $Error = ∣0.6931 - 0.6930∣ = 0.0001$

Conclusion: Simpson's $\frac{1}{3}$ Rule produces a significantly smaller error ( $0.0001$ ) compared to the Trapezoidal Rule ( $0.001$ ), making it more efficient and accurate for this integral.

Key Formulas or Methods Used

Step Size: $h = \frac{b - a}{n}$
Trapezoidal Rule: $\int_{a}^{b} f (x) d x \approx \frac{h}{2} [y_{0} + 2 (y_{1} + y_{2} + \dots + y_{n - 1}) + y_{n}]$
Simpson's $\frac{1}{3}$ Rule: $\int_{a}^{b} f (x) d x \approx \frac{h}{3} [y_{0} + 4 (y_{1} + y_{3} + \dots) + 2 (y_{2} + y_{4} + \dots) + y_{n}]$ (where $n$ is even)
Standard Integral: $\int \frac{1}{1 + x} d x = ln ∣1 + x ∣ + C$
Absolute Error: $Error = ∣ Exact Value - Approximate Value ∣$

Summary of Steps

Identify parameters: Set $a = 0$ , $b = 1$ , $n = 8$ , and calculate step size $h = 0.125$
Generate partition points: Calculate $x_{0}$ through $x_{8}$ in increments of $0.125$
Evaluate function: Compute $f (x_{i}) = \frac{1}{1 + x _{i}}$ for all 9 points
Apply Trapezoidal Rule: Use $\frac{h}{2} [y_{0} + 2 (sum of middle terms) + y_{8}]$ to get $\approx 0.6941$
Apply Simpson's Rule: Use $\frac{h}{3} [y_{0} + 4 (odd indices) + 2 (even indices) + y_{8}]$ to get $\approx 0.6930$
Calculate exact value: Integrate analytically to get $ln (2) \approx 0.6931$
Compare errors: Calculate absolute differences to verify Simpson's Rule is more accurate (error of $0.0001$ vs $0.001$ )

Question Statement

Evaluate $\int_{0}^{1} \frac{d x}{1 + x}$ using the following numerical methods with $n = 8$ subintervals:

Trapezoidal Rule
Simpson's $\frac{1}{3}$ Rule

Then compare both approximate results with the exact value of the integral.

Background and Explanation

Solution

Setup and Parameters

First, we identify the given parameters by comparing $\int_{0}^{1} \frac{d x}{1 + x}$ with the standard form $\int_{a}^{b} f (x) d x$ :

$a = 0, b = 1, n = 8, f (x) = \frac{1}{1 + x}$

Calculate the step size $h$ : $h = \frac{b - a}{n} = \frac{1 - 0}{8} = \frac{1}{8} = 0.125$

Calculating Partition Points ( $x_{i}$ )

Evaluating Function Values ( $f (x_{i})$ )

Calculate $f (x) = \frac{1}{1 + x}$ at each partition point:

$i$	$x_{i}$	$f (x_{i}) = \frac{1}{1 + x _{i}}$
0	0	$f (0) = \frac{1}{1 + 0} = 1$
1	0.125	$f (0.125) = \frac{1}{1.125} \approx 0.8889$
2	0.25	$f (0.25) = \frac{1}{1.25} = 0.8$
3	0.375	$f (0.375) = \frac{1}{1.375} \approx 0.7273$
4	0.5	$f (0.5) = \frac{1}{1.5} \approx 0.6667$
5	0.625	$f (0.625) = \frac{1}{1.625} \approx 0.6154$
6	0.75	$f (0.75) = \frac{1}{1.75} \approx 0.5714$
7	0.875	$f (0.875) = \frac{1}{1.875} \approx 0.5333$
8	1	$f (1) = \frac{1}{2} = 0.5$

Part (a): Trapezoidal Rule

The Trapezoidal Rule formula for $n$ subintervals is: $\int_{a}^{b} f (x) d x \approx \frac{h}{2} [f (x_{0}) + 2 \sum_{i = 1}^{n - 1} f (x_{i}) + f (x_{n})]$

Expanding for $n = 8$ : $\int_{0}^{1} \frac{d x}{1 + x} \approx \frac{h}{2} [f (x_{0}) + 2 (f (x_{1}) + f (x_{2}) + f (x_{3}) + f (x_{4}) + f (x_{5}) + f (x_{6}) + f (x_{7})) + f (x_{8})]$

Result: $\approx 0.6941$

Part (b): Simpson's $\frac{1}{3}$ Rule

(Note: Using the rounded values from the table: $4 (2.764) + 2 (2.0381) = 11.056 + 4.0762 = 15.1322$ , plus $1.5$ gives $16.6322$ , yielding $\frac{0.125}{3} \times 16.6322 \approx 0.6930$ )

Result: $\approx 0.6930$

Part (c): Exact Value and Comparison

Exact Integration: $\int_{0}^{1} \frac{d x}{1 + x} = [ln (1 + x)]_{0}^{1} = ln (2) - ln (1) = 0.6931 - 0 = 0.6931$

Error Analysis:

Trapezoidal Rule Error: $Error = ∣ Exact - Approximate ∣ = ∣0.6931 - 0.6941∣ = 0.001$

Simpson's $\frac{1}{3}$ Rule Error: $Error = ∣0.6931 - 0.6930∣ = 0.0001$

Conclusion: Simpson's $\frac{1}{3}$ Rule produces a significantly smaller error ( $0.0001$ ) compared to the Trapezoidal Rule ( $0.001$ ), making it more efficient and accurate for this integral.

Key Formulas or Methods Used

Step Size: $h = \frac{b - a}{n}$
Trapezoidal Rule: $\int_{a}^{b} f (x) d x \approx \frac{h}{2} [y_{0} + 2 (y_{1} + y_{2} + \dots + y_{n - 1}) + y_{n}]$
Simpson's $\frac{1}{3}$ Rule: $\int_{a}^{b} f (x) d x \approx \frac{h}{3} [y_{0} + 4 (y_{1} + y_{3} + \dots) + 2 (y_{2} + y_{4} + \dots) + y_{n}]$ (where $n$ is even)
Standard Integral: $\int \frac{1}{1 + x} d x = ln ∣1 + x ∣ + C$
Absolute Error: $Error = ∣ Exact Value - Approximate Value ∣$

Summary of Steps

Identify parameters: Set $a = 0$ , $b = 1$ , $n = 8$ , and calculate step size $h = 0.125$
Generate partition points: Calculate $x_{0}$ through $x_{8}$ in increments of $0.125$
Evaluate function: Compute $f (x_{i}) = \frac{1}{1 + x _{i}}$ for all 9 points
Apply Trapezoidal Rule: Use $\frac{h}{2} [y_{0} + 2 (sum of middle terms) + y_{8}]$ to get $\approx 0.6941$
Apply Simpson's Rule: Use $\frac{h}{3} [y_{0} + 4 (odd indices) + 2 (even indices) + y_{8}]$ to get $\approx 0.6930$
Calculate exact value: Integrate analytically to get $ln (2) \approx 0.6931$
Compare errors: Calculate absolute differences to verify Simpson's Rule is more accurate (error of $0.0001$ vs $0.001$ )

10.2 Q-4

Question Statement

Background and Explanation

Solution

Setup and Parameters

Calculating Partition Points ( $x_{i}$ )

Evaluating Function Values ( $f (x_{i})$ )

Part (a): Trapezoidal Rule

Part (b): Simpson's $\frac{1}{3}$ Rule

Part (c): Exact Value and Comparison

Key Formulas or Methods Used

Summary of Steps

10.2 Q-4

Question Statement

Background and Explanation

Solution

Setup and Parameters

Calculating Partition Points ( $x_{i}$ )

Evaluating Function Values ( $f (x_{i})$ )

Part (a): Trapezoidal Rule

Part (b): Simpson's $\frac{1}{3}$ Rule

Part (c): Exact Value and Comparison

Key Formulas or Methods Used

Summary of Steps