Question Statement

Compute the definite integral

${\int }_0^1\frac{dx}{2+x^2}$

using the following methods with $n=8$ subintervals:

• (a) The Trapezoidal Rule
• (b) Simpson's $\frac{1}{3}$ Rule

Also evaluate the integral (c) analytically to find the exact value, and (d) comment on the accuracy of the numerical outcomes compared to the exact solution.

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Background and Explanation

Numerical integration approximates definite integrals by evaluating the integrand at discrete points. The Trapezoidal Rule approximates the area under the curve using trapezoids (linear interpolation between points), while Simpson's $\frac{1}{3}$ Rule uses parabolic arcs (quadratic interpolation) for better accuracy. Both methods require equally spaced nodes, with Simpson's rule requiring an even number of subintervals.

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Solution

Setup: Parameters and Function Evaluation

First, identify the integration parameters by comparing ${\int }_0^1\frac{dx}{2+x^2}$ with ${\int }_a^{b}f(x)dx$:

$a=0,b=1,f(x)=\frac{1}{2+x^2},n=8$

Calculate the step size $h$:

$h=\frac{b-a}{n}=\frac{1-0}{8}=\frac{1}{8}=0.125$

Calculate the nodes $x_i$: $\begin{array}{rl}{x}_0& =0\\ x_1& =0+0.125=0.125\\ x_2& =0.125+0.125=0.25\\ x_3& =0.25+0.125=0.375\\ x_4& =0.375+0.125=0.5\\ x_5& =0.5+0.125=0.625\\ x_6& =0.625+0.125=0.75\\ x_7& =0.75+0.125=0.875\\ x_8& =0.875+0.125=1\end{array}$

Evaluate the function at each node: $\begin{array}{rl}f(x_0)& =f(0)=\frac{1}{2+0}=0.5\\ f(x_1)& =f(0.125)=\frac{1}{2+(0.125{)}^2}\approx 0.4961\\ f(x_2)& =f(0.25)=\frac{1}{2+(0.25{)}^2}\approx 0.4848\\ f(x_3)& =f(0.375)=\frac{1}{2+(0.375{)}^2}\approx 0.4672\\ f(x_4)& =f(0.5)=\frac{1}{2+(0.5{)}^2}\approx 0.4444\\ f(x_5)& =f(0.625)=\frac{1}{2+(0.625{)}^2}\approx 0.4183\\ f(x_6)& =f(0.75)=\frac{1}{2+(0.75{)}^2}\approx 0.3902\\ f(x_7)& =f(0.875)=\frac{1}{2+(0.875{)}^2}\approx 0.3616\\ f(x_8)& =f(1)=\frac{1}{2+1}\approx 0.3333\end{array}$

Part (a): Trapezoidal Rule

The Trapezoidal Rule formula for $n$ subintervals is:

${\int }_a^{b}f(x)dx\approx \frac{h}{2}\left[f(x_0)+2{\sum }_{i=1}^{n-1}f(x_i)+f(x_n)\right]$

Substituting the values:

$\begin{array}{rl}{\int }_0^1\frac{dx}{2+x^2}& \approx \frac{0.125}{2}[0.5+2(0.4961+0.4848+0.4672+0.4444+0.4183+0.3902+0.3616)+0.3333]\\ & =0.0625[0.5+2(3.0626)+0.3333]\\ & =0.0625[0.5+6.1252+0.3333]\\ & =0.0625(6.9585)\\ & \approx \boxed{0.4349}\end{array}$

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

Simpson's $\frac{1}{3}$ Rule (valid for even $n$) is:

${\int }_a^{b}f(x)dx\approx \frac{h}{3}\left[f(x_0)+4{\sum }_{\text{odd }i}f(x_i)+2{\sum }_{\text{even }i}f(x_i)+f(x_n)\right]$

Substituting the values (odd indices: 1,3,5,7; even indices: 2,4,6):

$\begin{array}{rl}{\int }_0^1\frac{dx}{2+x^2}& \approx \frac{0.125}{3}[0.5+4(0.4961+0.4672+0.4183+0.3616)+2(0.4848+0.4444+0.3902)+0.3333]\\ & \approx 0.0417[0.5+4(1.7432)+2(1.3194)+0.3333]\\ & =0.0417[0.5+6.9728+2.6388+0.3333]\\ & =0.0417(10.4449)\\ & \approx \boxed{0.4355}\end{array}$

Part (c): Analytical Evaluation

To find the exact value, evaluate the integral analytically using the standard form $\int \frac{1}{a^2+x^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)$:

$\begin{array}{rl}{\int }_0^1\frac{1}{2+x^2}dx& ={\int }_0^1\frac{1}{(\sqrt{2}{)}^2+x^2}dx\\ & ={\left[\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)\right]}_0^1\\ & =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)-\frac{1}{\sqrt{2}}{\tan }^{-1}(0)\\ & =\frac{1}{\sqrt{2}}(0.6155)-0\\ & \approx \boxed{0.4352}\end{array}$

Thus, the exact value of the integral is approximately 0.4352.

Part (d): Comments on Outcomes

Calculate the absolute error for each method:

Trapezoidal Rule Error: $\text{Error}=|\text{Exact}-\text{Approximate}|=|0.4352-0.4349|=0.0003$

Simpson's $\frac{1}{3}$ Rule Error: $\text{Error}=|0.4352-0.4355|=0.0003$

Comments:

• Both numerical methods yield results accurate to three decimal places, with an absolute error of approximately $0.0003$ (or $3\times 10^{-4}$).
• The Trapezoidal approximation (0.4349) underestimates the true value, while Simpson's approximation (0.4355) slightly overestimates it.
• For this particular integral with $n=8$, both methods demonstrate similar accuracy, though Simpson's rule is generally expected to be more accurate for smooth functions due to its higher-order error term ($O(h^4)$ vs $O(h^2)$). The comparable performance here results from the specific function behavior and step size used.

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Key Formulas or Methods Used

• Step size calculation: $h=\frac{b-a}{n}$
• Trapezoidal Rule: ${\int }_a^{b}f(x)dx\approx \frac{h}{2}\left[f(x_0)+2{\sum }_{i=1}^{n-1}f(x_i)+f(x_n)\right]$
• Simpson's $\frac{1}{3}$ Rule: ${\int }_a^{b}f(x)dx\approx \frac{h}{3}\left[f(x_0)+4{\sum }_{\text{odd }i}f(x_i)+2{\sum }_{\text{even }i}f(x_i)+f(x_n)\right]$ (requires even $n$)
• Standard integral: $\int \frac{1}{a^2+x^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C$
• Absolute error: $|\text{Exact Value}-\text{Approximate Value}|$

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Summary of Steps

1. Identify parameters: Set $a=0$, $b=1$, $n=8$, calculate $h=0.125$, and generate 9 equally spaced nodes $x_0$ through $x_8$.
2. Tabulate function values: Calculate $f(x_i)=\frac{1}{2+x_i^2}$ for all nodes from $x=0$ to $x=1$.
3. Apply Trapezoidal Rule: Use $\frac{h}{2}[y_0+2(y_1+y_2+y_3+y_4+y_5+y_6+y_7)+y_8]$ to obtain $\approx 0.4349$.
4. Apply Simpson's $\frac{1}{3}$ Rule: Use $\frac{h}{3}[y_0+4(y_1+y_3+y_5+y_7)+2(y_2+y_4+y_6)+y_8]$ to obtain $\approx 0.4355$.
5. Compute exact value: Integrate analytically to get $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)\approx 0.4352$.
6. Compare and comment: Calculate absolute errors ($\approx 0.0003$ for both) and analyze the accuracy of each method relative to the exact solution.