Question Statement

Evaluate the integral

${\int }_0^3\frac{x^2}{1+x}dx$

using the Trapezoidal rule and Simpson's $\frac{1}{3}$ rule by taking $n=10$.

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Background and Explanation

This problem requires numerical integration techniques to approximate a definite integral when an exact antiderivative may be difficult to use or when working with tabulated data. With $n=10$ subintervals, we divide the interval $[0,3]$ into equal segments of width $h$, calculate function values at each partition point, and apply weighted summation formulas.

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Solution

Setup and Parameters

Compare the given integral with the standard form ${\int }_a^{b}f(x)dx$:

$a=0,b=3,f(x)=\frac{x^2}{1+x},n=10$

Calculate the step size $h$:

$h=\frac{b-a}{n}=\frac{3-0}{10}=\frac{3}{10}=0.3$

Calculation of $x$ values

The partition points are calculated as $x_i=x_{i-1}+h$:

$\begin{array}{rl}{x}_0& =0\\ x_1& =0+\frac{3}{10}=\frac{3}{10}=0.3\\ x_2& =\frac{3}{10}+\frac{3}{10}=\frac{3}{5}=0.6\\ x_3& =\frac{3}{5}+\frac{3}{10}=\frac{9}{10}=0.9\\ x_4& =\frac{9}{10}+\frac{3}{10}=\frac{6}{5}=1.2\\ x_5& =\frac{6}{5}+\frac{3}{10}=\frac{3}{2}=1.5\\ x_6& =\frac{3}{2}+\frac{3}{10}=\frac{9}{5}=1.8\\ x_7& =\frac{9}{5}+\frac{3}{10}=\frac{21}{10}=2.1\\ x_8& =\frac{21}{10}+\frac{3}{10}=\frac{12}{5}=2.4\\ x_9& =\frac{12}{5}+\frac{3}{10}=\frac{27}{10}=2.7\\ x_{10}& =\frac{27}{10}+\frac{3}{10}=3\end{array}$

Calculation of $f(x)$ values

Evaluate $f(x)=\frac{x^2}{1+x}$ at each partition point:

$\begin{array}{rl}f(x_0)& =f(0)=\frac{0^2}{1+0}=0\\ f(x_1)& =f\left(\frac{3}{10}\right)=\frac{{\left(\frac{3}{10}\right)}^2}{1+\frac{3}{10}}=\frac{0.09}{1.3}\approx 0.0692\\ f(x_2)& =f\left(\frac{3}{5}\right)=\frac{{\left(\frac{3}{5}\right)}^2}{1+\frac{3}{5}}=\frac{0.36}{1.6}=0.225\\ f(x_3)& =f\left(\frac{9}{10}\right)=\frac{{\left(\frac{9}{10}\right)}^2}{1+\frac{9}{10}}=\frac{0.81}{1.9}\approx 0.4263\\ f(x_4)& =f\left(\frac{6}{5}\right)=\frac{{\left(\frac{6}{5}\right)}^2}{1+\frac{6}{5}}=\frac{1.44}{2.2}\approx 0.6545\\ f(x_5)& =f\left(\frac{3}{2}\right)=\frac{{\left(\frac{3}{2}\right)}^2}{1+\frac{3}{2}}=\frac{2.25}{2.5}=0.9\\ f(x_6)& =f\left(\frac{9}{5}\right)=\frac{{\left(\frac{9}{5}\right)}^2}{1+\frac{9}{5}}=\frac{3.24}{2.8}\approx 1.1571\\ f(x_7)& =f\left(\frac{21}{10}\right)=\frac{{\left(\frac{21}{10}\right)}^2}{1+\frac{21}{10}}=\frac{4.41}{3.1}\approx 1.4226\\ f(x_8)& =f\left(\frac{12}{5}\right)=\frac{{\left(\frac{12}{5}\right)}^2}{1+\frac{12}{5}}=\frac{5.76}{3.4}\approx 1.6941\\ f(x_9)& =f\left(\frac{27}{10}\right)=\frac{{\left(\frac{27}{10}\right)}^2}{1+\frac{27}{10}}=\frac{7.29}{3.7}\approx 1.9703\\ f(x_{10})& =f(3)=\frac{3^2}{1+3}=\frac{9}{4}=2.25\end{array}$

Part (a): Trapezoidal Rule

By the Trapezoidal rule:

${\int }_0^3\frac{x^2}{1+x}dx\approx \frac{h}{2}\left[f(x_0)+2\left(f(x_1)+f(x_2)+\cdots +f(x_9)\right)+f(x_{10})\right]$

Substituting the values:

$\begin{array}{rl}{\int }_0^3\frac{x^2}{1+x}dx& \approx \frac{0.3}{2}\left[0+2(0.0692+0.225+0.4263+0.6545+0.9+1.1571+1.4226+1.6941+1.9703)+2.25\right]\\ & \approx \frac{3}{20}\left[2(8.5191)+2.25\right]\\ & \approx \frac{3}{20}\left[17.0382+2.25\right]\\ & \approx \frac{3}{20}(19.2882)\\ & \approx \boxed{2.8932}\end{array}$

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

By Simpson's $\frac{1}{3}$ rule (valid since $n=10$ is even):

${\int }_0^3\frac{x^2}{1+x}dx\approx \frac{h}{3}\left[f(x_0)+4\left(f(x_1)+f(x_3)+f(x_5)+f(x_7)+f(x_9)\right)+2\left(f(x_2)+f(x_4)+f(x_6)+f(x_8)\right)+f(x_{10})\right]$

Substituting the values:

$\begin{array}{rl}{\int }_0^3\frac{x^2}{1+x}dx& \approx \frac{0.3}{3}\left[0+4(0.0692+0.4263+0.9+1.4226+1.9703)+2(0.225+0.6545+1.1571+1.6941)+2.25\right]\\ & \approx \frac{1}{10}\left[4(4.7884)+2(3.7307)+2.25\right]\\ & \approx \frac{1}{10}\left[19.1536+7.4614+2.25\right]\\ & \approx \frac{1}{10}(28.865)\\ & \approx \boxed{2.8865}\end{array}$

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Key Formulas or Methods Used

• Step size: $h=\frac{b-a}{n}$
• Trapezoidal Rule: ${\int }_a^{b}f(x)dx\approx \frac{h}{2}\left[y_0+2(y_1+y_2+\cdots +y_{n-1})+y_n\right]$
• Simpson's $\frac{1}{3}$ Rule: ${\int }_a^{b}f(x)dx\approx \frac{h}{3}\left[y_0+4(y_1+y_3+\cdots +y_{n-1})+2(y_2+y_4+\cdots +y_{n-2})+y_n\right]$ (where $n$ is even)

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Summary of Steps

1. Identify parameters: Determine $a=0$, $b=3$, $n=10$, and calculate $h=\frac{3-0}{10}=0.3$
2. Generate partition points: Calculate $x_0$ through $x_{10}$ using $x_i=x_{i-1}+h$
3. Evaluate function: Compute $f(x_i)=\frac{x_i^2}{1+x_i}$ for all 11 points
4. Apply Trapezoidal rule: Use $\frac{h}{2}$ multiplied by the sum of endpoints plus twice the sum of interior points
5. Apply Simpson's rule: Use $\frac{h}{3}$ multiplied by the weighted sum (alternating 4, 2, 4, 2... for interior points)
6. Compute final approximations: Trapezoidal $\approx 2.8932$, Simpson's $\approx 2.8865$