Question

In Exercises \(1-6,\) use the Trapezoidal Rule and Simpson's Rule to approximate the value of the definite integral for the given value of \(n .\) Round your answers to four decimal places and compare your results with the exact value of the definite integral. $$ \int_{0}^{2} x^{2} d x, \quad n=4 $$

Short answer

The approximate values of the integral using the Trapezoidal Rule is 2.375 and using Simpson's Rule is 2.3333, while the exact value of the integral is 2.6667.

Step-by-step solution

Apply the Trapezoial Rule

First apply the Trapezoidal Rule. A sequence of \(n+1\) equally spaced points in the interval from 0 to 2 will be defined first. Since \(n=4\), the points are \(x_0=0, x_1=0.5, x_2=1.0, x_3=1.5, x_4=2.0\). The value of the function \(f(x)\) at these points are: \(f(x_0)=0, f(x_1)=0.25, f(x_2)=1, f(x_3)=2.25, f(x_4)=4\). The Trapezoidal rule approximates the integral as \(\Delta x[(f(x_0)+f(x_4))/2+\sum_{i=1}^{n-1}f(x_i)] = 0.5[(0+4)/2+ (0.25+1+2.25)] = 2.375.\)

Apply Simpson's Rule

Second apply Simpson's Rule. In Simpson's Rule, the integral is approximated by the sum of the integral approximations in each subinterval. For \(n=4\), the integral approximations are formed by the intervals \([0, 1]\) and \([1, 2]\). The Simpsons's Rule approximates the integral as \(\Delta x/3 [ (f(x_0)+4f(x_1)+2f(x_2)) + (f(x_2)+4f(x_3)+f(x_4))]= 0.5/3[(0+4*0.25+2*1)+(1+4*2.25+4)]=2.3333.\)

Find the exact value

The exact value of the integral is found by evaluating \(\int_{0}^{2} x^{2} d x = [x^3/3]_0^2 = 8/3 = 2.6667\).

Compare the results

The approximation by the Trapezoidal Rule is 2.375, the approximation by Simpson's Rule is 2.3333, and the exact value is 2.6667. Therefore, in this case, Simpson's Rule gives an approximation closer to the exact value than the Trapezoidal Rule does.