Trapezoidal Rule

6.2 Trapezoidal Rule

As simplest polynomial interpolation, consider a linear approximation, i.e. the secant through the points \((a; f(a))\) and \((b; f(b))\). Then, to approximate the area under the curve, one calculates the area of the trapezoid with vertices \((a; 0)\), \((b; 0)\), \((b; f(b))\), \((a; f(a))\), which obviously is

\begin{equation*} T(f)=(b-a)\frac{f(a)+f(b)}{2}\,. \end{equation*}

(6.3)

If one splits the full interval \([a, b]\) into \(n\) subintervals of equal length, one can apply the trapezoidal rule for each of the subintervals, which significantly improves the approximation. To this end, we define

\begin{equation*}\label{eq:h_n} \begin{split} h_n&=\frac{b-a}{n}, \\ x_0&=a, \ x_1=a+h_n, \ x_2=a+2h_n, \ \ldots, \ x_n=b. \end{split} \end{equation*}

(6.4)

This allows to write

\begin{equation*} \begin{split} T_n(f)&=(x_1-x_0)\frac{f(x_0)+f(x_1)}{2}+(x_2-x_1)\frac{f(x_1)+f(x_2)}{2}+(x_3-x_2)\frac{f(x_2)+f(x_3)}{2}\\ &\quad +\ldots+(x_n-x_{n-1})\frac{f(x_{n-1})+f(x_n)}{2}\\ &=h_n\frac{f(x_0)+f(x_1)}{2}+h_n\frac{f(x_1)+f(x_2)}{2}+\ldots+h_n\frac{f(x_{n-1})+f(x_n)}{2}\\ &=\frac{h_n}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+\ldots+2f(x_{n-1})+f(x_n)\right]. \end{split} \end{equation*}

(6.5)

Theorem 1: error of the trapezoidal rule (without proof)
Let \(f\) be twice continuously differentiable on \([a,b]\). For the approximation of the integral of \(f\) over \([a,b]\), \(I(f)=\int_a^b f(x){\rm d}x\), by the trapezoidal rule for \(n\) sub intervals, \(T_n(f)\), it holds that

\begin{equation*} |I(f)-T_n(f)|=n\,\frac{h_n^3}{12}|f''(\epsilon_n)|= \frac{1}{12}\,\frac{(b-a)^3}{n^2}|f''(\epsilon_n)|;\ \ \ \epsilon_n\in[a,b]. \end{equation*}

(6.6)

It follows from Theorem 1 that doubling the number of sub intervals approximately quarters the absolute error of the trapezoidal rule.
Example:
Consider the following analytically determined integral of the function \(f(x)=\frac{1}{1+x}\):

\begin{equation*} I(f)=\int\limits_0^1\frac{{\rm d}x}{1+x}=\left.\ln(1+x)\rule{0mm}{1.0em}\right|_0^1=\ln(2)-\ln(1)=\ln(2)\approx 0.69314718. \end{equation*}

(6.7)

Applying the trapezoidal rule directly to this interval (\(a=0\), \(b=1\)) yields

\begin{equation*} T_1(f)=(1-0)\frac{\frac{1}{1+0}+\frac{1}{1+1}}{2}=\frac{3}{4}=0.75. \end{equation*}

(6.8)

This result has an absolute error of about \(\left|I(f)-T_1(f)\right|\approx0.057\) (relative error \(\approx 8\) %). To improve the approximation, we divide the interval \([a,b]\) into two subintervals, by which the error should roughly be reduced by a factor of \(1/4\):

\begin{equation*} \label{eq:T2example} \begin{split} T_2(f)&= \left(\frac{1}{2}-0\right)\frac{\frac{1}{1+0}+\frac{1}{1+\frac{1}{2}}}{2} +\left(1-\frac{1}{2}\right)\frac{\frac{1}{1+\frac{1}{2}}+\frac{1}{1+1}}{2}\\ &=\frac{1}{2}\left(\frac{{1}+\frac{2}{3}}{2} + \frac{\frac{2}{3}+\frac{1}{2}}{2}\right) =\frac{1}{2}\left(\frac{5}{6}+\frac{7}{12}\right) =0.708\bar{3} \end{split} \end{equation*}

(6.9)

This second approximation has an absolute error of \(\left|I(f)-T_2(f)\right|\approx0.015\) (rel. error \(\approx 2\) %), which indeed is roughly one quarter of the error of the first approximation (even a little less).