# Trapezoidal Rule – Algorithm, Implementation in C With Solved Examples

Numerical Methods & Algorithms / Sunday, October 21st, 2018

Table of Contents

At first we deduce the general integration formula based on Newton’s forward interpolation formula and after that we will use it to formulate Trapezoidal Rule and Simpson’s 1/3 rd rule.

The Newton’s forward interpolation formula for the equi-spaced points xi, i =0, 1, …, n, xi = x0 + ih is

$\phi \left( x \right)={{y}_{0}}+u\Delta {{y}_{0}}+\frac{u\left( u-1 \right)}{2!}{{\Delta }^{2}}{{y}_{0}}+\frac{u\left( u-1 \right)\left( u-2 \right)}{3!}{{\Delta }^{3}}{{y}_{0}}+….$

$where~~u=\frac{x-{{x}_{0}}}{h},~~h~~is~~the~~spacing.$

Let the interval [a, b] be divided into n equal subintervals such that a = x0 < x1 < x2 < … < xn = b. Then

$I=\int_{a}^{b}{f(x)dx=\int_{{{x}_{0}}}^{{{x}_{n}}}{\phi \left( x \right)}}dx$

$=\int_{{{x}_{0}}}^{{{x}_{n}}}{\left[ {{y}_{0}}+u\Delta {{y}_{0}}+\frac{u\left( u-1 \right)}{2!}{{\Delta }^{2}}{{y}_{0}}+\frac{u\left( u-1 \right)\left( u-2 \right)}{3!}{{\Delta }^{3}}{{y}_{0}}+…. \right]}dx$

Since x = x0 + uh, dx = hdu, when x = x0 then u = 0 and x = xn then u = n. Thus,

$I=\int_{0}^{n}{\left[ {{y}_{0}}+u\Delta {{y}_{0}}+\frac{{{u}^{2}}-u}{2!}{{\Delta }^{2}}{{y}_{0}}+\frac{{{u}^{3}}-3{{u}^{2}}+2u}{3!}{{\Delta }^{3}}{{y}_{0}}+…. \right]hdu}$

$=h\left[ {{y}_{0}}\left[ u \right]_{0}^{n}+\Delta {{y}_{0}}\left[ \frac{{{u}^{2}}}{2} \right]_{0}^{n}+\frac{{{\Delta }^{2}}{{y}_{0}}}{2!}\left[ \frac{{{u}^{3}}}{3}-\frac{{{u}^{2}}}{2} \right]_{0}^{n}+\frac{{{\Delta }^{3}}{{y}_{0}}}{3!}\left[ \frac{{{u}^{4}}}{4}-{{u}^{3}}+{{u}^{2}} \right]_{0}^{n}+…. \right]$

$=nh\left[ {{y}_{0}}+\frac{n}{2}\Delta {{y}_{0}}+\frac{2{{n}^{2}}-3n}{12}{{\Delta }^{2}}{{y}_{0}}+\frac{{{n}^{3}}-4{{n}^{2}}+4n}{24}{{\Delta }^{3}}{{y}_{0}}+…. \right]……..(1)$

From this formula, one can generate different integration formulae by substituting n = 1, 2, 3, …

# Trapezoidal Rule

Substituting n = 1 in the equation (1). In this case all differences higher than the first difference become zero. Then

$\int_{{{x}_{0}}}^{{{x}_{n}}}{f\left( x \right)dx=h\left[ {{y}_{0}}+\frac{1}{2}\Delta {{y}_{0}} \right]}$

$\Rightarrow \int_{{{x}_{0}}}^{{{x}_{n}}}{f\left( x \right)dx=h}\left[ {{y}_{0}}+\frac{1}{2}\left( {{y}_{1}}-{{y}_{0}} \right) \right]=\frac{h}{2}\left( {{y}_{0}}+{{y}_{1}} \right)……….(2)$

The formula (2) is known as the Trapezoidal Rule.

In this formula, the interval [a, b] is considered as a single interval, and it gives a very rough answer. But, if the interval [a, b] is divided into several subintervals and this formula is applied to each of these subintervals then a better approximate result may be obtained.

This formula is known as composite formula, deduced below.

## Composite Trapezoidal Rule

Let the interval [a, b] be divided into n equal subintervals by the points a = x0 < x1 < x2 < … < xn = b, where xi = x0 + ih, i = 1, 2, 3, …,n.

Applying the trapezoidal rule to each of the subintervals, one can find the composite formula as

$\int_{a}^{b}{f\left( x \right)dx=\int_{{{x}_{0}}}^{{{x}_{1}}}{f\left( x \right)dx}}+\int_{{{x}_{1}}}^{{{x}_{2}}}{f\left( x \right)dx}+….+\int_{{{x}_{n-1}}}^{{{x}_{n}}}{f\left( x \right)dx}$

$=\frac{h}{2}\left[ {{y}_{0}}+{{y}_{1}} \right]+\frac{h}{2}\left[ {{y}_{1}}+{{y}_{2}} \right]+….+\frac{h}{2}\left[ {{y}_{n-1}}+{{y}_{n}} \right]$

$\therefore \int_{a}^{b}{f\left( x \right)dx=}\frac{h}{2}\left[ {{y}_{0}}+2\left( {{y}_{1}}+{{y}_{2}}+….+{{y}_{n-1}} \right)+{{y}_{n}} \right]$

## Trapezoidal Rule Implementation in C

Output

Enter the values of a, b: 0 1
Enter the value of n: 100
The value of the integration is 0.25002

 Example 01

Find the value of

$\int_{1.2}^{1.6}{\left( x+\frac{1}{x} \right)dx}$

Taking 4 subintervals, correct up to four significant figures.

Solution:

$Let~~f\left( x \right)=\left( x+\frac{1}{x} \right)$

$Here~~{{x}_{0}}=1.2,~~{{x}_{n}}=1.6,~~n=4$

$\therefore h=\frac{1.6-1.2}{4}=0.1$

The tabulated values of f(x) for different values of x are given below:

 x 1.2 1.3 1.4 1.5 1.6 f(x) 2.03333 2.06923 2.11429 2.16667 2.225

By Trapezoidal Rule, we have

$\int_{a}^{b}{f\left( x \right)dx=}\frac{h}{2}\left[ {{y}_{0}}+2\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)+{{y}_{4}} \right]$

$\int_{1.2}^{1.6}{\left( x+\frac{1}{x} \right)dx}=\frac{0.1}{2}\left[ 2.033333+2\left( 2.069231+2.114286+2.166667 \right)+2.225 \right]$

$\therefore \int_{1.2}^{1.6}{\left( x+\frac{1}{x} \right)dx}=0.8477$

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