Question

How do you use the Trapezoidal rule with $n=4$ to approximate from $\left[ 2,3 \right]$ of $\dfrac{1}{{{\left( x-1 \right)}^{2}}}dx$?

Answer 1

In this question we have been given with an integral for which we have to approximate the value in the range $\left[ 2,3 \right]$ with $n=4$. We will use the trapezoidal rule which states that $\int\limits_{a}^{b}{f\left( x \right)dx}=\dfrac{\Delta x}{2}\left[ f\left( {{x}_{0}} \right)+2f\left( {{x}_{1}} \right)+2f\left( {{x}_{2}} \right)+....+2f\left( {{x}_{n-1}} \right)+f\left( {{x}_{n}} \right) \right]$, we will first calculate $\Delta x$ by using the formula $\Delta x=\dfrac{b-a}{n}$ and then find the sub intervals between point $\left[ 2,3 \right]$, we will then substitute the values in the given trapezoidal function to get the required expression.

Complete step by step solution:
We have the function given to us as $\dfrac{1}{{{\left( x-1 \right)}^{2}}}dx$ therefore, we can write it is as:
$\Rightarrow f\left( x \right)=\dfrac{1}{{{\left( x-1 \right)}^{2}}}$
Which we have to integrate from the points $\left[ 2,3 \right]$ with $n=4$.
Now we know that $\Delta x=\dfrac{b-a}{n}$, on substituting the value in the formula, we get:
$\Rightarrow \Delta x=\dfrac{3-2}{4}$
On simplifying, we get:
$\Rightarrow \Delta x=\dfrac{1}{4}$
Now the endpoints of the subintervals are found with $a=2$ adding $\Delta x$ to the points until we reach the final point ${{x}_{n}}=b=3$.
Therefore, we have:
$\Rightarrow {{x}_{0}}=2$
$\Rightarrow {{x}_{1}}=2+\dfrac{1}{4}=\dfrac{9}{4}$
$\Rightarrow {{x}_{2}}=\dfrac{9}{4}+\dfrac{1}{4}=\dfrac{10}{4}$
$\Rightarrow {{x}_{3}}=\dfrac{10}{4}+\dfrac{1}{4}=\dfrac{11}{4}$
$\Rightarrow {{x}_{4}}=\dfrac{11}{4}+\dfrac{1}{4}=\dfrac{12}{4}=3=b$
We know the trapezoidal rule has the formula as:
$\int\limits_{a}^{b}{f\left( x \right)dx}=\dfrac{\Delta x}{2}\left[ f\left( {{x}_{0}} \right)+2f\left( {{x}_{1}} \right)+2f\left( {{x}_{2}} \right)+....+2f\left( {{x}_{n-1}} \right)+f\left( {{x}_{n}} \right) \right]$
On substituting the values of $\Delta x$ and all the subintervals in the formula, we get:
$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{{{\left( x-1 \right)}^{2}}}}=\dfrac{\dfrac{1}{4}}{2}\left[ f\left( 2 \right)+2f\left( \dfrac{9}{4} \right)+2f\left( \dfrac{10}{4} \right)+2f\left( \dfrac{11}{4} \right)+f\left( 3 \right) \right]$
On simplifying, we get:
$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{{{\left( x-1 \right)}^{2}}}}=\dfrac{1}{8}\left[ f\left( 2 \right)+2f\left( \dfrac{9}{4} \right)+2f\left( \dfrac{10}{4} \right)+2f\left( \dfrac{11}{4} \right)+f\left( 3 \right) \right]\to \left( 1 \right)$
Now we have $f\left( 2 \right)$ as:
$\Rightarrow f\left( 2 \right)=\dfrac{1}{{{\left( 2-1 \right)}^{2}}}=\dfrac{1}{{{1}^{2}}}=1$
The value of $f\left( \dfrac{9}{4} \right)$ will be:
$\Rightarrow f\left( \dfrac{9}{4} \right)=\dfrac{1}{{{\left( \dfrac{9}{4}-1 \right)}^{2}}}=\dfrac{1}{{{\left( \dfrac{9-4}{4} \right)}^{2}}}=\dfrac{1}{{{\left( \dfrac{5}{4} \right)}^{2}}}=\dfrac{1}{\dfrac{25}{16}}=\dfrac{16}{25}$
The value of $f\left( \dfrac{10}{4} \right)$ will be:
$\Rightarrow f\left( \dfrac{10}{4} \right)=\dfrac{1}{{{\left( \dfrac{10}{4}-1 \right)}^{2}}}=\dfrac{1}{{{\left( \dfrac{10-4}{4} \right)}^{2}}}=\dfrac{1}{{{\left( \dfrac{6}{4} \right)}^{2}}}=\dfrac{1}{\dfrac{36}{16}}=\dfrac{16}{36}$
The value of $f\left( \dfrac{11}{4} \right)$ will be:
$\Rightarrow f\left( \dfrac{11}{4} \right)=\dfrac{1}{{{\left( \dfrac{11}{4}-1 \right)}^{2}}}=\dfrac{1}{{{\left( \dfrac{11-4}{4} \right)}^{2}}}=\dfrac{1}{{{\left( \dfrac{7}{4} \right)}^{2}}}=\dfrac{1}{\dfrac{49}{16}}=\dfrac{16}{49}$
The value of $f\left( \dfrac{11}{4} \right)$ will be:
$\Rightarrow f\left( 3 \right)=\dfrac{1}{{{\left( 3-1 \right)}^{2}}}=\dfrac{1}{{{\left( 2 \right)}^{2}}}=\dfrac{1}{4}$
On substituting the values in equation $\left( 1 \right)$, we get:
$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{{{\left( x-1 \right)}^{2}}}}=\dfrac{1}{8}\left[ 1+2\times \dfrac{16}{25}+2\times \dfrac{16}{36}+2\times \dfrac{16}{49}+\dfrac{1}{4} \right]$
On simplifying the fractions, we get:
$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{{{\left( x-1 \right)}^{2}}}}=\dfrac{1}{8}\left[ 1+\dfrac{32}{25}+\dfrac{8}{9}+\dfrac{32}{49}+\dfrac{1}{4} \right]$
On adding all the fractions, we get:
$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{{{\left( x-1 \right)}^{2}}}}=\dfrac{1}{8}\left[ \dfrac{179573}{44100} \right]$
On multiplying the terms in the right-hand side, we get:
$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{{{\left( x-1 \right)}^{2}}}}=\dfrac{179573}{352800}$
On dividing, we get:
$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{{{\left( x-1 \right)}^{2}}}}=0.50899$, which is the required solution.

Note: It is to be remembered that the value of the integral found after using the trapezoidal rule is an approximate value and not the real value. This implies that the answer found by directly integrating the function from the given limits will be somewhat different from the approximate value. The real value subtracted by the approximate value is known as the error.