# How do you use the trapezoidal rule with n=4 to approximate the area between the curve 1/(1 + x^2)  from 0 to 6?

Jul 28, 2015

Use the formula : $A r e a = \frac{h}{2} \left({y}_{1} + {y}_{n} + 2 \left({y}_{2} + {y}_{3} + \ldots + {y}_{n - 1}\right)\right)$
to obtain the result :
$A r e a = \frac{4314}{3145} \cong 1.37$

#### Explanation:

$h$ is the step length
We find the step length using the following formula : $h = \frac{b - a}{n - 1}$

$a$ is the minimum value of $x$ and $b$ is the maximum value of $x$. In our case $a = 0$ and $b = 6$

$n$ is the number of strips. Hence $n = 4$

$\implies h = \frac{6 - 0}{4 - 1} = 2$

So, the values of $x$ are $0 , 2 , 4 , 6$

$\text{NB :}$ Starting from $x = 0$ we add the step length $h = 2$ to get the next value of $x$ up to $x = 6$

In order to find ${y}_{1}$ up to ${y}_{n}$(or ${y}_{4}$) we plug-in each value of $x$ to get the corresponding $y$

For example : to get ${y}_{1}$ we plug-in $x = 0$ in $y = \frac{1}{1 + {x}^{2}}$

$\implies {y}_{1} = y = \frac{1}{1 + {\left(0\right)}^{2}} = 1$

For ${y}_{2}$ we plug-in $x = 2$ to have : ${y}_{2} = \frac{1}{1 + {\left(2\right)}^{2}} = \frac{1}{5}$

Similarly,

${y}_{3} = \frac{1}{1 + {\left(4\right)}^{2}} = \frac{1}{17}$

${y}_{4} = \frac{1}{1 + {\left(6\right)}^{2}} = \frac{1}{37}$

Next, we use the formula,

$A r e a = \frac{h}{2} \left({y}_{1} + {y}_{n} + 2 \left({y}_{2} + {y}_{3} + \ldots + {y}_{n - 1}\right)\right)$

$\implies A r e a = \frac{2}{2} \left[1 + \frac{1}{5} + 2 \left(\frac{1}{17} + \frac{1}{37}\right)\right] = \frac{3145 + 629 + 370 + 170}{3145} = \textcolor{b l u e}{\frac{4314}{3145}}$