Question

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?

Answer 1

Use the formula : `Area=h/2(y_1+y_n+2(y_2+y_3+...+y_(n-1)))`
to obtain the result :
`Area=4314/3145~=1.37`

Explanation:

`h` is the step length
We find the step length using the following formula : `h=(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`

`=>h=(6-0)/(4-1)=2`

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

`"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=1/(1+x^2)`

`=>y_1=y=1/(1+(0)^2)=1`

For `y_2` we plug-in `x=2` to have : `y_2=1/(1+(2)^2)=1/5`

Similarly,

`y_3=1/(1+(4)^2)=1/17`

`y_4=1/(1+(6)^2)=1/37`

Next, we use the formula,

`Area=h/2(y_1+y_n+2(y_2+y_3+...+y_(n-1)))`

`=>Area=2/2[1+1/5+2(1/17+1/37)]=(3145+629+370+170)/3145=color(blue)(4314/3145)`