# How do you use the trapezoidal rule with n=3 to approximate the area between the curve y=x^2 and the x-axis for 1 ≤ t ≤ 4?

Sep 3, 2015

$\text{Area"=22.125" square units}$

#### Explanation:

NB : Assuming that the interval is $\textcolor{b l u e}{1 \le x \le 4}$and not $\textcolor{red}{1 \le t \le 4}$

The trapezoidal rule says that the area can be found using the formula:
$\text{Area} = \frac{h}{2} \left({y}_{1} + {y}_{n} + 2 \left({y}_{2} + {y}_{3} + \ldots + {y}_{n - 1}\right)\right)$

where $h$ is the step length(the length of a single strip), ${y}_{1} , {y}_{2} , \ldots , {y}_{n}$ are the $y$ values corresponding to each $x$ value taken from the interval $\left[1 , 4\right]$

In this problem, since $n = 3$,
$\text{Area} = \frac{h}{2} \left({y}_{1} + {y}_{3} + 2 \left({y}_{2}\right)\right)$

First,
$h = \frac{{x}_{n} - {x}_{1}}{n - 1}$

${x}_{1}$ and ${x}_{n}$ are respectively the first and last $x$ values in the interval

$h = \frac{4 - 1}{3 - 1} = \frac{3}{2}$

Next,
${y}_{1} = {\left({x}_{1}\right)}^{2} = {\left(1\right)}^{2} = 1$
${y}_{2} = {\left(1 + \frac{3}{2}\right)}^{2} = {\left(\frac{5}{2}\right)}^{2} = \frac{25}{4}$
${y}_{3} = {\left(\frac{5}{2} + \frac{3}{2}\right)}^{2} = {\left(4\right)}^{2} = 16$

NB : In order to get the next $x$ we add $\frac{3}{2}$ to the previous.
Example : ${x}_{2} = 1 + \frac{3}{2} = \frac{5}{2}$

Finally,

$\text{Area} = \frac{\frac{3}{2}}{2} \left(1 + 16 + 2 \left(\frac{25}{4}\right)\right) \cong \textcolor{b l u e}{22.125}$