# How do you use the trapezoidal rule with n=4 to approximate the area between the curve y=sqrt(x+1) from 1 to 3?

Sep 3, 2015

$\text{Area"~=3.4439" square units}$

#### Explanation:

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

$A$ is the area, $h$ 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 $\left[1 , 3\right]$

In this problem, since $n = 4$,
$A = \frac{h}{2} \left({y}_{1} + {y}_{4} + 2 \left({y}_{2} + {y}_{3}\right)\right)$

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

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

${y}_{1} = \sqrt{{x}_{1} + 1} = \sqrt{1 + 1} = \sqrt{2}$
${y}_{2} = \sqrt{{x}_{2} + 1} = \sqrt{\left(1 + \frac{2}{3}\right) + 1} = \sqrt{\frac{5}{3} + 1} = \sqrt{\frac{8}{3}}$
${y}_{3} = \sqrt{{x}_{3} + 1} = \sqrt{\left(\frac{5}{3} + \frac{2}{3}\right) + 1} = \sqrt{\frac{7}{3} + 1} = \sqrt{\frac{10}{3}}$
${y}_{4} = \sqrt{{x}_{4} + 1} = \sqrt{\left(\frac{7}{3} + \frac{2}{3}\right) + 1} = \sqrt{3 + 1} = \sqrt{4} = 2$

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

Finally,

$A = \frac{\frac{2}{3}}{2} \left(\sqrt{2} + 2 + 2 \left(\sqrt{\frac{8}{3}} + \sqrt{\frac{10}{3}}\right)\right) \cong \textcolor{b l u e}{3.4439}$