How do I find arc length using the trapezoid rule? The question asks, "Use the trapezoid rule (when $n=8$) to approximate the arc length of the graph of $y=2x^3-2x+1$ from $A (0,1)$ to $B(2,13)$"
I first graphed this out and found the points to have $(0,1)$, $(0.25, 0.53125)$, $(0.5, 0.25), (0.75, 0.3437), (1,1), (1.25, 2.40625), (1.5, 4.75), (1.75, 8.21875), (2,13)$ as the points of each trapezoid, and found the distance between each point using the distance formula and got an answer something like $13.92$. This is not the correct answer to this problem. Please help! 
 A: There are two reasonable interpretations of this problem. More experienced students would likely say to do
$$s=\int ds=\int_0^2\sqrt{1+y^{\prime 2}(x)}dx\approx\frac h2\sum_{i=0}^8w_i\sqrt{1+(y^{\prime}(x_i))^2}\approx14.09254$$
Where $w_0=w_8=1$ and all other $w_i=2$.
But I betcha that the questioner is doing something like
$$s\approx\sum_{i=0}^7\sqrt{(x_{i+1}-x_i)^2+(y_{i+1}-y_i)^2}\approx13.92768$$
EDIT: That last sum could be rewritten as
$$s\approx\sum_{i=0}^7\sqrt{1+\left(\frac{y_{i+1}-y_i}{x_{i+1}-x_i}\right)^2}(x_{i+1}-x_i)$$
And as such could be considered to be the midpoint rule using central differences to approximate the derivative of the function. Normally the midpoint rule is a little more accurate than the trapezoidal rule with error in the opposite direction. The is indeed the case here as the exact value seems to be close to $13.964791$, perhaps validating the original questioner's intuition for the problem.
A: I agree with user5713492. 
Since the formula for determining the arc length,  $L$, can be given as:
$L= \int_a^b \!  \sqrt{1+(\frac{dy}{dx})^2} \, \mathrm{d}x$
, where $a=0$ and $b=2$. 
The derivative of $y$, that is $ \frac{dy}{dx}$ can first be determined as:
$ \frac{dy}{dx} = 6x^2 -2$
Now,  by letting the square-rooted term in the arc length formula be the function $g$ as follows and substituting for $\frac{dy}{dx}$ we have that, 
$g(x) =\sqrt{1+(6x^2 -2)^2} $
and therefore, 
$L= \int_a^b \!  g (x) \, \mathrm{d}x$
or put differently, 
$L= \int_a^b \!  \sqrt{1+(6x^2 -2)^2} \, \mathrm{d}x$
We can now apply the trapezoidal rule to integrate numerically on the interval $[a, b]$.
$L \approx \frac{b-a}{2} \sum_{k=1}^n (g(x_{k+1}) -g(x_k))$
Where $n=8$. The trapezoidal rule formula can be re-written as:
$L \approx \frac{b-a}{2n}[ g(x_1) +2g(x_2) + 2g(x_3) +\dots + 2g(x_7)+ g(x_8) ]$
