How do I find arc length using the trapezoid rule?

Question

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!

Answer 1

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.

Answer 2

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: $L \approx \frac{b-a}{2n}[ g(x_1) +2g(x_2) + 2g(x_3) +\dots + 2g(x_7)+ g(x_8) ]$