# How do you use the Trapezoidal Rule with n=60 to estimate the length of the curve y=sinx, with x greater or equal to 0 and x less than or equal to pi?

Oct 10, 2015

The length of the curve $y = f \left(x\right)$ from $a$ to $b$ is ${\int}_{a}^{b} \sqrt{1 - {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$

#### Explanation:

So we need to approximate ${\int}_{0}^{\pi} \sqrt{1 - {\cos}^{2} x} \mathrm{dx}$

using the trapezoidal rule. We'll use the formula.

T_60 = 1/2 Deltax(f(x_0)+2f(x_1)+2f(x_3)+* * * +2f(x_(n-1)+f(x_n))

In this case $f \left(x\right) = \sqrt{1 - {\cos}^{2} x} \mathrm{dx}$

$n = 60$ and $\Delta x = \frac{b - a}{n} = \frac{\pi - 0}{60} = \frac{\pi}{60}$

${x}_{0} = 0 , {x}_{1} = \frac{\pi}{60} , {x}_{2} = \frac{2 \pi}{60} , {x}_{3} = \frac{3 \pi}{60} , . . . {x}_{n - 1} = \frac{\left(n - 1\right) \pi}{60} , {x}_{n} = \pi$

Plug in the numbers and do the arithmetic.
(If permitted, use a computer spreadsheet for all that arithmetic.)

Note
${\int}_{0}^{\pi} \sqrt{1 - {\cos}^{2} x} \mathrm{dx} = {\int}_{0}^{\pi} \sqrt{{\sin}^{2} x} \mathrm{dx}$
$= {\int}_{0}^{\pi} \left\mid \sin x \right\mid \mathrm{dx}$
$= {\int}_{0}^{\pi} \sin x \mathrm{dx}$
$= 2$