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?

Question

Paisley Johnson · Accepted Answer

The length of the curve #y=f(x)# from #a# to #b# is #int_a^b sqrt(1-(f'(x))^2) dx#

So we need to approximate #int_0^pi sqrt(1-cos^2x) 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(x) = sqrt(1-cos^2x)dx#

#n=60# and #Deltax = (b-a)/n = (pi-0)/60 = pi/60#

#x_0 = 0, x_1=pi/60, x_2 = (2pi)/60, x_3=(3pi)/60, . . . x_(n-1)=((n-1)pi)/60, x_n = pi#

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

Note Your answer should be very close to the exact answer:

#int_0^pi sqrt(1-cos^2x) dx = int_0^pi sqrt(sin^2x) dx #

# = int_0^pi abs(sinx) dx #

# = int_0^pi sinx dx#

# = 2#

Christopher Agresta · Answer

undefined To use the Trapezoidal Rule with $ n = 60 $ to estimate the length of the curve $ y = \sin(x) $ from $ x = 0 $ to $ x = \pi $, follow these steps:

1. Divide the interval $[0, \pi]$ into $ n $ subintervals of equal width. Since $ n = 60 $, each subinterval width would be $ \Delta x = \frac{\pi - 0}{60} = \frac{\pi}{60} $.

2. Compute the function values at the endpoints of each subinterval. This means evaluating $ \sin(x) $ at $ x = 0, \frac{\pi}{60}, \frac{2\pi}{60}, \frac{3\pi}{60}, \ldots, \pi $.

3. Apply the Trapezoidal Rule formula to each subinterval:
$$ L_i = \frac{\Delta x}{2} \left( y_{i} + y_{i+1} \right) $$
where $ L_i $ is the length of the $ i $-th subinterval, $ y_{i} $ is the function value at the left endpoint, and $ y_{i+1} $ is the function value at the right endpoint.

4. Sum up all the lengths of the subintervals:
$$ L = L_1 + L_2 + \ldots + L_{60} $$

5. The total length of the curve is approximately equal to $ L $.